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Page 1 :

04, Laws of Motion, TOPIC 1, Newtons Laws of Motion and, Conservation of Momentum, 01 A ball of mass 0.15 kg is dropped, from a height 10 m, strikes the, ground and rebounds to the same, height. The magnitude of impulse, imparted to the ball is nearly, (g = 10 m/s 2 ), [NEET 2021], (a) 0, (c) 2.1 kg-m/s, , (b) 4.2 kg-m/s, (d) 1.4 kg-m/s, , Now, magnitude of the impulse imparted, to the ball,, |Impulse | = m | ∆v |, = 0.15 | 10 2 − (−10 2) |, = 0.15|2 × 10 2 |, = 4.2 kg -m / s, , 02 A truck is stationary and has a bob, suspended by a light string, in a, frame attached to the truck. The, truck, suddenly moves to the right, with an acceleration of a. The, pendulum will tilt, , Ans. (b), , From the above diagram (b) as the string, moves by an angle ofθ with the vertical, then the tangent angle is, ma, a, ⇒ θ = tan−1 , tanθ =, g, mg, , 03 A particle moving with velocity v is, acted by three forces shown by the, vector triangle PQR., The velocity of the particle will, [NEET (National) 2019], , P, , [NEET (Odisha) 2020], , Given, the mass of the ball dropped from, the height, m = 0.15 kg, The height from the ball dropped,, h = 10 m, We know that,, | Impulse | = m | ∆v |, where,, ∆v = v 2 − v 1, Here, v 2 is velocity reaches to the same, height,, v 1 is velocity just before striking to the, ground., For case (1), ball dropped from the 10m, height and strikes to the ground., Now, the velocity of the ball just before, striking to the ground is, v 1 = − 2gh, ⇒ v 1 = − 2(10)(10), ⇒ v 1 = − 10 2 m/s, For case (2), ball rebounds to the same, height., The velocity with which the ball just, reaches to the same height,, v 2 = 2gh, ⇒, , v 2 = 2(10)(10), , ⇒, , v 2 = 10 2 m/s, , (a) to the left and the angle of inclination, of the pendulum with the vertical is, g, sin−1 , a, (b) to the left and angle of inclination of, the pendulum with the vertical is, a, tan−1 , g, (c) to the left and angle of inclination of, the pendulum with the vertical is, a, sin1 − , g, (d) to the left and angle of inclination of, the pendulum with the vertical is, g, tan−1 , a, , R, , Q, , (a) decrease, (b) remain constant, (c) change according to the smallest, force QR, (d) increase, , Ans. (b), As the three forces are represented by, three sides of a triangle taken in order,, then they will be in equilibrium., P, , Ans. (b), As the truck move to the right, so the, bob will move to the left due to inertia of, rest with acceleration a., Thus, the given situation can be drawn, as, a, , ⇒, , θ, (a), , θ, , ma, mg, , (b), , R, , Q, , ⇒Fnet = FPQ + FQR + FRP = 0, dv, dv, Fnet = m × a = m = 0 ⇒ = 0, dt, dt, or v = constant, So, the velocity of particle remain, constant.

04, Laws of Motion, TOPIC 1, Newtons Laws of Motion and, Conservation of Momentum, 01 A ball of mass 0.15 kg is dropped, from a height 10 m, strikes the, ground and rebounds to the same, height. The magnitude of impulse, imparted to the ball is nearly, (g = 10 m/s 2 ), [NEET 2021], (a) 0, (c) 2.1 kg-m/s, , (b) 4.2 kg-m/s, (d) 1.4 kg-m/s, , Now, magnitude of the impulse imparted, to the ball,, |Impulse | = m | ∆v |, = 0.15 | 10 2 − (−10 2) |, = 0.15|2 × 10 2 |, = 4.2 kg -m / s, , 02 A truck is stationary and has a bob, suspended by a light string, in a, frame attached to the truck. The, truck, suddenly moves to the right, with an acceleration of a. The, pendulum will tilt, , Ans. (b), , From the above diagram (b) as the string, moves by an angle ofθ with the vertical, then the tangent angle is, ma, a, ⇒ θ = tan−1 , tanθ =, g, mg, , 03 A particle moving with velocity v is, acted by three forces shown by the, vector triangle PQR., The velocity of the particle will, [NEET (National) 2019], , P, , [NEET (Odisha) 2020], , Given, the mass of the ball dropped from, the height, m = 0.15 kg, The height from the ball dropped,, h = 10 m, We know that,, | Impulse | = m | ∆v |, where,, ∆v = v 2 − v 1, Here, v 2 is velocity reaches to the same, height,, v 1 is velocity just before striking to the, ground., For case (1), ball dropped from the 10m, height and strikes to the ground., Now, the velocity of the ball just before, striking to the ground is, v 1 = − 2gh, ⇒ v 1 = − 2(10)(10), ⇒ v 1 = − 10 2 m/s, For case (2), ball rebounds to the same, height., The velocity with which the ball just, reaches to the same height,, v 2 = 2gh, ⇒, , v 2 = 2(10)(10), , ⇒, , v 2 = 10 2 m/s, , (a) to the left and the angle of inclination, of the pendulum with the vertical is, g, sin−1 , a, (b) to the left and angle of inclination of, the pendulum with the vertical is, a, tan−1 , g, (c) to the left and angle of inclination of, the pendulum with the vertical is, a, sin1 − , g, (d) to the left and angle of inclination of, the pendulum with the vertical is, g, tan−1 , a, , R, , Q, , (a) decrease, (b) remain constant, (c) change according to the smallest, force QR, (d) increase, , Ans. (b), As the three forces are represented by, three sides of a triangle taken in order,, then they will be in equilibrium., P, , Ans. (b), As the truck move to the right, so the, bob will move to the left due to inertia of, rest with acceleration a., Thus, the given situation can be drawn, as, a, , ⇒, , θ, (a), , θ, , ma, mg, , (b), , R, , Q, , ⇒Fnet = FPQ + FQR + FRP = 0, dv, dv, Fnet = m × a = m = 0 ⇒ = 0, dt, dt, or v = constant, So, the velocity of particle remain, constant.

Page 2 :

04 A rigid ball of mass m strikes a rigid, wall at 60° and gets reflected, without loss of speed as shown in, the figure. The value of impulse, imparted by the wall on the ball will, be, [NEET 2016], , Ans. (c), The area underF- t graph gives change in, momentum., 1, For 0 to 2 s, ∆ p1 = × 2 × 6, 2, = 6 kg - m/s, For 4 to 8 s, ∆ p3 = 4 × 3 = 12 kg - m/s, , v, , m ′g, , ⇒, , So, total change in momentum for 0 to 8 s, 60°, , ∆ pnet = ∆ p1 + ∆ p2 + ∆ p3, , 60°, , = (+ 6 − 6 + 12), , (a) mv, , (b) 2mv, , (c) mv /2, , (d) mv /3, , Ans. (a), As we know that, impulse is imparted, due to change in perpendicular, components of momentum of ball., J = ∆ p = mv f − mv i, , = mv cos60 °− (− mv cos60 ° ), 1, = 2mv cos 60 ° = 2mv × = mv, 2, , 05 A bullet of mass 10 g moving, horizontal with a velocity of 400 m/s, strikes a wood block of mass 2 kg, which is suspended by light, inextensible string of length 5 m. As, result, the centre of gravity of the, block found to rise a vertical, distance of 10 cm. The speed of the, bullet after it emerges of, horizontally from the block will be, , NOTE, Graphs on negative axis gives − ve, momentum., , 07 A balloon with mass m is, descending down with an, acceleration a (where, a < g). How, much mass should be removed, from it so that it starts moving up, with an acceleration a?, [CBSE AIPMT 2014], , 2ma, 2ma, ma, ma, (b), (c), (d), (a), g+a, g −a, g+a, g −a, , Ans. (a), When, the balloon is descending down, with acceleration a, B, , (b) 80 m/s, (d) 160 m/s, , ⇒, , a, mg, , (0.01) × 400 + 0 = 2v + (0.01) v ′ ...(i), , Also velocity v of the block just after the, collision is, ...(ii), , ⇒ From Eqs. (i) and (ii), we have, v′ ≈ 120 m/s, , 06 The force F acting on a particle of, mass m is indicated by the, force-time graph shown below., The change in momentum of the, particle over the time interval, from 0 to 8 s is [CBSE AIPMT 2014], (a) 24 N-s (b) 20 N-s (c) 12 N-s (d) 6 N-s, , So, mass removed (m − m′), B − m′ g = m′ × a, mg − B = m × a, B − m′g = m′ × a, mg − m′g = ma + m′ a, (mg − ma) = m′ (g + a), ⇒, , m (g − a) = m′ (g + a), m(g − a), m′ =, g+a, , [Qp = mv], , 3, , ⇒, , p 3 = − (12 i + 16 j), p 3 = (12) 2 + (16) 2, = 144 + 256, , Let the new mass of the balloon be m′., , Solving Eqs. (i) and (ii),, , p1 + p2 + p3 = 0, , 1 × 12 i + 2 × 8 j + p3 = 0, 12 $i + 16$j + p = 0, , ∴, , …(i), , [B → Buoyant force], Here, we should assume that while, removing same mass the volume of, balloon and hence,buoyant force will not, change., , So,, , (b) 5 kg, (d)17 kg, , Ans. (b), Concept Momentum is conserved, before and after collision., We have,, , mg − B = m × a, , According to the law of conservation of, momentum, pi = p f, , v = 2gh = 2 × 10 × 0.1= 2, , (a) 3 kg, (c) 7 kg, , So, from free body diagram, , Ans. (c), , ⇒, , 08 An explosion breaks a rock into, three parts in a horizontal plane., Two of them go off at right angles, to each other. The first part of, mass 1 kg moves with a speed of 12, ms −1 and the second part of mass, 2 kg moves with 8 ms −1 speed. If, the third part flies off with 4 ms −1, speed, then its mass is [NEET 2013], , ∴, , [NEET 2016], , (a) 100 m/s, (c) 120 m/s, , , (g − a) , m 1 −, , (g + a) , (g + a) − (g − a) , =m , , (g + a), , , , g +a − g +a , 2ma, ⇒ m, ⇒ ∆m = g + a, g+a, , , , = 12 kg - m /s= 12 N-s, , v, , a, , So, mass removed = m − m′, , For 2 to 4 s, ∆ p2 = 2 × − 3 = − 6 kg - m/s, , m, , B, , …(ii), , = 20 kg-m/s, Now,, ⇒, , p 3 = m3 v3, p 20, m3 = 3 =, = 5 kg, v3, 4, , 09 Two spheres A and B of masses m 1, and m2 respectively collide. A is at, rest initially and B is moving with, velocity v along x-axis. After, v, collision, B has a velocity in a, 2, direction perpendicular to the, original direction. The mass A, moves after collision in the, direction, [CBSE AIPMT 2012], (a) same as that of B, (b) opposite to that of B

04 A rigid ball of mass m strikes a rigid, wall at 60° and gets reflected, without loss of speed as shown in, the figure. The value of impulse, imparted by the wall on the ball will, be, [NEET 2016], , Ans. (c), The area underF- t graph gives change in, momentum., 1, For 0 to 2 s, ∆ p1 = × 2 × 6, 2, = 6 kg - m/s, For 4 to 8 s, ∆ p3 = 4 × 3 = 12 kg - m/s, , v, , m ′g, , ⇒, , So, total change in momentum for 0 to 8 s, 60°, , ∆ pnet = ∆ p1 + ∆ p2 + ∆ p3, , 60°, , = (+ 6 − 6 + 12), , (a) mv, , (b) 2mv, , (c) mv /2, , (d) mv /3, , Ans. (a), As we know that, impulse is imparted, due to change in perpendicular, components of momentum of ball., J = ∆ p = mv f − mv i, , = mv cos60 °− (− mv cos60 ° ), 1, = 2mv cos 60 ° = 2mv × = mv, 2, , 05 A bullet of mass 10 g moving, horizontal with a velocity of 400 m/s, strikes a wood block of mass 2 kg, which is suspended by light, inextensible string of length 5 m. As, result, the centre of gravity of the, block found to rise a vertical, distance of 10 cm. The speed of the, bullet after it emerges of, horizontally from the block will be, , NOTE, Graphs on negative axis gives − ve, momentum., , 07 A balloon with mass m is, descending down with an, acceleration a (where, a < g). How, much mass should be removed, from it so that it starts moving up, with an acceleration a?, [CBSE AIPMT 2014], , 2ma, 2ma, ma, ma, (b), (c), (d), (a), g+a, g −a, g+a, g −a, , Ans. (a), When, the balloon is descending down, with acceleration a, B, , (b) 80 m/s, (d) 160 m/s, , ⇒, , a, mg, , (0.01) × 400 + 0 = 2v + (0.01) v ′ ...(i), , Also velocity v of the block just after the, collision is, ...(ii), , ⇒ From Eqs. (i) and (ii), we have, v′ ≈ 120 m/s, , 06 The force F acting on a particle of, mass m is indicated by the, force-time graph shown below., The change in momentum of the, particle over the time interval, from 0 to 8 s is [CBSE AIPMT 2014], (a) 24 N-s (b) 20 N-s (c) 12 N-s (d) 6 N-s, , So, mass removed (m − m′), B − m′ g = m′ × a, mg − B = m × a, B − m′g = m′ × a, mg − m′g = ma + m′ a, (mg − ma) = m′ (g + a), ⇒, , m (g − a) = m′ (g + a), m(g − a), m′ =, g+a, , [Qp = mv], , 3, , ⇒, , p 3 = − (12 i + 16 j), p 3 = (12) 2 + (16) 2, = 144 + 256, , Let the new mass of the balloon be m′., , Solving Eqs. (i) and (ii),, , p1 + p2 + p3 = 0, , 1 × 12 i + 2 × 8 j + p3 = 0, 12 $i + 16$j + p = 0, , ∴, , …(i), , [B → Buoyant force], Here, we should assume that while, removing same mass the volume of, balloon and hence,buoyant force will not, change., , So,, , (b) 5 kg, (d)17 kg, , Ans. (b), Concept Momentum is conserved, before and after collision., We have,, , mg − B = m × a, , According to the law of conservation of, momentum, pi = p f, , v = 2gh = 2 × 10 × 0.1= 2, , (a) 3 kg, (c) 7 kg, , So, from free body diagram, , Ans. (c), , ⇒, , 08 An explosion breaks a rock into, three parts in a horizontal plane., Two of them go off at right angles, to each other. The first part of, mass 1 kg moves with a speed of 12, ms −1 and the second part of mass, 2 kg moves with 8 ms −1 speed. If, the third part flies off with 4 ms −1, speed, then its mass is [NEET 2013], , ∴, , [NEET 2016], , (a) 100 m/s, (c) 120 m/s, , , (g − a) , m 1 −, , (g + a) , (g + a) − (g − a) , =m , , (g + a), , , , g +a − g +a , 2ma, ⇒ m, ⇒ ∆m = g + a, g+a, , , , = 12 kg - m /s= 12 N-s, , v, , a, , So, mass removed = m − m′, , For 2 to 4 s, ∆ p2 = 2 × − 3 = − 6 kg - m/s, , m, , B, , …(ii), , = 20 kg-m/s, Now,, ⇒, , p 3 = m3 v3, p 20, m3 = 3 =, = 5 kg, v3, 4, , 09 Two spheres A and B of masses m 1, and m2 respectively collide. A is at, rest initially and B is moving with, velocity v along x-axis. After, v, collision, B has a velocity in a, 2, direction perpendicular to the, original direction. The mass A, moves after collision in the, direction, [CBSE AIPMT 2012], (a) same as that of B, (b) opposite to that of B

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1, (c) θ = tan−1 to the x-axis, 2, −1 −1, (d) θ = tan to the x-axis, 2, Ans. (c), Here, p i = m2 vi + m1 × 0, p f = m2, , v$, j + m1 × v1, 2, , Law of conservation of momentum, pi = pf, v, m2 v$i = m2 $j + m1 × v1, 2, m, m v, v1 = 2 v $i + 2 $j, m1, m1 2, , a = 1ms−2, , = 10 2 kg, , Given, m = 0.5 kg, v = 12 m/s, t = 0.25 s, θ = 30 °, 2 × 0.5 × 12 sin 30 °, Hence, F =, = 24 N, 0.25, , 12 A 0.5 kg ball moving with a speed, of 12 m/s strikes a hard wall at an, angle of 30° with the wall. It is, reflected with the same speed and, at the same angle. If the ball is in, contact with the wall for 0.25 s,, the average force acting on the, wall is, [CBSE AIPMT 2006], , 13 A block of mass m is placed on a, smooth wedge of inclination θ. The, whole system is accelerated, horizontally, so that the block does, not slip on the wedge. The force, exerted by the wedge on the block, (g is acceleration due to gravity), will be, [CBSE AIPMT 2004], , ∴, , m=, , 10 2, 1, , 30°, , v/2, , A (m1), rest, , Ans. (d), , x, θ, A u, (m1), , Let an acceleration to the wedge be, given towards left, then the block (being, in non-inertial frame) has a pseudo, acceleration to the right because of, which the block is not slipping, , From this equation, we can find, y 1, 1, tanθ = = , θ = tan−1 to the x-axis., 2, x 2, , (a) 48 N, , (b) 24 N, , The vectorOA represents the, momentum of the object before the, collision, and the vectorOB that after the, collision. The vector AB represents the, change in momentum of the object ∆p., B, , O, , 60°, , 3 0°, , m1 r1 = m2 r2, mr, 0.5 × 10, r2 = 1 1 =, = 0.1, m2, 50, , So, distance travelled by the man will be, 10 + 0.1 = 10.1 m, , 11 A body, under the action of a force, F = 6i$ − 8$j + 10 k$ , acquires an, acceleration of 1 ms −2 . The mass of, this body must be, , A, , As the magnitudes ofOA and OB are, equal, the components ofOA and OB, along the wall are equal and in the same, direction, while those perpendicular to, the wall are equal and opposite. Thus,, the change in momentum is only due to, the change in direction of the, perpendicular components., Hence,, ∆p = OB sin 30 ° – (– OA sin 30 ° ), , [CBSE AIPMT 2009], , (b) 10 kg, (d) 10 2 kg, , Ans. (d), Here, F = 6$i − 8$j + 10k$, |F | = 36 + 64 + 100 = 10 2 N, , = mv sin 30 ° – (–mv sin 30 ° ), = 2 mv sin 30 °, Its time rate will appear in the form of, average force acting on the wall., ∴, , ∴ mg sin θ = apseudo cos θ, , F × t = 2mv sin 30 ° or, 2mv sin 30 °, F=, t, , apseudo =, , mg sin θ, cos θ, , 14 An object of mass 3 kg is at rest. If, a force F = (6 t 2 i$ + 4t $j) N is applied, on the object, then the velocity of, the object at t = 3 s is, [CBSE AIPMT 2002], , 60°, , (d) 20 m, , As, mr = constant, , (a) 2 10 kg, (c) 20 kg, , (d) 96 N, , ⇒, , Ans. (b), , ⇒, , (c) 12 N, , Ans. (b), , 30°, , 10 A man of 50 kg mass is standing in, a gravity free space at a height of, 10 m above the floor. He throws a, stone of 0.5 kg mass downwards, with a speed 2 ms −1 . When the, stone reaches the floor, the, distance of the man above the floor, will be, [CBSE AIPMT 2010], (a) 9.9 m (b) 10.1 m (c) 10 m, , (b) mg sinθ, mgsinθ, (d), cos θ, , B, 30°, , B, (m 2 ), , (a) mg cos θ, (c) mg, , y, , v, , [QF = ma], , (a) 18 $i + 3$j, (c) 3$i + 18 $j, , (b) 18 $i + 6 $j, (d) 18 $i + 4 $j, , Ans. (b), According to Newton’s 2nd law, force, applied on an object is equal to rate of, change of momentum., dp, i.e., F=, dt, dv, …(i), or, F=m, dt, Given, m = 3 kg, t = 3 s,F = (6 t 2 $i + 4t $j) N, Substituting these values in Eq. (i), we get, dv, (6 t 2 $i + 4 t $j) = 3, dt, 1 2$, or, dv = (6 t i + 4 t$j) dt, 3, Now, taking integration of both sides, we, get, t 1, 2$, $, ∫ dv = ∫ 0 3 (6t i + 4 tj ) dt, 1 t, v = ∫ (6t 2 $i + 4 t$j ) dt, 3 0

1, (c) θ = tan−1 to the x-axis, 2, −1 −1, (d) θ = tan to the x-axis, 2, Ans. (c), Here, p i = m2 vi + m1 × 0, p f = m2, , v$, j + m1 × v1, 2, , Law of conservation of momentum, pi = pf, v, m2 v$i = m2 $j + m1 × v1, 2, m, m v, v1 = 2 v $i + 2 $j, m1, m1 2, , a = 1ms−2, , = 10 2 kg, , Given, m = 0.5 kg, v = 12 m/s, t = 0.25 s, θ = 30 °, 2 × 0.5 × 12 sin 30 °, Hence, F =, = 24 N, 0.25, , 12 A 0.5 kg ball moving with a speed, of 12 m/s strikes a hard wall at an, angle of 30° with the wall. It is, reflected with the same speed and, at the same angle. If the ball is in, contact with the wall for 0.25 s,, the average force acting on the, wall is, [CBSE AIPMT 2006], , 13 A block of mass m is placed on a, smooth wedge of inclination θ. The, whole system is accelerated, horizontally, so that the block does, not slip on the wedge. The force, exerted by the wedge on the block, (g is acceleration due to gravity), will be, [CBSE AIPMT 2004], , ∴, , m=, , 10 2, 1, , 30°, , v/2, , A (m1), rest, , Ans. (d), , x, θ, A u, (m1), , Let an acceleration to the wedge be, given towards left, then the block (being, in non-inertial frame) has a pseudo, acceleration to the right because of, which the block is not slipping, , From this equation, we can find, y 1, 1, tanθ = = , θ = tan−1 to the x-axis., 2, x 2, , (a) 48 N, , (b) 24 N, , The vectorOA represents the, momentum of the object before the, collision, and the vectorOB that after the, collision. The vector AB represents the, change in momentum of the object ∆p., B, , O, , 60°, , 3 0°, , m1 r1 = m2 r2, mr, 0.5 × 10, r2 = 1 1 =, = 0.1, m2, 50, , So, distance travelled by the man will be, 10 + 0.1 = 10.1 m, , 11 A body, under the action of a force, F = 6i$ − 8$j + 10 k$ , acquires an, acceleration of 1 ms −2 . The mass of, this body must be, , A, , As the magnitudes ofOA and OB are, equal, the components ofOA and OB, along the wall are equal and in the same, direction, while those perpendicular to, the wall are equal and opposite. Thus,, the change in momentum is only due to, the change in direction of the, perpendicular components., Hence,, ∆p = OB sin 30 ° – (– OA sin 30 ° ), , [CBSE AIPMT 2009], , (b) 10 kg, (d) 10 2 kg, , Ans. (d), Here, F = 6$i − 8$j + 10k$, |F | = 36 + 64 + 100 = 10 2 N, , = mv sin 30 ° – (–mv sin 30 ° ), = 2 mv sin 30 °, Its time rate will appear in the form of, average force acting on the wall., ∴, , ∴ mg sin θ = apseudo cos θ, , F × t = 2mv sin 30 ° or, 2mv sin 30 °, F=, t, , apseudo =, , mg sin θ, cos θ, , 14 An object of mass 3 kg is at rest. If, a force F = (6 t 2 i$ + 4t $j) N is applied, on the object, then the velocity of, the object at t = 3 s is, [CBSE AIPMT 2002], , 60°, , (d) 20 m, , As, mr = constant, , (a) 2 10 kg, (c) 20 kg, , (d) 96 N, , ⇒, , Ans. (b), , ⇒, , (c) 12 N, , Ans. (b), , 30°, , 10 A man of 50 kg mass is standing in, a gravity free space at a height of, 10 m above the floor. He throws a, stone of 0.5 kg mass downwards, with a speed 2 ms −1 . When the, stone reaches the floor, the, distance of the man above the floor, will be, [CBSE AIPMT 2010], (a) 9.9 m (b) 10.1 m (c) 10 m, , (b) mg sinθ, mgsinθ, (d), cos θ, , B, 30°, , B, (m 2 ), , (a) mg cos θ, (c) mg, , y, , v, , [QF = ma], , (a) 18 $i + 3$j, (c) 3$i + 18 $j, , (b) 18 $i + 6 $j, (d) 18 $i + 4 $j, , Ans. (b), According to Newton’s 2nd law, force, applied on an object is equal to rate of, change of momentum., dp, i.e., F=, dt, dv, …(i), or, F=m, dt, Given, m = 3 kg, t = 3 s,F = (6 t 2 $i + 4t $j) N, Substituting these values in Eq. (i), we get, dv, (6 t 2 $i + 4 t $j) = 3, dt, 1 2$, or, dv = (6 t i + 4 t$j) dt, 3, Now, taking integration of both sides, we, get, t 1, 2$, $, ∫ dv = ∫ 0 3 (6t i + 4 tj ) dt, 1 t, v = ∫ (6t 2 $i + 4 t$j ) dt, 3 0

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(given), t =3s, 1 3, 2$, $, v = ∫ (6 t i + 4 t j ) dt, 3 0, , but, ∴, , 3, , 1 6 t $ 4 t $ , v= , i+, j, 3 3, 2 0, 1, v = [2(3) 3 $i + 2(3) 2 $j ], 3, 1, v = [54$i + 18$j ], 3, v = 18$i + 6$j, 3, , or, or, or, or, , 2, , 15 A player takes 0.1 s in catching a, ball of mass 150 g moving with, velocity of 20 m/s. The force, imparted by the ball on the hands, of the player is [CBSE AIPMT 2001], (a) 0.3 N (b) 3 N, , (c) 30 N, , (d) 300 N, , Ans. (c), Force imparted = rate of change of, momentum, p − p2, m(v1 − v2 ), ∆p, or F = 1, or F =, F=, ∆t, ∆t, ∆t, Here, mass of body m = 150 g = 0.150 kg,, v1 = 20 m/s,, , Equating the momentum of the system, along OA and OB to zero, we get, m × 30 − 3 m × v cosθ = 0 ...(i), and m × 30 − 3 m × v sinθ = 0, , From Eqs. (i) and (ii), we get, 3 mv cos θ = 3 mv sin θ or cos θ = sin θ, ∴, θ = 45°, Thus,, ∠AOC = ∠BOC, = 180 ° − 45° = 135°, Putting the value ofθ in Eq. (i), we get, 3 mv, 30 m = 3 mv cos 45° =, 2, v = 10 2 m/s, ∴, The third piece will go with a velocity of, 10 2 m/s in a direction making an angle, of 135° with either piece., , 16 1 kg body explodes into three, fragments. The ratio of their, masses is 1 : 1 : 3. The fragments of, same mass move perpendicular to, each other with speeds 30 m/s,, while the heavier part remains in, the initial direction. The speed of, heavier part is [CBSE AIPMT 2001], 10, (a), m/s, 2, (c) 20 2 m/s, , (b) 10 2 m/s, (d) 30 2 m/s, , 18 A ball of mass 3 kg moving with a, speed of 100 m/s, strikes a wall at, an angle 60° (as shown in figure)., The ball rebounds at the same, speed and remains in contact with, the wall for 0.2 s, the force exerted, by the ball on the wall is, [CBSE AIPMT 2000], , 60°, , 3, , or, , 3 mv3 = (m × 30) 2 + (m × 30) 2, , or, , 30 2, v3 =, = 10 2 m/s, 3, , 17 A particle of mass 1 kg is thrown, vertically upwards with speed, 100 m/s. After 5 s, it explodes into, two parts. One part of mass 400 g, comes back with speed 25 m/s, what, is the speed of other part just after, explosion?, [CBSE AIPMT 2000], (a) 100 m/s upwards, (b) 600 m/s upwards, (c) 100 m/s downwards, (d) 300 m/s upwards, , 60°, , (a) 1500 3 N, (c) 300 3 N, , (b) 1500 N, (d) 300 N, , Ans. (a), Concept Apply 2nd law of motion i.e.,, rate of change of momentum is equal to, force applied., The vectorOA represents the, momentum of the wall, before the, collision and the vectorOB that after the, collision. The vector AB represents the, change in momentum of the ball ∆P., As, the magnitude ofOA and OB are, equal the components ofOA and AB, along the wall are equal and in the same, direction, while those perpendicular to, the wall are equal and opposite. Thus,, the change in momentum is only due to, the change in direction of the, perpendicular components., , Ans. (a), , A, , According to 1st equation of motion,, velocity of particle after 5 s, , 60°, , v = u − gt, , 60°, , v = 100 − 10 × 5, , Ans. (b), Concept Apply conservation of, momentum with direction., Let u be the velocity andθ the direction, of the third piece as shown., m, , B, , θ, O, , C, , 0.6 v2 = 60 or v2 =, , Alternative, The square of momentum of third piece, is equal to sum of squares of momentum, first and second pieces., p2 = p12 + p22, 3, or, p = p12 + p22, , v2 = 0, Time taken, ∆t = 0.1 s, 0.150 × (20 − 0), = 30 N, F=, ∴, 0.1, , ...(ii), , 60, = 100 m/s, 0.6, As v2 is positive, therefore the other, part will move upwards with a velocity of, 100 m/s., , or, , 3m, , A, m, , = 100 − 50 = 50 m/s (upwards), Applying conservation of linear, momentum gives, …(i), Mv = m1 v1 + m2 v2, Taking upward direction positive, the, velocity v1 will be negative., ∴ v1 = − 25 m/s, v = 50 m/s, Also, M = 1 kg, m1 = 400 g = 0.4 kg, and m2 = (M − m1 ) = 1 − 0.4 = 0.6 kg, Thus, Eq. (i) becomes,, 1 × 50 = 0.4 × (− 25) + 0.6 v2, or, 50 = − 10 + 0.6 v2, , O, 60°, B, , Hence, ∆P = OA sin60 ° − (− OB sin 60 ° ), = mv sin 60 ° + mv sin 60 °, = 2 mv sin 60 ° = 2 × 3 × 100 ×, = 300 3 kg-m/s, The force exerted on the wall, , F=, , ∆P 300 3, = 1500 3 N, =, ∆t, 02, ., , 3, 2

(given), t =3s, 1 3, 2$, $, v = ∫ (6 t i + 4 t j ) dt, 3 0, , but, ∴, , 3, , 1 6 t $ 4 t $ , v= , i+, j, 3 3, 2 0, 1, v = [2(3) 3 $i + 2(3) 2 $j ], 3, 1, v = [54$i + 18$j ], 3, v = 18$i + 6$j, 3, , or, or, or, or, , 2, , 15 A player takes 0.1 s in catching a, ball of mass 150 g moving with, velocity of 20 m/s. The force, imparted by the ball on the hands, of the player is [CBSE AIPMT 2001], (a) 0.3 N (b) 3 N, , (c) 30 N, , (d) 300 N, , Ans. (c), Force imparted = rate of change of, momentum, p − p2, m(v1 − v2 ), ∆p, or F = 1, or F =, F=, ∆t, ∆t, ∆t, Here, mass of body m = 150 g = 0.150 kg,, v1 = 20 m/s,, , Equating the momentum of the system, along OA and OB to zero, we get, m × 30 − 3 m × v cosθ = 0 ...(i), and m × 30 − 3 m × v sinθ = 0, , From Eqs. (i) and (ii), we get, 3 mv cos θ = 3 mv sin θ or cos θ = sin θ, ∴, θ = 45°, Thus,, ∠AOC = ∠BOC, = 180 ° − 45° = 135°, Putting the value ofθ in Eq. (i), we get, 3 mv, 30 m = 3 mv cos 45° =, 2, v = 10 2 m/s, ∴, The third piece will go with a velocity of, 10 2 m/s in a direction making an angle, of 135° with either piece., , 16 1 kg body explodes into three, fragments. The ratio of their, masses is 1 : 1 : 3. The fragments of, same mass move perpendicular to, each other with speeds 30 m/s,, while the heavier part remains in, the initial direction. The speed of, heavier part is [CBSE AIPMT 2001], 10, (a), m/s, 2, (c) 20 2 m/s, , (b) 10 2 m/s, (d) 30 2 m/s, , 18 A ball of mass 3 kg moving with a, speed of 100 m/s, strikes a wall at, an angle 60° (as shown in figure)., The ball rebounds at the same, speed and remains in contact with, the wall for 0.2 s, the force exerted, by the ball on the wall is, [CBSE AIPMT 2000], , 60°, , 3, , or, , 3 mv3 = (m × 30) 2 + (m × 30) 2, , or, , 30 2, v3 =, = 10 2 m/s, 3, , 17 A particle of mass 1 kg is thrown, vertically upwards with speed, 100 m/s. After 5 s, it explodes into, two parts. One part of mass 400 g, comes back with speed 25 m/s, what, is the speed of other part just after, explosion?, [CBSE AIPMT 2000], (a) 100 m/s upwards, (b) 600 m/s upwards, (c) 100 m/s downwards, (d) 300 m/s upwards, , 60°, , (a) 1500 3 N, (c) 300 3 N, , (b) 1500 N, (d) 300 N, , Ans. (a), Concept Apply 2nd law of motion i.e.,, rate of change of momentum is equal to, force applied., The vectorOA represents the, momentum of the wall, before the, collision and the vectorOB that after the, collision. The vector AB represents the, change in momentum of the ball ∆P., As, the magnitude ofOA and OB are, equal the components ofOA and AB, along the wall are equal and in the same, direction, while those perpendicular to, the wall are equal and opposite. Thus,, the change in momentum is only due to, the change in direction of the, perpendicular components., , Ans. (a), , A, , According to 1st equation of motion,, velocity of particle after 5 s, , 60°, , v = u − gt, , 60°, , v = 100 − 10 × 5, , Ans. (b), Concept Apply conservation of, momentum with direction., Let u be the velocity andθ the direction, of the third piece as shown., m, , B, , θ, O, , C, , 0.6 v2 = 60 or v2 =, , Alternative, The square of momentum of third piece, is equal to sum of squares of momentum, first and second pieces., p2 = p12 + p22, 3, or, p = p12 + p22, , v2 = 0, Time taken, ∆t = 0.1 s, 0.150 × (20 − 0), = 30 N, F=, ∴, 0.1, , ...(ii), , 60, = 100 m/s, 0.6, As v2 is positive, therefore the other, part will move upwards with a velocity of, 100 m/s., , or, , 3m, , A, m, , = 100 − 50 = 50 m/s (upwards), Applying conservation of linear, momentum gives, …(i), Mv = m1 v1 + m2 v2, Taking upward direction positive, the, velocity v1 will be negative., ∴ v1 = − 25 m/s, v = 50 m/s, Also, M = 1 kg, m1 = 400 g = 0.4 kg, and m2 = (M − m1 ) = 1 − 0.4 = 0.6 kg, Thus, Eq. (i) becomes,, 1 × 50 = 0.4 × (− 25) + 0.6 v2, or, 50 = − 10 + 0.6 v2, , O, 60°, B, , Hence, ∆P = OA sin60 ° − (− OB sin 60 ° ), = mv sin 60 ° + mv sin 60 °, = 2 mv sin 60 ° = 2 × 3 × 100 ×, = 300 3 kg-m/s, The force exerted on the wall, , F=, , ∆P 300 3, = 1500 3 N, =, ∆t, 02, ., , 3, 2

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19 The force on a rocket moving with, a velocity 300 m/s is 210 N. The, rate of consumption of fuel of, rocket is, [CBSE AIPMT 1999], (a) 0.7 kg/s, (c) 0.07 kg/s, , (b) 1.4 kg/s, (d) 10.7 kg/s, , Concept Whenever there is change in, dm, the mass w.r.t. time, apply F = − v, dt, Thrust force on the rocket, dm , Ft = vr −, , dt , , (upwards), , Rate of combustion of fuel, dm Ft, −, =, dt v r, , –1, , (a) 127.5kg s, (c) 185.5kg s–1, , F = 600 − 2 × 10 5 t, When, bullet leaves the barrel, the force, on the bullet becomes zero., 600, So, 600 − 2 × 10 5 t = 0 ⇒ t =, 2 × 10 5, , t, , I = ∫ Fdt, , (b) 187.5kg s, (d) 137.5kg s–1, , Ans. (b), , 3 × 10, 0, , −3, , (600 − 2 × 10 5 t) dt, 3 × 10 –3, , , 2 × 10 5 t 2 , = 600 t −, , 2, , 0, , = 600 × 3 × 10 −3 − 10 5 × (3 × 10 −3 ) 2, = 1.8 − 0.9 = 0.9 N-s, Alternative, As obtained in previous method, the, time taken by bullet when it leaves the, barrel, t = 3 × 10 −3 s, , Thrust force on the rocket, dm , Ft = vr −, , dt , , (upwards), , Weight of the rocket, (downwards), , Net force on the rocket, Fnet = Ft − w, −dm , ⇒, ma = vr , − mg, dt , −dm = m (g + a), , , dt , vr, , ∴ Rate of gas ejected per second, 5000 (10 + 20) 5000 × 30, =, =, 800, 800, = 187.5 kg s, , –1, , 21 A bullet is fired from a gun. The, force on the bullet is given by, F = 600 − 2 × 10 5 t, where, F is in newton and t in, second. The force on the bullet, , Let F1 and F2 denote the forces at the, time of firing of bullets i.e. at t = 0 and at, the time of leaving the bullet i.e. at, t = 3 × 10 −3 s., F1 = 600 − 2 × 10 5 × 0 = 600 N, F2 = 600 − 2 × 10 5 × 3 × 10 −3 = 0, Mean value of force, 600 + 0, 1, F = (F1 + F2 ) =, = 300 N, 2, 2, Thus, impulse = F × t, = 300 × 3 × 10 −3 = 0.9 N - s, , 22 A 10 N force is applied on a body, produces an acceleration of, 1 m /s2 . The mass of the body is, [CBSE AIPMT 1996], , (a) 5 kg, (c) 15 kg, , or force F = ma, Here, F = 10 N, a = 1m / s2 ⇒ ∴ m =, , F 10, =, = 10 kg, 1, a, , 23 A ball of mass 150 g moving with an, acceleration 20 m/s2 is hit by a, force, which acts on it for 0.1 s. The, impulsive force is, [CBSE AIPMT 1996], , (a) 0.5 N-s, (c) 0.3 N-s, , (b) 0.1 N-s, (d) 1.2 N-s, , Ans. (c), Impulse of a force, which is the product, of average force during impact and the, time for, which the impact lasts is, measured by the total change in linear, momentum produced during the impact., Impulse I = Fav × t = p 2 − p 1, , 0, , –1, , w = mg, , Concept To calculate impulse first of all, calculate the time during which force, becomes zero., We have given,, , =∫, , multiplying mass of the body and the, acceleration produced in it., , (b) Zero, (d) 1.8 N-s, , Then, average impulse imparted to the, bullet, , Ft = 210 N, , 20 A 5000 kg rocket is set for vertical, firing. The exhaust speed is, 800 ms −1 . To give an initial upward, acceleration of 20 m/s 2 , the, amount of gas ejected per second, to supply the needed thrust will be, [CBSE AIPMT 1998], (g = 10 ms −2 ), , ⇒, , (a) 8 N-s, (c) 0.9 N-s, , = 3 × 10 −3 s, , vr = 300 m/s, dm 210, −, =, = 0.7 kg/s, dt 300, , ∴, , [CBSE AIPMT 1998], , Ans. (c), , Ans. (a), , Given,, , becomes zero as soon as it leaves, the barrel. What is the average, impulse imparted to the bullet?, , (b) 10 kg, (d) 20 kg, , Ans. (b), According to second law of motion,, magnitude of force can be calculated by, , 150, kg, 1000, , Here, Mass = 150 g =, , 150, × 20 = 3 N, 1000, , ∴, , F=, , ∴, , I = F ⋅∆t = 3 × 0.1 = 0.3 N-s, , 24 If the force on a rocket moving with, a velocity of 300 m/s is 345 N, then, the rate of combustion of the fuel, is, [CBSE AIPMT 1995], (a) 0.55 kg/s, (c) 1.15 kg/s, , (b) 0.75 kg/s, (d) 2.25 kg/s, , Ans. (c), Thrust on the rocket is the force with, which the rocket moves upwards. Thrust, dm, on rocket at timet is given by F = − u, dt, The negative sign indicates that thrust, on the rocket is in a direction opposite to, the direction of escaping gases., Here, velocity of the rocket u = 300 m/s, and force F = 345 N, ∴ Rate of combustion of fuel, dm F 345, − , = 1.15 kg/s, = =, dt u 300, , 25 A satellite in a force free space, sweeps stationary interplanetary, dM, dust at a rate. = αv. The, dt , acceleration of satellite is, [CBSE AIPMT 1994], , 2αv2, αv2, αv2, (a) −, (c) −, (d) − αv2, (b) −, M, 2M, M

19 The force on a rocket moving with, a velocity 300 m/s is 210 N. The, rate of consumption of fuel of, rocket is, [CBSE AIPMT 1999], (a) 0.7 kg/s, (c) 0.07 kg/s, , (b) 1.4 kg/s, (d) 10.7 kg/s, , Concept Whenever there is change in, dm, the mass w.r.t. time, apply F = − v, dt, Thrust force on the rocket, dm , Ft = vr −, , dt , , (upwards), , Rate of combustion of fuel, dm Ft, −, =, dt v r, , –1, , (a) 127.5kg s, (c) 185.5kg s–1, , F = 600 − 2 × 10 5 t, When, bullet leaves the barrel, the force, on the bullet becomes zero., 600, So, 600 − 2 × 10 5 t = 0 ⇒ t =, 2 × 10 5, , t, , I = ∫ Fdt, , (b) 187.5kg s, (d) 137.5kg s–1, , Ans. (b), , 3 × 10, 0, , −3, , (600 − 2 × 10 5 t) dt, 3 × 10 –3, , , 2 × 10 5 t 2 , = 600 t −, , 2, , 0, , = 600 × 3 × 10 −3 − 10 5 × (3 × 10 −3 ) 2, = 1.8 − 0.9 = 0.9 N-s, Alternative, As obtained in previous method, the, time taken by bullet when it leaves the, barrel, t = 3 × 10 −3 s, , Thrust force on the rocket, dm , Ft = vr −, , dt , , (upwards), , Weight of the rocket, (downwards), , Net force on the rocket, Fnet = Ft − w, −dm , ⇒, ma = vr , − mg, dt , −dm = m (g + a), , , dt , vr, , ∴ Rate of gas ejected per second, 5000 (10 + 20) 5000 × 30, =, =, 800, 800, = 187.5 kg s, , –1, , 21 A bullet is fired from a gun. The, force on the bullet is given by, F = 600 − 2 × 10 5 t, where, F is in newton and t in, second. The force on the bullet, , Let F1 and F2 denote the forces at the, time of firing of bullets i.e. at t = 0 and at, the time of leaving the bullet i.e. at, t = 3 × 10 −3 s., F1 = 600 − 2 × 10 5 × 0 = 600 N, F2 = 600 − 2 × 10 5 × 3 × 10 −3 = 0, Mean value of force, 600 + 0, 1, F = (F1 + F2 ) =, = 300 N, 2, 2, Thus, impulse = F × t, = 300 × 3 × 10 −3 = 0.9 N - s, , 22 A 10 N force is applied on a body, produces an acceleration of, 1 m /s2 . The mass of the body is, [CBSE AIPMT 1996], , (a) 5 kg, (c) 15 kg, , or force F = ma, Here, F = 10 N, a = 1m / s2 ⇒ ∴ m =, , F 10, =, = 10 kg, 1, a, , 23 A ball of mass 150 g moving with an, acceleration 20 m/s2 is hit by a, force, which acts on it for 0.1 s. The, impulsive force is, [CBSE AIPMT 1996], , (a) 0.5 N-s, (c) 0.3 N-s, , (b) 0.1 N-s, (d) 1.2 N-s, , Ans. (c), Impulse of a force, which is the product, of average force during impact and the, time for, which the impact lasts is, measured by the total change in linear, momentum produced during the impact., Impulse I = Fav × t = p 2 − p 1, , 0, , –1, , w = mg, , Concept To calculate impulse first of all, calculate the time during which force, becomes zero., We have given,, , =∫, , multiplying mass of the body and the, acceleration produced in it., , (b) Zero, (d) 1.8 N-s, , Then, average impulse imparted to the, bullet, , Ft = 210 N, , 20 A 5000 kg rocket is set for vertical, firing. The exhaust speed is, 800 ms −1 . To give an initial upward, acceleration of 20 m/s 2 , the, amount of gas ejected per second, to supply the needed thrust will be, [CBSE AIPMT 1998], (g = 10 ms −2 ), , ⇒, , (a) 8 N-s, (c) 0.9 N-s, , = 3 × 10 −3 s, , vr = 300 m/s, dm 210, −, =, = 0.7 kg/s, dt 300, , ∴, , [CBSE AIPMT 1998], , Ans. (c), , Ans. (a), , Given,, , becomes zero as soon as it leaves, the barrel. What is the average, impulse imparted to the bullet?, , (b) 10 kg, (d) 20 kg, , Ans. (b), According to second law of motion,, magnitude of force can be calculated by, , 150, kg, 1000, , Here, Mass = 150 g =, , 150, × 20 = 3 N, 1000, , ∴, , F=, , ∴, , I = F ⋅∆t = 3 × 0.1 = 0.3 N-s, , 24 If the force on a rocket moving with, a velocity of 300 m/s is 345 N, then, the rate of combustion of the fuel, is, [CBSE AIPMT 1995], (a) 0.55 kg/s, (c) 1.15 kg/s, , (b) 0.75 kg/s, (d) 2.25 kg/s, , Ans. (c), Thrust on the rocket is the force with, which the rocket moves upwards. Thrust, dm, on rocket at timet is given by F = − u, dt, The negative sign indicates that thrust, on the rocket is in a direction opposite to, the direction of escaping gases., Here, velocity of the rocket u = 300 m/s, and force F = 345 N, ∴ Rate of combustion of fuel, dm F 345, − , = 1.15 kg/s, = =, dt u 300, , 25 A satellite in a force free space, sweeps stationary interplanetary, dM, dust at a rate. = αv. The, dt , acceleration of satellite is, [CBSE AIPMT 1994], , 2αv2, αv2, αv2, (a) −, (c) −, (d) − αv2, (b) −, M, 2M, M

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Ans. (b), Thrust on the satellite is the force with, which the satellite moves upwards in, space. It is given by, dm, F =− u, dt, Here, initial velocity, u = v, rate of change of mass, dm, = αv, dt, As we know that,, dm, F = −v, = − v (αv) = − αv2, dt, F, αv2, Acceleration = = −, M, M, , 26 Physical independence of force is a, consequence of [CBSE AIPMT 1991], (a) third law of motion, (b) second law of motion, (c) first law of motion, (d) All of these, Ans. (c), According to Newton’s first law of, motion, a body continues to be in a state, of rest or of uniform motion, unless it is, acted upon by an external force to, change the state. Hence, Newton’s first, law of motion is related to physical, independence of force., , 27 A particle of mass m is moving with, a uniform velocity v 1 . It is given an, impulse such that its velocity, becomes v2 . The impulse is equal, to, [CBSE AIPMT 1990], (a) m [| v2 | − | v1 |], (c) m (v1 + v2 ), , 1, (b) m (v22 − v12 ), 2, (d) m (v2 − v1 ), , Ans. (d), Concept Impulse of a force can be, calculated as the product of large force, applied to the small time to which force, act., dp, i.e., F=, dt, ⇒, , F ⋅dt = dp, , impulse = p 2 = p1, ⇒, Impulse of a force, which is the product, of average force during impact and the, time for which the impact lasts, is, measured by the total change in linear, momentum produced during the impact., Here,, Impulse,, , p 1 = mv1 , p 2 = mv2, I = mv2 − mv1 = m ( v2 − v1 ), , 28 A 600 kg rocket is set for a vertical, firing. If the exhaust speed is 1000, ms −1 , the mass of the gas ejected, per second to supply the thrust, needed to overcome the weight of, rocket is, [CBSE AIPMT 1990], (a) 117.6kg s−1, (c) 6kg s−1, , (b) 58.6kg s−1, (d) 76.4kg s−1, , From the above free body diagram, the, relation for acceleration of the given, system can be given as, m − m1 , a= 2, ⋅g, m1 + m2 , 6 − 4, g, =, ×g=, 5, 4 + 6, Hence, correct option is (b)., , Ans. (c), Thrust on the rocket is the force with, which the rocket moves upwards., Thrust on the rocket at timet is given by, dm, F =−u, dt, where, u is relative velocity of exhaust, dm, gases with respect to the rocket., is, dt, rate of combustion of fuel at that instant., dm, dm mg, = mg ⇒ −, =, ∴ F =−u, dt, dt, u, , 30 A block of mass m is placed on a, smooth inclined wedge ABC of, inclination θ as shown in the figure., The wedge is given an acceleration, a towards the right. The relation, between a and θ for the block to, remain stationary on the wedge is, [NEET 2018], , A, , Here, m = 600 kg, u = 1000 ms−1, dm 600 × 10, = 6 kg s−1, ∴, −, =, dt, 1000, , Equilibrium of a Particle and, Common Forces in Mechanics, 29 Two bodies of mass 4 kg and 6 kg, are tied to the ends of a massless, string. The string passes over a, pulley which is frictionless, (see figure). The acceleration of the, system in terms of acceleration, due to gravity g is, [NEET (Sep.) 2020], , 4 kg, 6 kg, , (b) g/5, , a, q, C, , B, , (a) a = g cos θ, , TOPIC 2, , (a) g/2, , m, , (c) g/10, , (d) g, , Ans. (b), Given, m1 = 4 kg, m2 = 6 kg and a = ?, a, , 4 kg, m 1g, , 6 kg, m2 g, , a, , (c) a =, , (b) a =, , g, cosec θ, , g, sinθ, , (d) a = g tanθ, , Ans. (d), According to the question, the FBD of, the given condition will be, A, , R cosq, , R, , q, , ma, (Pseudo, force), B, , R sinq, q, mg q, , a, , C, , Since, the wedge is accelerating towards, right witha, thus a pseudo force acts in, the left direction in order to keep the, block stationary. As, the system is in, equilibrium., ∴, ΣFx = 0, or, ΣFy = 0, ⇒, R sinθ = ma, or, …(i), mg sinθ = ma, Similarly,, R cosθ = mg, or mg cosθ = mg, …(ii), Dividing Eq. (i) by Eq (ii), we get, mg sin θ ma, =, mg cos θ mg, a, tanθ =, ⇒, g

Ans. (b), Thrust on the satellite is the force with, which the satellite moves upwards in, space. It is given by, dm, F =− u, dt, Here, initial velocity, u = v, rate of change of mass, dm, = αv, dt, As we know that,, dm, F = −v, = − v (αv) = − αv2, dt, F, αv2, Acceleration = = −, M, M, , 26 Physical independence of force is a, consequence of [CBSE AIPMT 1991], (a) third law of motion, (b) second law of motion, (c) first law of motion, (d) All of these, Ans. (c), According to Newton’s first law of, motion, a body continues to be in a state, of rest or of uniform motion, unless it is, acted upon by an external force to, change the state. Hence, Newton’s first, law of motion is related to physical, independence of force., , 27 A particle of mass m is moving with, a uniform velocity v 1 . It is given an, impulse such that its velocity, becomes v2 . The impulse is equal, to, [CBSE AIPMT 1990], (a) m [| v2 | − | v1 |], (c) m (v1 + v2 ), , 1, (b) m (v22 − v12 ), 2, (d) m (v2 − v1 ), , Ans. (d), Concept Impulse of a force can be, calculated as the product of large force, applied to the small time to which force, act., dp, i.e., F=, dt, ⇒, , F ⋅dt = dp, , impulse = p 2 = p1, ⇒, Impulse of a force, which is the product, of average force during impact and the, time for which the impact lasts, is, measured by the total change in linear, momentum produced during the impact., Here,, Impulse,, , p 1 = mv1 , p 2 = mv2, I = mv2 − mv1 = m ( v2 − v1 ), , 28 A 600 kg rocket is set for a vertical, firing. If the exhaust speed is 1000, ms −1 , the mass of the gas ejected, per second to supply the thrust, needed to overcome the weight of, rocket is, [CBSE AIPMT 1990], (a) 117.6kg s−1, (c) 6kg s−1, , (b) 58.6kg s−1, (d) 76.4kg s−1, , From the above free body diagram, the, relation for acceleration of the given, system can be given as, m − m1 , a= 2, ⋅g, m1 + m2 , 6 − 4, g, =, ×g=, 5, 4 + 6, Hence, correct option is (b)., , Ans. (c), Thrust on the rocket is the force with, which the rocket moves upwards., Thrust on the rocket at timet is given by, dm, F =−u, dt, where, u is relative velocity of exhaust, dm, gases with respect to the rocket., is, dt, rate of combustion of fuel at that instant., dm, dm mg, = mg ⇒ −, =, ∴ F =−u, dt, dt, u, , 30 A block of mass m is placed on a, smooth inclined wedge ABC of, inclination θ as shown in the figure., The wedge is given an acceleration, a towards the right. The relation, between a and θ for the block to, remain stationary on the wedge is, [NEET 2018], , A, , Here, m = 600 kg, u = 1000 ms−1, dm 600 × 10, = 6 kg s−1, ∴, −, =, dt, 1000, , Equilibrium of a Particle and, Common Forces in Mechanics, 29 Two bodies of mass 4 kg and 6 kg, are tied to the ends of a massless, string. The string passes over a, pulley which is frictionless, (see figure). The acceleration of the, system in terms of acceleration, due to gravity g is, [NEET (Sep.) 2020], , 4 kg, 6 kg, , (b) g/5, , a, q, C, , B, , (a) a = g cos θ, , TOPIC 2, , (a) g/2, , m, , (c) g/10, , (d) g, , Ans. (b), Given, m1 = 4 kg, m2 = 6 kg and a = ?, a, , 4 kg, m 1g, , 6 kg, m2 g, , a, , (c) a =, , (b) a =, , g, cosec θ, , g, sinθ, , (d) a = g tanθ, , Ans. (d), According to the question, the FBD of, the given condition will be, A, , R cosq, , R, , q, , ma, (Pseudo, force), B, , R sinq, q, mg q, , a, , C, , Since, the wedge is accelerating towards, right witha, thus a pseudo force acts in, the left direction in order to keep the, block stationary. As, the system is in, equilibrium., ∴, ΣFx = 0, or, ΣFy = 0, ⇒, R sinθ = ma, or, …(i), mg sinθ = ma, Similarly,, R cosθ = mg, or mg cosθ = mg, …(ii), Dividing Eq. (i) by Eq (ii), we get, mg sin θ ma, =, mg cos θ mg, a, tanθ =, ⇒, g

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or, a = g tanθ, ∴The relation betweena and g for the, block to remain stationary on the wedge, is a = g tanθ., , 31 Two blocks A and B of masses 3m, and m respectively are connected, by a massless and inextensible, string. The whole system is, suspended by a massless spring as, shown in figure. The magnitudes of, acceleration of A and B, immediately after the string is cut,, are respectively, [NEET 2017], , 32 Three blocks A, B and C of masses 4, kg, 2 kg and 1 kg respectively, are in, contact on a frictionless surface,, as shown. If a force of 14 N is, applied on the 4 kg block, then the, contact force between A and B is, , 34 The mass of a lift is 2000 kg. When, the tension in the supporting cable, is 28000 N, then its acceleration is, [CBSE AIPMT 2009], , (a) 30 ms−2 downwards, (b) 4 ms−2 upwards, (c) 4 ms−2 downwards, (d) 14 ms−2 upwards, , [CBSE AIPMT 2015], , A, , (a) 2 N, , (b) 6 N, , B, , Ans. (b), , C, , (c) 8 N, , Here, lift is accelerating upward at the, rate of a., , (d) 18 N, , Ans. (b), , R, , Given, mA = 4 kg, , a, , mB = 2 kg ⇒ mC = 1 kg, a, A, , 3m, , mg, , F, A, , B m, , g, (a) g,, 3, , g, (b) , g, 3, , (c) g, g, , g g, (d) ,, 3 3, , B, , C, , So, total mass (M) = 4 + 2 + 1 = 7 kg, , Initially system, is in equilibrium with a, total weight of 4mg over spring., , [Qg = 10 ms −2 ], , a, , 4 kg, , F − F ′ = 4a, , ∴, , B m, , ⇒, , kx = 4mg, , When string is cut at the location as, shown above., Free body diagram for m is, , m, , So,force on mass m = mg, ∴ Acceleration of mass, m = g, , mg, , For mass 3m; free body diagram is, kx=4mg, , a, , 3m, , F ′ = 14 − 4 × 2 ⇒ F ′ = 6 N, , 33 A person of mass 60 kg is inside a, lift of mass 940 kg and presses the, button on control panel. The lift, starts moving upwards with an, acceleration 1.0 m/s 2 . Ifg = 10m / s2 ,, the tension in the supporting, cable is, [CBSE AIPMT 2011], (a) 9680 N, (c) 1200 N, , ⇒ a=, , F′, , kx, (3m+m), , 4 mg, , 28000 − 20000 = 2000a, , FBDof block A,, F, , Cutting, plane, , R − mg = ma, , F = Ma ⇒ 14 = 7a ⇒a = 2 m/s 2, , Now,, , Ans. (b), , A 3m, , Hence, equation of motion is, written as, , 8000, = 4ms−2 upwards, 2000, , 35 Three forces acting on a body are, shown in the figure. To have the, resultant force only along the, y-direction, the magnitude of the, minimum additional force needed, is, [CBSE AIPMT 2008], y, 1N, 4N, 30°, 60°, , (b) 11000 N, (d) 8600 N, , 30°, , Ans. (b), , 2N, , Total mass (m), = Mass of lift + Mass of person, , If a = acceleration of block of mass, 3m, then, , = 940 + 60 = 1000 kg, , Ans. (a), y, , ⇒, , So, accelerations for blocks A and B are, g, and a B = g, aA =, 3, , 4 cos 30° + 1 sin 60°, 1N, , T − mg = ma, 4N, a = 1 m/s2, , 30°, , Fnet = 4mg − 3mg, g, 3m⋅a A = mg or a A =, 3, , 3, (c), N (d) 3 N, 4, , (a) 0.5 N (b) 1.5 N, , So, from the free body diagram, 3 mg, , x, , m = 1000 kg, , 60°, , x', , 30°, , 2N, , mg, , Hence, T − 1000 × 10 = 1000 × 1, T = 11000 N, , x, , 1 cos 60° + 2 sin 30°, , y'

or, a = g tanθ, ∴The relation betweena and g for the, block to remain stationary on the wedge, is a = g tanθ., , 31 Two blocks A and B of masses 3m, and m respectively are connected, by a massless and inextensible, string. The whole system is, suspended by a massless spring as, shown in figure. The magnitudes of, acceleration of A and B, immediately after the string is cut,, are respectively, [NEET 2017], , 32 Three blocks A, B and C of masses 4, kg, 2 kg and 1 kg respectively, are in, contact on a frictionless surface,, as shown. If a force of 14 N is, applied on the 4 kg block, then the, contact force between A and B is, , 34 The mass of a lift is 2000 kg. When, the tension in the supporting cable, is 28000 N, then its acceleration is, [CBSE AIPMT 2009], , (a) 30 ms−2 downwards, (b) 4 ms−2 upwards, (c) 4 ms−2 downwards, (d) 14 ms−2 upwards, , [CBSE AIPMT 2015], , A, , (a) 2 N, , (b) 6 N, , B, , Ans. (b), , C, , (c) 8 N, , Here, lift is accelerating upward at the, rate of a., , (d) 18 N, , Ans. (b), , R, , Given, mA = 4 kg, , a, , mB = 2 kg ⇒ mC = 1 kg, a, A, , 3m, , mg, , F, A, , B m, , g, (a) g,, 3, , g, (b) , g, 3, , (c) g, g, , g g, (d) ,, 3 3, , B, , C, , So, total mass (M) = 4 + 2 + 1 = 7 kg, , Initially system, is in equilibrium with a, total weight of 4mg over spring., , [Qg = 10 ms −2 ], , a, , 4 kg, , F − F ′ = 4a, , ∴, , B m, , ⇒, , kx = 4mg, , When string is cut at the location as, shown above., Free body diagram for m is, , m, , So,force on mass m = mg, ∴ Acceleration of mass, m = g, , mg, , For mass 3m; free body diagram is, kx=4mg, , a, , 3m, , F ′ = 14 − 4 × 2 ⇒ F ′ = 6 N, , 33 A person of mass 60 kg is inside a, lift of mass 940 kg and presses the, button on control panel. The lift, starts moving upwards with an, acceleration 1.0 m/s 2 . Ifg = 10m / s2 ,, the tension in the supporting, cable is, [CBSE AIPMT 2011], (a) 9680 N, (c) 1200 N, , ⇒ a=, , F′, , kx, (3m+m), , 4 mg, , 28000 − 20000 = 2000a, , FBDof block A,, F, , Cutting, plane, , R − mg = ma, , F = Ma ⇒ 14 = 7a ⇒a = 2 m/s 2, , Now,, , Ans. (b), , A 3m, , Hence, equation of motion is, written as, , 8000, = 4ms−2 upwards, 2000, , 35 Three forces acting on a body are, shown in the figure. To have the, resultant force only along the, y-direction, the magnitude of the, minimum additional force needed, is, [CBSE AIPMT 2008], y, 1N, 4N, 30°, 60°, , (b) 11000 N, (d) 8600 N, , 30°, , Ans. (b), , 2N, , Total mass (m), = Mass of lift + Mass of person, , If a = acceleration of block of mass, 3m, then, , = 940 + 60 = 1000 kg, , Ans. (a), y, , ⇒, , So, accelerations for blocks A and B are, g, and a B = g, aA =, 3, , 4 cos 30° + 1 sin 60°, 1N, , T − mg = ma, 4N, a = 1 m/s2, , 30°, , Fnet = 4mg − 3mg, g, 3m⋅a A = mg or a A =, 3, , 3, (c), N (d) 3 N, 4, , (a) 0.5 N (b) 1.5 N, , So, from the free body diagram, 3 mg, , x, , m = 1000 kg, , 60°, , x', , 30°, , 2N, , mg, , Hence, T − 1000 × 10 = 1000 × 1, T = 11000 N, , x, , 1 cos 60° + 2 sin 30°, , y'

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Breaking all the forces in x-y axis., otal force along (+ x) axis, = (1 cos60 ° + 2 sin30 ° ), along (−x) axis = (4 sin30 ° ) along (+ y), axis = (4 cos30 ° + 1 sin60 ° ) along (−y), axis = (2 cos30 ° ), ⇒ Net force along x-axis, = − (1 cos60 ° + 2 sin30 ° ) + 4 sin30 °, 1, 1, 1, ⇒ − + 2 × + 4 ×, 2, 2, 2, −3, 1, + 2= +, 2, 2, Net force along y-axis, , ⇒, , 4×, , 3 5 3 2 3, 3, 3, +, −2×, =, −, 2, 2, 2, 2, 2, , 3 3, 2, To have, resultant only in y-axis we must, 1, have N force towards +x-axis, so that, 2, it can compensate the net force of −x axis., =, , 36 A monkey of mass 20 kg is holding, a vertical rope. The rope will not, break, when a mass of 25 kg is, suspended from it but will break, if, the mass exceeds 25 kg. What is, the maximum acceleration with, which the monkey can climb up, along the rope? (Take g = 10 m / s2 ), [CBSE AIPMT 2003], , (a) 25 m/s 2, (c) 5 m/s 2, , 37 A man weighs 80 kg. He stands on, a weighing scale in a lift which is, moving upwards with a uniform, acceleration of 5 m / s2. What, would be the reading on the scale?, (Take g = 10 m / s2 ), , (b) 2 .5 m/s 2, (d) 10 m/s 2, , (a) 800 N, (c) Zero, , Maximum bearable tension in the rope, T = 25 × 10 = 250 N, , [CBSE AIPMT 2000], , M1, , Acceleration of lift,a = 5 m/s 2, When, lift is moving upwards, the reading, of weighing scale will be equal to R., , a, , From the figure,, T − mg = ma or acceleration a =, , T − mg, m, , (c), , g, 3, , T, , T, , (d), , g, 4, , Mg, , The equation of motion gives, R − Mg = Ma or, R = Mg + Ma = M (g + a), , a, , M1, M2, , M1 g, M2g, , As M2 > M1 , so mass M2 moves down and, mass M1 moves up with the same, acceleration a (say). The arrangement of, the motion is represented in the figure., , R = 80 (10 + 5) = 80 × 15 = 1200 N, , According to free body diagram of mass, M2 , is, , 38 A lift of mass 1000 kg is moving, upwards with an acceleration of 1, m / s 2 . The tension developed in, the string, which is connected to, lift is (g = 9.8 m / s2 ), , According to free body diagram of mass, M1 , is, , (b) 10800 N, (d) 10000 N, , Ans. (b), , mg, , g, 2, , Ans. (c), Concept In the case of masses hanging, from a pulley by a string, the tension in, whole string is same, say equal to T., , a, , R, , (a) 9800 N, (c) 11000 N, , a, , (b), , g, , [CBSE AIPMT 2002], , Monkey, , M2, , (a) g, , Mass of man M = 80 kg, , T, , g = 10 m / s2 ,, , are connected at the ends of an, inextensible string passing over a, frictionless pulley as shown. When, masses are released, then, acceleration of masses will be, , (b) 1200 N, (d) 400 N, , Ans. (b), , ∴, , Ans. (b), , Given, mass m = 20 kg,, , 39 Two masses M 1 = 5 kg,M2 = 10 kg, , [CBSE AIPMT 2003], , = 4 cos30 ° + 1 sin60 ° − 2 cos30 °, ⇒, , Taking, T = 250 N, 250 − 20 × 10 50, Hence, a =, =, 20, 20, 2, = 2 .5 m/s, , When, lift move upwards, with same acceleration,, T, then according to free, body diagram of the left, a, T − mg = ma, or T = m (g + a), mg, Given,, m = 1000 kg, a = 1 m /s 2 ,, g = 9.8 m /s 2, Thus, T = 1000 (9.8 + 1) = 1000 × 10.8, = 10800 N, , M2 g − T = M2 a, , T − M1 g = M1a, , …(i), , …(ii), , Adding Eqs. (i) and (ii), we get, (M2 g − T ) + (T − M1 g) = (M1 + M2 ) a, (M2 − M1 ) g = (M1 + M2 ) a, M − M1 , ⇒, a= 2, g, M1 + M2 , Given, M1 = 5 kg, M2 = 10 kg, 10 − 5 , 5, g, 2, Hence, a = , g = g = m/s, 15, 3, 5 + 10 , Alternative, Acceleration,, (F ) system (10 − 5) × g g, a = net, =, = m/s 2, Net mass, 5 + 10, 3, In a mass-pulley system, the tension, in the string is always towards the pulley.

Breaking all the forces in x-y axis., otal force along (+ x) axis, = (1 cos60 ° + 2 sin30 ° ), along (−x) axis = (4 sin30 ° ) along (+ y), axis = (4 cos30 ° + 1 sin60 ° ) along (−y), axis = (2 cos30 ° ), ⇒ Net force along x-axis, = − (1 cos60 ° + 2 sin30 ° ) + 4 sin30 °, 1, 1, 1, ⇒ − + 2 × + 4 ×, 2, 2, 2, −3, 1, + 2= +, 2, 2, Net force along y-axis, , ⇒, , 4×, , 3 5 3 2 3, 3, 3, +, −2×, =, −, 2, 2, 2, 2, 2, , 3 3, 2, To have, resultant only in y-axis we must, 1, have N force towards +x-axis, so that, 2, it can compensate the net force of −x axis., =, , 36 A monkey of mass 20 kg is holding, a vertical rope. The rope will not, break, when a mass of 25 kg is, suspended from it but will break, if, the mass exceeds 25 kg. What is, the maximum acceleration with, which the monkey can climb up, along the rope? (Take g = 10 m / s2 ), [CBSE AIPMT 2003], , (a) 25 m/s 2, (c) 5 m/s 2, , 37 A man weighs 80 kg. He stands on, a weighing scale in a lift which is, moving upwards with a uniform, acceleration of 5 m / s2. What, would be the reading on the scale?, (Take g = 10 m / s2 ), , (b) 2 .5 m/s 2, (d) 10 m/s 2, , (a) 800 N, (c) Zero, , Maximum bearable tension in the rope, T = 25 × 10 = 250 N, , [CBSE AIPMT 2000], , M1, , Acceleration of lift,a = 5 m/s 2, When, lift is moving upwards, the reading, of weighing scale will be equal to R., , a, , From the figure,, T − mg = ma or acceleration a =, , T − mg, m, , (c), , g, 3, , T, , T, , (d), , g, 4, , Mg, , The equation of motion gives, R − Mg = Ma or, R = Mg + Ma = M (g + a), , a, , M1, M2, , M1 g, M2g, , As M2 > M1 , so mass M2 moves down and, mass M1 moves up with the same, acceleration a (say). The arrangement of, the motion is represented in the figure., , R = 80 (10 + 5) = 80 × 15 = 1200 N, , According to free body diagram of mass, M2 , is, , 38 A lift of mass 1000 kg is moving, upwards with an acceleration of 1, m / s 2 . The tension developed in, the string, which is connected to, lift is (g = 9.8 m / s2 ), , According to free body diagram of mass, M1 , is, , (b) 10800 N, (d) 10000 N, , Ans. (b), , mg, , g, 2, , Ans. (c), Concept In the case of masses hanging, from a pulley by a string, the tension in, whole string is same, say equal to T., , a, , R, , (a) 9800 N, (c) 11000 N, , a, , (b), , g, , [CBSE AIPMT 2002], , Monkey, , M2, , (a) g, , Mass of man M = 80 kg, , T, , g = 10 m / s2 ,, , are connected at the ends of an, inextensible string passing over a, frictionless pulley as shown. When, masses are released, then, acceleration of masses will be, , (b) 1200 N, (d) 400 N, , Ans. (b), , ∴, , Ans. (b), , Given, mass m = 20 kg,, , 39 Two masses M 1 = 5 kg,M2 = 10 kg, , [CBSE AIPMT 2003], , = 4 cos30 ° + 1 sin60 ° − 2 cos30 °, ⇒, , Taking, T = 250 N, 250 − 20 × 10 50, Hence, a =, =, 20, 20, 2, = 2 .5 m/s, , When, lift move upwards, with same acceleration,, T, then according to free, body diagram of the left, a, T − mg = ma, or T = m (g + a), mg, Given,, m = 1000 kg, a = 1 m /s 2 ,, g = 9.8 m /s 2, Thus, T = 1000 (9.8 + 1) = 1000 × 10.8, = 10800 N, , M2 g − T = M2 a, , T − M1 g = M1a, , …(i), , …(ii), , Adding Eqs. (i) and (ii), we get, (M2 g − T ) + (T − M1 g) = (M1 + M2 ) a, (M2 − M1 ) g = (M1 + M2 ) a, M − M1 , ⇒, a= 2, g, M1 + M2 , Given, M1 = 5 kg, M2 = 10 kg, 10 − 5 , 5, g, 2, Hence, a = , g = g = m/s, 15, 3, 5 + 10 , Alternative, Acceleration,, (F ) system (10 − 5) × g g, a = net, =, = m/s 2, Net mass, 5 + 10, 3, In a mass-pulley system, the tension, in the string is always towards the pulley.

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Trolly, , 40 A mass of 1 kg is suspended by a, thread. It is, 1. lifted up with an acceleration, 4.9 m/s 2 ,, 2. lowered with an acceleration, 4.9 m/s 2 ., The ratio of the tensions is, [CBSE AIPMT 1998], , (a) 3 : 1, , (b) 1 : 3, , (c) 1 : 2, , (d) 2 : 1, , Ans. (a), (i) When, mass is lifted upwards with an, acceleration a, then according to free, body diagram, , 2 kg, Block, , (a) 1.25 m/s 2, (c) 1.66 m/s 2, Ans. (a), , (b) 1.50 m/s 2, (d) 1.00 m/s 2, , The given situation is shown in the, following diagram., a, , T1, , 10 kg, , T, , a, , T a, , mg, 2 kg, , T1 − mg = ma ⇒ T1 = mg + ma, T1 = m(g + a), Substituting the values, we obtain, , If ‘a’ be the acceleration of the system, then, equation of motion of 10 kg trolly,, , . ) = 14.7 N, T1 = (1) (9.8 + 49, , ∴, , (ii) When, mass is lowered downwards, with an acceleration a, then, mg − T2 = ma, ⇒ T2 = mg − ma = m(g − a), Substituting the values, we have, T2, , a, , mg, , T2 = (1) (9.8 − 49, . ) = 49, . N, Then, ratio of tensions, T1 14.7 3, =, = ⇒ T1 :T2 = 3 : 1, T2, 49, ., 1, , TOPIC 3, Friction, 41 Calculate the acceleration of the, block and trolly system shown in, the figure. The coefficient of, kinetic friction between the trolly, and the surface is 0.05. (g = 10, m/s 2 , mass of the string is, negligible and no other friction, exists)., [NEET (Oct.) 2020], , 2π, TA, 2π, For particle B,ωB =, TB, ωA 2 π TB TB, ∴, =, ×, =, ωB TA 2 π TA, 1, [QTA = TB (given)], = or 1 : 1, 1, , For particle A,ωA =, , 10 kg, , 43 A body of mass m is kept on a, rough horizontal surface (coefficient, of friction = µ). Horizontal force is, applied on the body, but it does not, move. The resultant of normal, reaction and the frictional force, acting on the object is given F,, where F is, [NEET (Odisha) 2019], (a) |F | = mg + µ mg, (b) |F | = µmg, (c) |F | ≤ mg 1 + µ 2, (d) |F | = mg, Ans. (c), The situation can be drawn as, , T − µ R = 10a, , F, , T − 0.05 × 10 g = 10a, [Qµ = 0.05, R = 10 g], ⇒ T − 0.05 × 10 × 10 = 10a, … (i), ⇒, T − 5 = 10a, Equation of motion of 2kg block,, 2g − T = 2a, 2 × 10 − T = 2a, … (ii), 20 − T = 2a, Adding Eqs. (i) and (ii), we have, 20 − 5 = 12a, ⇒, 15 = 12a, 15 5, ⇒, . ms− 2, a = = = 125, 12 4, , N, , ⇒, , 42 Two particles A and B are moving in, uniform circular motion in concentric, circles of radii rA and rB with speed, v A and v B respectively. Their time, period of rotation is the same. The, ratio of angular speed of A to that, of B will be, [NEET (National) 2019], (a) v A :v B (b) rB : rA, , (c) 1 : 1, , (d) rA: rB, , Ans. (c), The angular speed of a particle in a, uniform circular motion is given by, angle of circle, ω=, Time, 2π, ω = , whereT is the time period, T, of rotation, , f, , FH, , mg, , The frictional force,f = µN = µ mg, [QN = mg], From Free body diagram (FBD), the, resultant force is, |F | = N 2 + f 2, = (mg) 2 + (µmg) 2, = mg 1 + µ 2, This is the minimum force required to move, the object. But as the body is not moving, ∴, , |F | ≤ mg 1 − µ 2, , 44 Which one of the following, statements is incorrect?, [NEET 2018], , (a) Frictional force opposes the relative, motion, (b) Limiting value of static friction is, directly proportional to normal, reaction, (c) Rolling friction is smaller than sliding, friction, (d) Coefficient of sliding friction has, dimensions of length

Trolly, , 40 A mass of 1 kg is suspended by a, thread. It is, 1. lifted up with an acceleration, 4.9 m/s 2 ,, 2. lowered with an acceleration, 4.9 m/s 2 ., The ratio of the tensions is, [CBSE AIPMT 1998], , (a) 3 : 1, , (b) 1 : 3, , (c) 1 : 2, , (d) 2 : 1, , Ans. (a), (i) When, mass is lifted upwards with an, acceleration a, then according to free, body diagram, , 2 kg, Block, , (a) 1.25 m/s 2, (c) 1.66 m/s 2, Ans. (a), , (b) 1.50 m/s 2, (d) 1.00 m/s 2, , The given situation is shown in the, following diagram., a, , T1, , 10 kg, , T, , a, , T a, , mg, 2 kg, , T1 − mg = ma ⇒ T1 = mg + ma, T1 = m(g + a), Substituting the values, we obtain, , If ‘a’ be the acceleration of the system, then, equation of motion of 10 kg trolly,, , . ) = 14.7 N, T1 = (1) (9.8 + 49, , ∴, , (ii) When, mass is lowered downwards, with an acceleration a, then, mg − T2 = ma, ⇒ T2 = mg − ma = m(g − a), Substituting the values, we have, T2, , a, , mg, , T2 = (1) (9.8 − 49, . ) = 49, . N, Then, ratio of tensions, T1 14.7 3, =, = ⇒ T1 :T2 = 3 : 1, T2, 49, ., 1, , TOPIC 3, Friction, 41 Calculate the acceleration of the, block and trolly system shown in, the figure. The coefficient of, kinetic friction between the trolly, and the surface is 0.05. (g = 10, m/s 2 , mass of the string is, negligible and no other friction, exists)., [NEET (Oct.) 2020], , 2π, TA, 2π, For particle B,ωB =, TB, ωA 2 π TB TB, ∴, =, ×, =, ωB TA 2 π TA, 1, [QTA = TB (given)], = or 1 : 1, 1, , For particle A,ωA =, , 10 kg, , 43 A body of mass m is kept on a, rough horizontal surface (coefficient, of friction = µ). Horizontal force is, applied on the body, but it does not, move. The resultant of normal, reaction and the frictional force, acting on the object is given F,, where F is, [NEET (Odisha) 2019], (a) |F | = mg + µ mg, (b) |F | = µmg, (c) |F | ≤ mg 1 + µ 2, (d) |F | = mg, Ans. (c), The situation can be drawn as, , T − µ R = 10a, , F, , T − 0.05 × 10 g = 10a, [Qµ = 0.05, R = 10 g], ⇒ T − 0.05 × 10 × 10 = 10a, … (i), ⇒, T − 5 = 10a, Equation of motion of 2kg block,, 2g − T = 2a, 2 × 10 − T = 2a, … (ii), 20 − T = 2a, Adding Eqs. (i) and (ii), we have, 20 − 5 = 12a, ⇒, 15 = 12a, 15 5, ⇒, . ms− 2, a = = = 125, 12 4, , N, , ⇒, , 42 Two particles A and B are moving in, uniform circular motion in concentric, circles of radii rA and rB with speed, v A and v B respectively. Their time, period of rotation is the same. The, ratio of angular speed of A to that, of B will be, [NEET (National) 2019], (a) v A :v B (b) rB : rA, , (c) 1 : 1, , (d) rA: rB, , Ans. (c), The angular speed of a particle in a, uniform circular motion is given by, angle of circle, ω=, Time, 2π, ω = , whereT is the time period, T, of rotation, , f, , FH, , mg, , The frictional force,f = µN = µ mg, [QN = mg], From Free body diagram (FBD), the, resultant force is, |F | = N 2 + f 2, = (mg) 2 + (µmg) 2, = mg 1 + µ 2, This is the minimum force required to move, the object. But as the body is not moving, ∴, , |F | ≤ mg 1 − µ 2, , 44 Which one of the following, statements is incorrect?, [NEET 2018], , (a) Frictional force opposes the relative, motion, (b) Limiting value of static friction is, directly proportional to normal, reaction, (c) Rolling friction is smaller than sliding, friction, (d) Coefficient of sliding friction has, dimensions of length

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From Eq. (ii), T = m2 (g − a), , Ans. (d), The opposing force that comes into play, when one body is actually sliding over, the surface of the other body is called, sliding friction., The coefficient of sliding is given as, N, µS =, Fsliding, where,N is the normal reaction andFsliding, is the sliding force., As, the dimensions ofN and Fsliding are, same. Thus,µ S is a dimensionless, quantity., Hence, statement(d) is incorrect., , 45 A block A of mass m 1 rests on a, horizontal table. A light string, connected to it passes over a, frictionless pulley at the edge of, table and from its other end, another block B of mass m2 is, suspended. The coefficient of, kinetic friction between the block, and the table is µ k . When the block, A is sliding on the table, the tension, in the string is [CBSE AIPMT 2015], (a), , (c), , (m2 + µ k m1 )g, (m1 + m2 ), , (b), , m1 m 2 (1 + µ k ) g, (m 1 + m 2 ), , (d), , (m 2 − µ k m 1 )g, (m 1 + m 2 ), , m1 m 2 (1 − µ k ) g, (m 1 + m 2 ), , (m − µ k m1 ) , = m2 1 − 2, g, m1 + m2 , , m m (1 + µ k ), g, T= 1 2, m1 + m2, , 46 A plank with a box on it at one end, is gradually raised about the other, end. As the angle of inclination with, the horizontal reaches 30°, the box, starts to slip and slides 4.0 m down, the plank in 4.0 s. The coefficients, of static and kinetic friction, between the box and the plank will, be, respectively [CBSE AIPMT 2015], , g (1 − gµ), 9, g(1 − 2 µ), (c), 3, , m3, , θ, , (a) 0.6 and 0.6, (c) 0.5 and 0.6, , 2 gµ, 3, g (1 − 2 µ), (d), 2, , (b), , (a), , (b) 0.6 and 0.5, (d) 0.4 and 0.3, , Ans. (c), First of all consider the forces on the, blocks, , Ans. (b), , a, , Given a plank with a box on its one end is, gradually raised about the end having, angle of inclination is 30°, the box starts, to slip and slides down 4 m the plank in 4, s as shown in figure., , T1, , 2 T2 T3, , m, , 3, , m, , µmg, a, , µmg, , T1, , m 1, , a, , mg, , mg, , m1, , θ, , fk, , For the Ist block,, , [Qm1 = m2 = m3 ], , mg − T1 = m × a, , T − m1a = fk, , …(i), T, , The coefficient of static friction,, 1, µ s = tan30 ° =, = 0.6, 3, , m2, , So, distance covered by a plank,, 1, s = ut + at 2, 2, , FBDof block B,, , a, , u = 0 and a = g (sinθ − µ cos θ), , Here,, ∴, , m 2g, , m2 g − T = m2 a, Adding Eqs. (i) and (ii), we get, m2 g − m1a = m2 a + fk, ⇒, , m2, , P, , mg, , FBDof block A,, , ⇒, , [CBSE AIPMT 2014], , m1, , Ans. (c), , T, , 47 A system consists of three masses, m 1 , m2 and m3 connected by a, string passing over a pulley P. The, mass m 1 hangs freely and m2 and, m3 are on a rough horizontal table, (the coefficient of friction = µ). The, pulley is frictionless and of, negligible mass. The downward, acceleration of mass m 1 is (Assume,, m 1 = m2 = m3 = m), , m2 g − m1a = m2 a + µ k m1 g, (m − µ k m1 ) g, a= 2, m1 + m2, , …(ii), , ⇒, ⇒, ⇒, , 1, g (sin30 ° − µ k cos 30 ° ) (4) 2, 2, 1, 3, 0.5 = 10 × − µ K × 10 ×, 2, 2, , 4=, , 5 3 µ K = 4.5, µ K = 0.51, , Thus, coefficient of kinetic friction, between the box and the plank is 0.51., , …(i), , Let us consider 2nd and 3rd block as a, system., So, T1 − 2µ mg = 2m × a, , …(ii), , Solving Eqs. (i) and (ii),, ⇒, , mg − T1 = m × a, , ⇒ T1 − 2µ mg = 2m × a, Addiing Eqs. (i) and (ii), mg (1 − 2µ ) = 3m × a ⇒ a =, , g, (1 − 2µ ), 3, , 48 Three blocks with masses m,2m, and 3m are connected by strings,, as shown in the figure. After an, upward force F is applied on block, m, the masses move upward at, constant speed v. What is the net

From Eq. (ii), T = m2 (g − a), , Ans. (d), The opposing force that comes into play, when one body is actually sliding over, the surface of the other body is called, sliding friction., The coefficient of sliding is given as, N, µS =, Fsliding, where,N is the normal reaction andFsliding, is the sliding force., As, the dimensions ofN and Fsliding are, same. Thus,µ S is a dimensionless, quantity., Hence, statement(d) is incorrect., , 45 A block A of mass m 1 rests on a, horizontal table. A light string, connected to it passes over a, frictionless pulley at the edge of, table and from its other end, another block B of mass m2 is, suspended. The coefficient of, kinetic friction between the block, and the table is µ k . When the block, A is sliding on the table, the tension, in the string is [CBSE AIPMT 2015], (a), , (c), , (m2 + µ k m1 )g, (m1 + m2 ), , (b), , m1 m 2 (1 + µ k ) g, (m 1 + m 2 ), , (d), , (m 2 − µ k m 1 )g, (m 1 + m 2 ), , m1 m 2 (1 − µ k ) g, (m 1 + m 2 ), , (m − µ k m1 ) , = m2 1 − 2, g, m1 + m2 , , m m (1 + µ k ), g, T= 1 2, m1 + m2, , 46 A plank with a box on it at one end, is gradually raised about the other, end. As the angle of inclination with, the horizontal reaches 30°, the box, starts to slip and slides 4.0 m down, the plank in 4.0 s. The coefficients, of static and kinetic friction, between the box and the plank will, be, respectively [CBSE AIPMT 2015], , g (1 − gµ), 9, g(1 − 2 µ), (c), 3, , m3, , θ, , (a) 0.6 and 0.6, (c) 0.5 and 0.6, , 2 gµ, 3, g (1 − 2 µ), (d), 2, , (b), , (a), , (b) 0.6 and 0.5, (d) 0.4 and 0.3, , Ans. (c), First of all consider the forces on the, blocks, , Ans. (b), , a, , Given a plank with a box on its one end is, gradually raised about the end having, angle of inclination is 30°, the box starts, to slip and slides down 4 m the plank in 4, s as shown in figure., , T1, , 2 T2 T3, , m, , 3, , m, , µmg, a, , µmg, , T1, , m 1, , a, , mg, , mg, , m1, , θ, , fk, , For the Ist block,, , [Qm1 = m2 = m3 ], , mg − T1 = m × a, , T − m1a = fk, , …(i), T, , The coefficient of static friction,, 1, µ s = tan30 ° =, = 0.6, 3, , m2, , So, distance covered by a plank,, 1, s = ut + at 2, 2, , FBDof block B,, , a, , u = 0 and a = g (sinθ − µ cos θ), , Here,, ∴, , m 2g, , m2 g − T = m2 a, Adding Eqs. (i) and (ii), we get, m2 g − m1a = m2 a + fk, ⇒, , m2, , P, , mg, , FBDof block A,, , ⇒, , [CBSE AIPMT 2014], , m1, , Ans. (c), , T, , 47 A system consists of three masses, m 1 , m2 and m3 connected by a, string passing over a pulley P. The, mass m 1 hangs freely and m2 and, m3 are on a rough horizontal table, (the coefficient of friction = µ). The, pulley is frictionless and of, negligible mass. The downward, acceleration of mass m 1 is (Assume,, m 1 = m2 = m3 = m), , m2 g − m1a = m2 a + µ k m1 g, (m − µ k m1 ) g, a= 2, m1 + m2, , …(ii), , ⇒, ⇒, ⇒, , 1, g (sin30 ° − µ k cos 30 ° ) (4) 2, 2, 1, 3, 0.5 = 10 × − µ K × 10 ×, 2, 2, , 4=, , 5 3 µ K = 4.5, µ K = 0.51, , Thus, coefficient of kinetic friction, between the box and the plank is 0.51., , …(i), , Let us consider 2nd and 3rd block as a, system., So, T1 − 2µ mg = 2m × a, , …(ii), , Solving Eqs. (i) and (ii),, ⇒, , mg − T1 = m × a, , ⇒ T1 − 2µ mg = 2m × a, Addiing Eqs. (i) and (ii), mg (1 − 2µ ) = 3m × a ⇒ a =, , g, (1 − 2µ ), 3, , 48 Three blocks with masses m,2m, and 3m are connected by strings,, as shown in the figure. After an, upward force F is applied on block, m, the masses move upward at, constant speed v. What is the net

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force on the block of mass 2m ?, (g is the acceleration due to, gravity)., [NEET 2013], F, , 50 A block of mass m is in contact, with the cart C as shown in the, figure., , v, , m, m, , C, 2m, 3m, , (a) Zero, (c) 3 mg, , (b) 2 mg, (d) 6 mg, , Ans. (a), Since, all the blocks are moving with, constant velocity and we know that, if, velocity is constant, acceleration of the, body becomes zero. Hence, the net, force on all the blocks will be zero., , 49 The upper half of an inclined plane, of inclination θ is perfectly smooth, while lower half is rough. A block, starting from rest at the top of the, plane will again come to rest at the, bottom, if the coefficient of friction, between the block and lower half, of the plane is given by [NEET 2013], 1, (a) µ =, tan θ, 2, (b) µ =, tan θ, , The coefficient of static friction, between the block and the cart is, µ. The acceleration α of the cart, that will prevent the block from, falling satisfies [CBSE AIPMT 2010], mg, (a) α >, µ, g, (c) α ≥, µ, , µmg = ma, v, v, µmg = m or t =, t, µg, , i.e., or, , 52 The coefficient of static friction,, µ s , between block A of mass 2 kg, and the table as shown in the, figure, is 0.2. What would be the, maximum mass value of block B, so, that the two blocks do not move ?, The string and the pulley are, assumed to be smooth and, massless (g = 10 m/s 2 ), [CBSE AIPMT 2004], , g, (b) α >, µm, g, (d) α <, µ, , Ans. (c), When, a cart moves with some, acceleration towards right, then a, pseudo force (mα) acts on block towards, left. This force (mα) is action force by a, block on cart, , (a) 2.0 kg, (c) 0.2 kg, Ans. (d), , Let the mass of the block B be M., , F r = µR, , ma, , m, , (b) 4.0 kg, (d) 0.4 kg, , Fs =, , R, , mg, , Now, block will remain static w.r.t. cart,, if frictional force µR ≥ mg, , (c) µ = 2 tan θ, (d) µ = tan θ, , ⇒, , Ans. (c), Concept Net work done by the block in, going from top to bottom of the inclined, plane, must be equal to the work done by, frictional force., L/, , R, , 2, L / in θ, s, mg, , θ, , θ, , 2, , µ mα ≥ mg, g, α≥, µ, , ⇒, , [as R = mα], , T − Mg = 0 ⇒ T = Mg, , …(i), , T = fs, , [CBSE AIPMT 2007], , where,fs = frictional force, = µ s R = µ s mg, ∴, , T = µ s mg, , (a), , v, gµ, , (b), , gµ, v, , (c), , g, v, , (d), , v, g, , Ans. (a), Block B will come to rest, if force applied, to it will vanish due to frictional force, acting between block B and surface, i.e., frictional force = force applied, , …(ii), , Thus, from Eqs. (i) and (ii), we have, Mg = µ s mg or M = µ s m, Given, µ s = 0.2, m = 2 kg, ∴, , v, , mg, , µ = 2 tan θ, , In equilibrium,, If blocks do not move, then, , 51 A block B is pushed momentarily, along a horizontal surface with an, initial velocity v. If µ is the, coefficient of sliding friction, between B and the surface, block B, will come to rest after a time, , mg cos θ, , The block may be stationary, when, L, mg sin θ⋅L = µ mg cosθ, 2, mg sin θ⋅L, or, µ=, L, mg cos θ⋅, 2, sin θ, =2, = 2 tan θ, cos θ, , Mg, , M = 0.2 × 2 = 0.4 kg, , 53 A block of mass 10 kg is placed on a, rough horizontal surface having, coefficient of friction µ = 0.5 . If a, horizontal force of 100 N is applied, on it, then the acceleration of the, block will be (Takeg = 10 m/s 2 ), [CBSE AIPMT 2002], , (a) 15 m/s 2, (c) 5 m/s 2, , (b) 10 m/s 2, (d) 05, . m/s 2

force on the block of mass 2m ?, (g is the acceleration due to, gravity)., [NEET 2013], F, , 50 A block of mass m is in contact, with the cart C as shown in the, figure., , v, , m, m, , C, 2m, 3m, , (a) Zero, (c) 3 mg, , (b) 2 mg, (d) 6 mg, , Ans. (a), Since, all the blocks are moving with, constant velocity and we know that, if, velocity is constant, acceleration of the, body becomes zero. Hence, the net, force on all the blocks will be zero., , 49 The upper half of an inclined plane, of inclination θ is perfectly smooth, while lower half is rough. A block, starting from rest at the top of the, plane will again come to rest at the, bottom, if the coefficient of friction, between the block and lower half, of the plane is given by [NEET 2013], 1, (a) µ =, tan θ, 2, (b) µ =, tan θ, , The coefficient of static friction, between the block and the cart is, µ. The acceleration α of the cart, that will prevent the block from, falling satisfies [CBSE AIPMT 2010], mg, (a) α >, µ, g, (c) α ≥, µ, , µmg = ma, v, v, µmg = m or t =, t, µg, , i.e., or, , 52 The coefficient of static friction,, µ s , between block A of mass 2 kg, and the table as shown in the, figure, is 0.2. What would be the, maximum mass value of block B, so, that the two blocks do not move ?, The string and the pulley are, assumed to be smooth and, massless (g = 10 m/s 2 ), [CBSE AIPMT 2004], , g, (b) α >, µm, g, (d) α <, µ, , Ans. (c), When, a cart moves with some, acceleration towards right, then a, pseudo force (mα) acts on block towards, left. This force (mα) is action force by a, block on cart, , (a) 2.0 kg, (c) 0.2 kg, Ans. (d), , Let the mass of the block B be M., , F r = µR, , ma, , m, , (b) 4.0 kg, (d) 0.4 kg, , Fs =, , R, , mg, , Now, block will remain static w.r.t. cart,, if frictional force µR ≥ mg, , (c) µ = 2 tan θ, (d) µ = tan θ, , ⇒, , Ans. (c), Concept Net work done by the block in, going from top to bottom of the inclined, plane, must be equal to the work done by, frictional force., L/, , R, , 2, L / in θ, s, mg, , θ, , θ, , 2, , µ mα ≥ mg, g, α≥, µ, , ⇒, , [as R = mα], , T − Mg = 0 ⇒ T = Mg, , …(i), , T = fs, , [CBSE AIPMT 2007], , where,fs = frictional force, = µ s R = µ s mg, ∴, , T = µ s mg, , (a), , v, gµ, , (b), , gµ, v, , (c), , g, v, , (d), , v, g, , Ans. (a), Block B will come to rest, if force applied, to it will vanish due to frictional force, acting between block B and surface, i.e., frictional force = force applied, , …(ii), , Thus, from Eqs. (i) and (ii), we have, Mg = µ s mg or M = µ s m, Given, µ s = 0.2, m = 2 kg, ∴, , v, , mg, , µ = 2 tan θ, , In equilibrium,, If blocks do not move, then, , 51 A block B is pushed momentarily, along a horizontal surface with an, initial velocity v. If µ is the, coefficient of sliding friction, between B and the surface, block B, will come to rest after a time, , mg cos θ, , The block may be stationary, when, L, mg sin θ⋅L = µ mg cosθ, 2, mg sin θ⋅L, or, µ=, L, mg cos θ⋅, 2, sin θ, =2, = 2 tan θ, cos θ, , Mg, , M = 0.2 × 2 = 0.4 kg, , 53 A block of mass 10 kg is placed on a, rough horizontal surface having, coefficient of friction µ = 0.5 . If a, horizontal force of 100 N is applied, on it, then the acceleration of the, block will be (Takeg = 10 m/s 2 ), [CBSE AIPMT 2002], , (a) 15 m/s 2, (c) 5 m/s 2, , (b) 10 m/s 2, (d) 05, . m/s 2

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Ans. (c), , NOTE, Coefficient of kinetic friction between any, two surfaces in contact is equal to the, tangent of the angle of inclination between, them., , Free body diagram of block is, N, , a, F, f = mN, , mg, Y, X, , From Newton’s second law along X-axis, ΣFx = ma, i.e., F − f = ma, or, F − µ mg = ma, F − µmg, or, a=, m, Given, F = 100 N,µ = 0.5, m = 10 kg,, g = 10 m /s 2, Substituting, the values in the above, relation for acceleration of block,, (100) − (0.5) (10) (10), = 5 m /s 2, a=, (10), , 54 A block has been placed on an, inclined plane with the slope angle, θ, block slides down the plane at, constant speed. The coefficient of, kinetic friction is equal to, [CBSE AIPMT 1993], , (a) sin θ, (c) g, , (b) cos θ, (d) tan θ, , Ans. (d), Angle of repose or angle of sliding is, defined as the minimum angle of, inclination of a plane with the horizontal,, such that a body placed on the plane just, begins to slide down., R, , tio, (fric f, , mg sin θ, A, , f, nal, , o rc, , e), , B, , θ, mg, , θ, , mg cos θ, C, , AB is an inclined plane such that a body, placed on it just begins to slide down, ∠BAC = θ = angle of repose, In equilibrium,, f = mg sinθ, and, ∴, i.e., , R = mg cosθ, f mg sinθ, =, = tanθ, R mg cosθ, µ = tanθ, , 55 Consider, a car moving along a, straight horizontal road with a, speed of 72 km/h. If the coefficient, of static friction between the tyres, and the road is 0.5, the shortest, distance in which the car can be, stopped is (Take g = 10m / s2 ), , (a) 0.80, (c) 0.25, , (b) 0.75, (d) 0.33, , Ans. (b), When, a plane is inclined to the, horizontal at an angleθ, which is greater, than the angle of repose, the body, placed on the inclined plane slides down, with an accelerationa., , R, f, , na, ctio, (fri B, , l fo, , ), rce, , [CBSE AIPMT 1992], , (a) 30 m (b) 40 m (c) 72 m, , (d) 20 m, , mg sin θ, , Ans. (b), When, static friction is present, then, acceleration of body is given bya = − µg, Here, initial velocity, 5, u = 72 km h−1 = 72 × = 20 m / s, 18, Final velocity v = 0, ∴, a = −µ g = − 0.5 × 10 = − 5 m / s2, Now, from third equation of motion,, i.e., v2 = u 2 + 2as, v2 − u 2 0 − (20) 2, = 40 m, s=, =, 2 × (− 5), 2a, , 56 A heavy uniform chain lies on, horizontal table top. If the coefficient, of friction between the chain and, the table surface is 0.25, then the, maximum friction of the length of, the chain that can hang over one, edge of the table is, (a) 20%, , (b) 25%, , (c) 35%, , (d) 15%, , The force of friction should balance the, weight of chain hanging. If M is the mass, of whole chain of lengthL and x is the, length of chain hanging to balance, then, M, M, µ (L − x) g = xg, L, L, or, µ (L − x) = x, 0.25 L, µL, (As, µ = 0.25), or, x=, =, µ+1, 125, ., x 1 1, L, ∴, x = or = = × 100 = 20%, L 5 5, 5, , 57 Starting from rest, a body slides, down a 45° inclined plane in twice, the time it takes to slide down the, same distance in the absence of, friction. The coefficient of friction, between the body and the inclined, plane is, [CBSE AIPMT 1988], , θ, , A, , mg, , mg cos θ, C, , As, it is clear from figure, …(i), R = mg cosθ, Net force on the body down the inclined, plane which means it is sliding, downwards, …(ii), F = mg sinθ − f, i.e., F = ma = mg sinθ − µR, (f = µR), ∴, ma = mg sinθ − µ mg cosθ, = mg (sinθ − µ cosθ), Hence, a = g (sinθ − µ cosθ), ∴Time taken by body to slide down the, plane, t1 =, , 2s, 2s, =, a, g (sinθ − µ cosθ), , When friction is absent, then time taken, to slide down the plane, t2 =, , [CBSE AIPMT 1991], , Ans. (a), , θ, , ∴, or, , 2s, ⇒ Q t 1 = 2t 2, g sinθ, , (given), , t 12 = 4t 22, 2s, 2s × 4, =, g (sinθ − µ cosθ) g sinθ, , or, or, , ∴, , sinθ = 4 sinθ − 4µ cosθ, 3, 3, µ = tanθ = tan45°, 4, 4, , 3, µ = = 0.75, 4, , TOPIC 4, Dynamics of Circular Motion, 58 A block of mass 10 kg is in contact, against the inner wall of a hollow, cylindrical drum of radius 1 m. The, coefficient of friction between the, block and the inner wall of the, cylinder is 0.1. The minimum, angular velocity needed for the

Ans. (c), , NOTE, Coefficient of kinetic friction between any, two surfaces in contact is equal to the, tangent of the angle of inclination between, them., , Free body diagram of block is, N, , a, F, f = mN, , mg, Y, X, , From Newton’s second law along X-axis, ΣFx = ma, i.e., F − f = ma, or, F − µ mg = ma, F − µmg, or, a=, m, Given, F = 100 N,µ = 0.5, m = 10 kg,, g = 10 m /s 2, Substituting, the values in the above, relation for acceleration of block,, (100) − (0.5) (10) (10), = 5 m /s 2, a=, (10), , 54 A block has been placed on an, inclined plane with the slope angle, θ, block slides down the plane at, constant speed. The coefficient of, kinetic friction is equal to, [CBSE AIPMT 1993], , (a) sin θ, (c) g, , (b) cos θ, (d) tan θ, , Ans. (d), Angle of repose or angle of sliding is, defined as the minimum angle of, inclination of a plane with the horizontal,, such that a body placed on the plane just, begins to slide down., R, , tio, (fric f, , mg sin θ, A, , f, nal, , o rc, , e), , B, , θ, mg, , θ, , mg cos θ, C, , AB is an inclined plane such that a body, placed on it just begins to slide down, ∠BAC = θ = angle of repose, In equilibrium,, f = mg sinθ, and, ∴, i.e., , R = mg cosθ, f mg sinθ, =, = tanθ, R mg cosθ, µ = tanθ, , 55 Consider, a car moving along a, straight horizontal road with a, speed of 72 km/h. If the coefficient, of static friction between the tyres, and the road is 0.5, the shortest, distance in which the car can be, stopped is (Take g = 10m / s2 ), , (a) 0.80, (c) 0.25, , (b) 0.75, (d) 0.33, , Ans. (b), When, a plane is inclined to the, horizontal at an angleθ, which is greater, than the angle of repose, the body, placed on the inclined plane slides down, with an accelerationa., , R, f, , na, ctio, (fri B, , l fo, , ), rce, , [CBSE AIPMT 1992], , (a) 30 m (b) 40 m (c) 72 m, , (d) 20 m, , mg sin θ, , Ans. (b), When, static friction is present, then, acceleration of body is given bya = − µg, Here, initial velocity, 5, u = 72 km h−1 = 72 × = 20 m / s, 18, Final velocity v = 0, ∴, a = −µ g = − 0.5 × 10 = − 5 m / s2, Now, from third equation of motion,, i.e., v2 = u 2 + 2as, v2 − u 2 0 − (20) 2, = 40 m, s=, =, 2 × (− 5), 2a, , 56 A heavy uniform chain lies on, horizontal table top. If the coefficient, of friction between the chain and, the table surface is 0.25, then the, maximum friction of the length of, the chain that can hang over one, edge of the table is, (a) 20%, , (b) 25%, , (c) 35%, , (d) 15%, , The force of friction should balance the, weight of chain hanging. If M is the mass, of whole chain of lengthL and x is the, length of chain hanging to balance, then, M, M, µ (L − x) g = xg, L, L, or, µ (L − x) = x, 0.25 L, µL, (As, µ = 0.25), or, x=, =, µ+1, 125, ., x 1 1, L, ∴, x = or = = × 100 = 20%, L 5 5, 5, , 57 Starting from rest, a body slides, down a 45° inclined plane in twice, the time it takes to slide down the, same distance in the absence of, friction. The coefficient of friction, between the body and the inclined, plane is, [CBSE AIPMT 1988], , θ, , A, , mg, , mg cos θ, C, , As, it is clear from figure, …(i), R = mg cosθ, Net force on the body down the inclined, plane which means it is sliding, downwards, …(ii), F = mg sinθ − f, i.e., F = ma = mg sinθ − µR, (f = µR), ∴, ma = mg sinθ − µ mg cosθ, = mg (sinθ − µ cosθ), Hence, a = g (sinθ − µ cosθ), ∴Time taken by body to slide down the, plane, t1 =, , 2s, 2s, =, a, g (sinθ − µ cosθ), , When friction is absent, then time taken, to slide down the plane, t2 =, , [CBSE AIPMT 1991], , Ans. (a), , θ, , ∴, or, , 2s, ⇒ Q t 1 = 2t 2, g sinθ, , (given), , t 12 = 4t 22, 2s, 2s × 4, =, g (sinθ − µ cosθ) g sinθ, , or, or, , ∴, , sinθ = 4 sinθ − 4µ cosθ, 3, 3, µ = tanθ = tan45°, 4, 4, , 3, µ = = 0.75, 4, , TOPIC 4, Dynamics of Circular Motion, 58 A block of mass 10 kg is in contact, against the inner wall of a hollow, cylindrical drum of radius 1 m. The, coefficient of friction between the, block and the inner wall of the, cylinder is 0.1. The minimum, angular velocity needed for the

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cylinder to keep the block, stationary when the cylinder is, vertical and rotating about its axis,, will be (g = 10 m/s 2 ), [NEET (National) 2019], , 10, (a), rad/s, 2π, (c) 10π rad/s, , (b) T +, , (a)T, (c) T −, , mv 2, l, , Ans. (a), , (d) 10 rad/s, , peg, , Ans. (b), radius of cylinder, r = 1 m, coefficient of friction,µ = 0.1., The given situation can be as shown in, the figure given below., , fl, N, mg, , From the above figure, it can be, concluded that the block will be, stationary when the limiting friction (fl ) is, equal to or greater than the downward, force or weight of block, i.e., …(i), , Also, the magnitude of limiting friction, between two bodies is directly, proportional to the normal reaction (N), between them, i.e., fl ∝ N or fl = µ N, , …(ii), , From Eqs. (i) and (ii), we get, µN ≥ mg or µ( mrω2 ) ≥ mg, [QN = mrω2 ], ⇒ ω≥, , Speed of the particle is v. As the particle, is in uniform circular motion, net force, on the particle must be equal to, centripetal force which is provided by, the tension (T )., ∴ Net force = Centripetal force, mv 2, ⇒, =T, l, , ω, , fl ≥ mg, , particle, v, l, , Given, mass of cylinder m = 10 kg,, , g, rµ, , 60 A uniform circular disc of radius 50, cm at rest is free to turn about an, axis which is perpendicular to its, plane and passes through its, centre. It is subjected to a torque, which produces a constant angular, acceleration of 2.0 rad s −2 . Its net, acceleration in ms −2 at the end of, 2.0 s is a approximately [NEET 2016], (a) 7.0, (c) 3.0, , a = (8) 2 + (1) 2 = 65 ⇒ a ≈ 8 m/s 2, , 61 A car is negotiating a curved road, of radius R. The road is banked at, angle θ. The coefficient of friction, between the tyres of the car and, the road is µ s . The maximum safe, velocity on this road is [NEET 2016], µ + tanθ , , (a) gR s, 1 − µ s tanθ , (b), , g µ s + tanθ , , , R 1 − µ s tanθ , , (c), , g, R2, , µ + tanθ , , (d) gR 2 s, 1 − µ s tanθ , Ans. (a), According to question, a car is, negotiating a curved road of radius R., The road is banked at angleθ and the, coefficient of friction between the tyres, of car and the road isµ s .So, this given, situation can be drawn as shown in, figure below., Ncosθ, , Ans. (d), , 59 One end of the string of length l is, connected to a particle of mass m, and the other end is connected to a, small peg on a smooth horizontal, table. If the particle moves in circle, with speed v, the net force on the, particle (directed towards center) will, be (T represents the tension in the, string), [NEET 2017], , θ, , Nsinθ, , According to given question, a uniform, circular disc of radius 50 cm at rest is, free to turn about an axis having, perpendicular to its plane and passes, through its centre. This situation can be, shown by the figure given below:, 0.5m, , Fl cosθ, fl, θ, , ∴ Angular acceleration,α = 2 rad s −2, (given), Angular speed,ω = αt = 4 rad s −1, Q Centripetal acceleration,a c = ω2 r, = (4) 2 × 0.5, = 16 × 0.5, a c = 8 m/s 2, , Therefore, the net acceleration at the, end of 2.0 s is given by, , fl sinθ, , Considering the case of vertical, equilibrium, N cos θ = mg + fl sin θ, mg = N cos θ − fl sin θ, , …(i), , Considering the case of horizontal, equilibrium,, mv 2, …(ii), N sin θ + fl cos θ =, R, Divide eqs. (i) and (ii), we get, v 2 sin θ + µ s cos θ, =, Rg cos θ − µ s sin θ, ⇒, , sinθ + µ s cos θ , v = Rg , , cos θ − µ s sin θ , , ⇒, , tan θ + µ s , v = Rg , , 1 − µ s tan θ , , Q Linear acceleration at the end of2s,, a t = αr = 2 × 0.5 ⇒ a t = 1 m/s 2, , θ, , mg, , ⇒, , g, 10, =, = 10 rad/s, rµ, 1 × 0.1, , µ s + tanθ , , , 1 − µ s tanθ , , (b) 6.0, (d) 8.0, , Thus, the minimum angular velocity is, ωmin =, , a = a c2 + a t2, , (d) Zero, , Consider the string of length l connected, to a particle as shown in the figure., , (b) 10 rad/s, , mrω2, , mv 2, l, , [fl ∝µ s ]

cylinder to keep the block, stationary when the cylinder is, vertical and rotating about its axis,, will be (g = 10 m/s 2 ), [NEET (National) 2019], , 10, (a), rad/s, 2π, (c) 10π rad/s, , (b) T +, , (a)T, (c) T −, , mv 2, l, , Ans. (a), , (d) 10 rad/s, , peg, , Ans. (b), radius of cylinder, r = 1 m, coefficient of friction,µ = 0.1., The given situation can be as shown in, the figure given below., , fl, N, mg, , From the above figure, it can be, concluded that the block will be, stationary when the limiting friction (fl ) is, equal to or greater than the downward, force or weight of block, i.e., …(i), , Also, the magnitude of limiting friction, between two bodies is directly, proportional to the normal reaction (N), between them, i.e., fl ∝ N or fl = µ N, , …(ii), , From Eqs. (i) and (ii), we get, µN ≥ mg or µ( mrω2 ) ≥ mg, [QN = mrω2 ], ⇒ ω≥, , Speed of the particle is v. As the particle, is in uniform circular motion, net force, on the particle must be equal to, centripetal force which is provided by, the tension (T )., ∴ Net force = Centripetal force, mv 2, ⇒, =T, l, , ω, , fl ≥ mg, , particle, v, l, , Given, mass of cylinder m = 10 kg,, , g, rµ, , 60 A uniform circular disc of radius 50, cm at rest is free to turn about an, axis which is perpendicular to its, plane and passes through its, centre. It is subjected to a torque, which produces a constant angular, acceleration of 2.0 rad s −2 . Its net, acceleration in ms −2 at the end of, 2.0 s is a approximately [NEET 2016], (a) 7.0, (c) 3.0, , a = (8) 2 + (1) 2 = 65 ⇒ a ≈ 8 m/s 2, , 61 A car is negotiating a curved road, of radius R. The road is banked at, angle θ. The coefficient of friction, between the tyres of the car and, the road is µ s . The maximum safe, velocity on this road is [NEET 2016], µ + tanθ , , (a) gR s, 1 − µ s tanθ , (b), , g µ s + tanθ , , , R 1 − µ s tanθ , , (c), , g, R2, , µ + tanθ , , (d) gR 2 s, 1 − µ s tanθ , Ans. (a), According to question, a car is, negotiating a curved road of radius R., The road is banked at angleθ and the, coefficient of friction between the tyres, of car and the road isµ s .So, this given, situation can be drawn as shown in, figure below., Ncosθ, , Ans. (d), , 59 One end of the string of length l is, connected to a particle of mass m, and the other end is connected to a, small peg on a smooth horizontal, table. If the particle moves in circle, with speed v, the net force on the, particle (directed towards center) will, be (T represents the tension in the, string), [NEET 2017], , θ, , Nsinθ, , According to given question, a uniform, circular disc of radius 50 cm at rest is, free to turn about an axis having, perpendicular to its plane and passes, through its centre. This situation can be, shown by the figure given below:, 0.5m, , Fl cosθ, fl, θ, , ∴ Angular acceleration,α = 2 rad s −2, (given), Angular speed,ω = αt = 4 rad s −1, Q Centripetal acceleration,a c = ω2 r, = (4) 2 × 0.5, = 16 × 0.5, a c = 8 m/s 2, , Therefore, the net acceleration at the, end of 2.0 s is given by, , fl sinθ, , Considering the case of vertical, equilibrium, N cos θ = mg + fl sin θ, mg = N cos θ − fl sin θ, , …(i), , Considering the case of horizontal, equilibrium,, mv 2, …(ii), N sin θ + fl cos θ =, R, Divide eqs. (i) and (ii), we get, v 2 sin θ + µ s cos θ, =, Rg cos θ − µ s sin θ, ⇒, , sinθ + µ s cos θ , v = Rg , , cos θ − µ s sin θ , , ⇒, , tan θ + µ s , v = Rg , , 1 − µ s tan θ , , Q Linear acceleration at the end of2s,, a t = αr = 2 × 0.5 ⇒ a t = 1 m/s 2, , θ, , mg, , ⇒, , g, 10, =, = 10 rad/s, rµ, 1 × 0.1, , µ s + tanθ , , , 1 − µ s tanθ , , (b) 6.0, (d) 8.0, , Thus, the minimum angular velocity is, ωmin =, , a = a c2 + a t2, , (d) Zero, , Consider the string of length l connected, to a particle as shown in the figure., , (b) 10 rad/s, , mrω2, , mv 2, l, , [fl ∝µ s ]

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62 A car of mass 1000 kg negotiates a, banked curve of radius 90 m on a, frictionless road. If the banking, angle is 45°, the speed of the car is, , Ffrictional ≥ Fcentripetal, i.e., , µ mg ≥ mω2 r, µg, ≥ ω2, r, , [CBSE AIPMT 2012], , (a) 20 ms −1, (b) 30 ms −1, (c) 5 ms −1, (d) 10 ms −1, Ans. (b), The angle of banking, tanθ =, , v2, rg, , θ = 45°, , Given,, , Radius of banked curve road, r = 90 m and g = 10 m/s 2, ⇒, , tan45° =, , 2, , v, 90 × 10, , v = 90 × 10 × tan45°, = 90 × 10 × 1 = 30 m/s, , 63 A gramophone record is revolving, with an angular velocity ω. A coin is, placed at a distance r from the, centre of the record. The static, coefficient of friction is µ. The coin, will revolve with the record if, [CBSE AIPMT 2010], , (a) r = µgω, (c) r ≤, , µg, ω2, , 2, , ω2, (b) r <, µg, µg, (d) r ≥ 2, ω, , Ans. (c), When the disc spins, the frictional force, between the gramophone record and, coin is µ mg., The coin will revolve with record, if, , 64 A ball of mass 0.25 kg attached to, the end of a string of length 1.96 m, is moving in a horizontal circle. The, string will break if the tension is, more than 25 N. What is the, maximum speed with which the, ball can be moved ?, [CBSE AIPMT 1998], , (a) 14 m/s, (c) 3.92 m/s, , (b) 3 m/s, (d) 5 m/s, , Ans. (a), For a ball to move in horizontal circle, the, ball should satisfy the condition, Tension in the string = Centripetal force, Mv2, ⇒, Tmax = max, R, T ⋅R, ...(i), vmax = max, ⇒, M, Making substitution, we obtain, 25 × 196, ., vmax =, = 196 = 14 m/s, 0.25, In a vertical circle, the tension at the, highest point is zero and at lowest point is, maximum., , 65 What will be the maximum speed, of a car on a road turn of radius 30, m, if the coefficient of friction, between the tyres and the road is, 0.4 ? (Take g = 9.8 m / s2 ), [CBSE AIPMT 1995], , (a) 10.84 m/s, (b) 9.84 m/s, (c) 8.84 m/s, (d) 6.84 m/s, , Ans. (a), When a vehicle goes round a curved, road, it requires some centripetal force., While rounding the curve, the wheels of, the vehicle have a tendency to leave the, curved path and regain the straight line, path. Force of friction between the, wheels and the road opposes this, tendency of the wheels. This force (of, friction) therefore, acts towards the, centre of the circular track and provides, the necessary centripetal force., If v is the velocity of the vehicle while, rounding the curve, the centripetal force, mv2, required =, r, As, this force is provided only by the, force of friction,, mv2, ≤ µmg ⇒ v2 ≤ µrg ⇒ v≤ µrg, r, ∴, , vmax = µrg, , Here, radius of curved road r = 30 m,, coefficient of friction µ = 0.4, ∴, , vmax = 0.4 × 30 × 9.8, = 10.84 m/s, , 66 Two racing cars of masses m and, 4m are moving in circles of radii r, and 2r respectively. If their speeds, are such that each makes a, complete circle in the same time,, then the ratio of the angular, speeds of the first to the second, car is, [CBSE AIPMT 1995], (a) 8 : 1, , (b) 4 : 1, , (c) 2 : 1, , (d) 1 : 1, , Ans. (d), As both cars take the same time to, 2π, complete the circle and as ω = ,, t, therefore ratio of angular speeds of the, cars will be 1 : 1.

62 A car of mass 1000 kg negotiates a, banked curve of radius 90 m on a, frictionless road. If the banking, angle is 45°, the speed of the car is, , Ffrictional ≥ Fcentripetal, i.e., , µ mg ≥ mω2 r, µg, ≥ ω2, r, , [CBSE AIPMT 2012], , (a) 20 ms −1, (b) 30 ms −1, (c) 5 ms −1, (d) 10 ms −1, Ans. (b), The angle of banking, tanθ =, , v2, rg, , θ = 45°, , Given,, , Radius of banked curve road, r = 90 m and g = 10 m/s 2, ⇒, , tan45° =, , 2, , v, 90 × 10, , v = 90 × 10 × tan45°, = 90 × 10 × 1 = 30 m/s, , 63 A gramophone record is revolving, with an angular velocity ω. A coin is, placed at a distance r from the, centre of the record. The static, coefficient of friction is µ. The coin, will revolve with the record if, [CBSE AIPMT 2010], , (a) r = µgω, (c) r ≤, , µg, ω2, , 2, , ω2, (b) r <, µg, µg, (d) r ≥ 2, ω, , Ans. (c), When the disc spins, the frictional force, between the gramophone record and, coin is µ mg., The coin will revolve with record, if, , 64 A ball of mass 0.25 kg attached to, the end of a string of length 1.96 m, is moving in a horizontal circle. The, string will break if the tension is, more than 25 N. What is the, maximum speed with which the, ball can be moved ?, [CBSE AIPMT 1998], , (a) 14 m/s, (c) 3.92 m/s, , (b) 3 m/s, (d) 5 m/s, , Ans. (a), For a ball to move in horizontal circle, the, ball should satisfy the condition, Tension in the string = Centripetal force, Mv2, ⇒, Tmax = max, R, T ⋅R, ...(i), vmax = max, ⇒, M, Making substitution, we obtain, 25 × 196, ., vmax =, = 196 = 14 m/s, 0.25, In a vertical circle, the tension at the, highest point is zero and at lowest point is, maximum., , 65 What will be the maximum speed, of a car on a road turn of radius 30, m, if the coefficient of friction, between the tyres and the road is, 0.4 ? (Take g = 9.8 m / s2 ), [CBSE AIPMT 1995], , (a) 10.84 m/s, (b) 9.84 m/s, (c) 8.84 m/s, (d) 6.84 m/s, , Ans. (a), When a vehicle goes round a curved, road, it requires some centripetal force., While rounding the curve, the wheels of, the vehicle have a tendency to leave the, curved path and regain the straight line, path. Force of friction between the, wheels and the road opposes this, tendency of the wheels. This force (of, friction) therefore, acts towards the, centre of the circular track and provides, the necessary centripetal force., If v is the velocity of the vehicle while, rounding the curve, the centripetal force, mv2, required =, r, As, this force is provided only by the, force of friction,, mv2, ≤ µmg ⇒ v2 ≤ µrg ⇒ v≤ µrg, r, ∴, , vmax = µrg, , Here, radius of curved road r = 30 m,, coefficient of friction µ = 0.4, ∴, , vmax = 0.4 × 30 × 9.8, = 10.84 m/s, , 66 Two racing cars of masses m and, 4m are moving in circles of radii r, and 2r respectively. If their speeds, are such that each makes a, complete circle in the same time,, then the ratio of the angular, speeds of the first to the second, car is, [CBSE AIPMT 1995], (a) 8 : 1, , (b) 4 : 1, , (c) 2 : 1, , (d) 1 : 1, , Ans. (d), As both cars take the same time to, 2π, complete the circle and as ω = ,, t, therefore ratio of angular speeds of the, cars will be 1 : 1.

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