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HSPTA MALAPPURAM, PHYSOL-The Solution for Learning Physics, , Question Bank, CHAPTER 04- MOTION IN A PLANE, Each question scores One, 1 The physical quantities having only magnitude but no direction are called as.........., Ans: Scalars, 2 The physical quantities having both magnitude and direction are called as............, Ans: Vectors, 3 The vectors having zero magnitude are called as...................., Ans: Null vectors or Zero vectors., 4 The angle between, a)450, , ⃗ = ^i + ^j and B, ⃗ = ^i − ^j is ...................., A, , b) 600 c) 900 d)1800, , Ans; 900, 5 Identify the scalar quantity from the following alternatives., (i) Momentum, (ii) Work, (iii)Torque, (iv) Acceleration, Ans: Work., 6 A ball is dropped through the window of a train travelling with high velocity, to a man standing, near the track. The ball............., i.Falls down vertically, ii.Moves straight horizontally, iii.Follows an elliptical path, iv.Follows a parabolic path, Ans: Follows a parabolic path., 7 At the top of a projectile vertical velocity of the object will be.........., Ans: zero., 8 At the top of a projectile, angle between velocity and acceleration is...., a)0° b) 45° c)60° d) 90°., Ans: zero.90°, 9 From a height an object A is dropped, at the same time another object B is thrown horizontally with, a velocity 20m/s from the same point. Which one will reach the ground first?, , Prepared by Higher Secondary Physics Teachers Association Malappuram

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Ans: both will reach the ground at the same time, 10 An object is projected with a velocity u at an angle 20° with horizontal. To get the same range, another object projected from the same point with same velocity at an angle of........, Ans :70°, 11 Maximum range of a projectile is 1.6 m. Then the velocity of projection will be..... (g=10m/s2), Ans : 4 m/s., 12 If two vectors can be connected by the equation b= 4a. Then a and b will be............Vectors?, Ans: Colinear vectors, 13 The physical quantity which is constant at any point in projectile motion is.........., a)Velocity b)Acceleration c)Kinetic energy d)Linear momentum, Ans : Acceleration, Each question scores Two, 1 A boy throws a ball of mass 200 g with a velocity 20 ms -1 at an angle of 40° with the horizontal., What is the kinetic energy of the ball at the highest point of the trajectory?, Ans: At the highest point Vx=ucosθ and Vy=0., Given m=0.2 kg u=20 m/s and θ=400., 1, mV 2x, KE at the highest point =, 2, 1, m(ucos θ )2, =, 2, 1, x 0.2 x (20 cos 40)2, =, 2, = 23.47 J, A particle is projected up into the air from the point with a speed of 20 m/s at an angle of projection, 2 30°. What is the maximum height reached by it.?, θ=300., u2 sin 2 θ, H=, 2g, 202 sin2 30, H=, 2 x 9.8, , Ans: Given u=20m/s,, , ie, , H=5.1 m, , 3 A football is kicked into the air vertically upwards. What is its, (a) acceleration at the highest point, (b) velocity at the highest point?, Ans: (a) At highest point the acceleration remains same as acceleration due to gravity, (b) At highest point, velocity becomes zero, 4 If horizontal range is equal to 4 times maximum height. Find the angle of projection?, , Prepared by Higher Secondary Physics Teachers Association Malappuram

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Ans:, , 4H, R, Here R = 4H, tan θ=, , So θ=45, , 5 A ball is thrown with a velocity u at an angle θ at the same time a boy running towards the ball, from the point projection with a velocity u/2. Can the boy be able to catch the ball?, u, 2, Cosθ= 1/2, θ=60°., , Ans : Yes. When ucosθ =, , 6 Why electric current is a scalar quantity?, Ans : Electric current will not obey the law of Vector Algebra.(Vector Addition Laws) So it can be, considered as a scalar, 7 A ball is projected with a velocity 30m/s. Find the maximum range?(take g=10m/s2), u2, g, R= 30x30/10, R=900/10, R=90m, Ans : R =, , 8 A ball thrown by one player is caught by another player in 5 seconds then calculate the maximum, height reached by the ball (g= 10 m/s2), given 2usinθ /g= 5, usinθ= 25, Now., , H = u2Sin2 θ/ 2g, = 252/2 x 10 = 31.25 m, , 9 A food packet is released from a plane flying horizontally, reaches the surface of the earth in 10, second. Calculate the height from which the packet is dropped, - h = - 1/2 gt2, h = 1/2 x 10 X 100 = 500m, Each question scores Three, 1 (a) Give the expression to find the unit vector of a given vector, ⃗ = 4 ^i−3 ^j+ k^, (b) Find the unit vector of A, Ans: (a) The unit vector of, ⃗, ^ A, A=, |⃗, A|, (b), , ⃗, A, , ⃗ ,, A, , ⃗ = 4 ^i−3 ^j+ k^, A, A|=√ A 2x + A2y + A 2z, Here | ⃗, |⃗, A|=√ 4 2+(−3)2 +12, |⃗, A|=√ 16+9+1=√26, Prepared by Higher Secondary Physics Teachers Association Malappuram

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There fore, 2, , ^, ^ ^, ⃗, ^ A = 4 i−3 j+ k, A=, |⃗, A|, √ 26, , The position vector ⃗r, , of a particle P located in an x-y plane is shown in figure., , a) Redraw the figure by showing the rectangular components., b) Write the position vector in terms of rectangular components., c) Write an equation to find the magnitude of the resultant of two vectors A and B, Ans: a), , yj, , b) ⃗r =x ^i+ y ^j, 2, 2, c) R= √ A + B +2 ABcos θ, 3, , A stone is thrown with the help of a sling with initial velocity ‘u’ at an angle 'θ' from the horizontal.θ'θ' from the horizontal. from the horizontal., , a) Working of sling is based on........... law of vector addition., b) Derive the expression for the maximum height reached by the stone., Ans:, a)Parallelogram law of vector addition., , Prepared by Higher Secondary Physics Teachers Association Malappuram

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2 gH =u2 sin 2 θ, u2 sin 2 θ, 2g, Expression for Time of flight (T):, 1 2, We have S=ut + at, 2, Taking vertical components;, 1, S y =u y t + a y t 2, 2, Here Sy=0, uy=usinθ , ay=-g and t=T, time of flight., 1, 2, Therefore 0=u sin θ T − g T, 2, 1, 2, g T =usin θ T, 2, 1, g T =usin θ, 2, 2 u sin θ, Time of flight T =, g, b) For Maximum horizontal range , angle of projection θ=450., Maximum Height ,, , H=, , 2 a)A man throws a stone up into air at an angle 'θ' from the horizontal.θ'θ' from the horizontal. with the horizontal. Draw the path of the projectile, and mark directions of velocity and acceleration at the highest position., b) Derive an expression for the maximum height reached by the stone., Ans: a), , b)Expression for Maximum height(H):, We have V 2=u2 +2 as, Taking the vertical components;, V 2y =u 2y + 2a y s y, Here Vy=0, uy=usinθ , ay=-g and Sy =H, Therefore, 0=(usin θ )2 −2 gH, 2 gH =u2 sin 2 θ, u2 sin 2 θ, Maximum Height , H=, 2g, 3 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown, with a speed of 40 m/s can go without hitting the ceiling of the hall?, Ans: Given H=25m , u= 40 m/s., u2 sin 2 θ 402 sin 2 θ, H=, =, =25 m, 2g, 9.8, 25 x 9.8, sin 2 θ=, 402, Prepared by Higher Secondary Physics Teachers Association Malappuram

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Therefore θ=33.60., 2, u2 sin 2 θ 40 sin 2(33.6), Range R=, =, =150.5 m, g, 9.8, 4 A ball is thrown horizontally from the top of a tower with a velocity of 40 m/s. Take g=10 m/s2., (a)Find the horizontal and vertical displacement after 1,2,3,4,5 seconds, then plot the path of, motion of the ball., (b)If the ball reaches the ground in 4 s , find the height of the tower., Ans: (a) Horizontal displacement S x =ux t, 1 2, Vertical displacement S y = g t, 2, , because ax=0, because uy=0 ay=g, , t, , 1, , 2, , 3, , 4, , 5, , Sx, , 40, , 80, , 120, , 160, , 200, , Sy, , 5, , 20, , 45, , 80, , 125, , (b) Hight of the tower ,, , 1, 1, H= g t 2 = x 10 x 4 2, 2, 2, , H=80m., Each question scores Five, 1 A projectile is any body that is given an initial velocity and then follows a path determined entirely, by the effects of gravitational acceleration and air resistance., a) The path of a projectile is........., i) straight line, ii) parabola, iii) circle, iv) semi circle, b) Derive an expression for time to reach maximum height and hence the time of flight of a, projectile., c) A baseball leaves a bat with an initial speed of 37 m/s at an angle of 53.1 o . Find the position of, the ball when t = 2s (treat baseball as a projectile and g = 9.8 m/s 2 ), Ans: a)(ii) parabola., b)Expression for time to reach maximum height:, , At the maximum height , Vy=0 , uy=usinθ and ay=-g, We have v y =u y −g t, 0=usinθ-gt, usin θ, Therefore time to reach maximum height , t=, g, , Prepared by Higher Secondary Physics Teachers Association Malappuram

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Expression for Time of flight (T):, 1 2, We have S=ut + at, 2, Taking vertical components;, 1, S y =u y t + a y t 2, 2, Here Sy=0, uy=usinθ , ay=-g and t=T, time of flight., 1, 2, Therefore 0=u sin θ T − g T, 2, 1, g T 2=usin θ T, 2, 1, g T =usin θ, 2, 2 u sin θ, Time of flight T =, g, c) Here u= 37 m/s, θ= 53.1o g=9.8 m/s2, ux = ucos θ=37 cos (53.1)=37x 0.6=22.2m/s, uy=usin θ=37 sin (53.1)=37x0.79=29.59 m/s., The x-coordinate is given by, 1, S x =ux t+ a x t 2=u x t, ( ax=0), 2, Therefore at t=2s, x=22.2 x 2=44.4 m, The y-coordinate is given by, 1, S y =u y t + a y t 2, 2, 1, y=29.59 x 2− 9.8 x 22 (ay=-g=-9.8 m/s2), 2, y=39.6m, Therefore the position of the ball when t = 2s is given by (44.4, 39.6), 2 A body is projected into air at an angle θ with the horizontal., a) What is the trajectory followed by this projectile?, i) Ellipse, ii) Parabola, iii) Straight line, iv) Circle, b) Give a mathematical proof for your answer., c) Trajectory of a body in a projectile motion is given by y=x−, , x2, 80, , where x and y are in meters., , Find maximum height of this projectile., Ans:, a)Parabola., b) Here ux=ucosθ , uy=usinθ , ay=-g, Let ‘x’ be the horizontal distance covered in a time ‘t’ , then, x, t=, -----(1), ucos θ, Let ‘y’ be the vertical distance covered in a time ‘t’ , then, 1, y=u sin θ t− g t 2, 2, , Prepared by Higher Secondary Physics Teachers Association Malappuram

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x, 1, x 2, − g(, ), u cos θ 2 u cos θ, g, y=(tan θ ) x−( 2 2 ) x 2, Therefore, 2 u cos θ, The above equation is similar to, y=ax−b x 2 and represents a parabola., g, x2, 2, c)Given y=x−, and We have y=(tan θ ) x−( 2 2 ) x, 80, 2 u cos θ, Comparing, we get, tanθ=1 Therefore θ=450., g, 1, And, =, 2, 2, 80, 2 u cos θ, 2, u, That is, =80, g, u2 sin 2 θ, We have Maximum height, H=, 2g, 80, H=, =20 m., 2x 2, y=u sin θ, , Prepared by Higher Secondary Physics Teachers Association Malappuram