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KINEMATICS, , MOTION IN TWO DIMENSION, 10. SCALARS AND VECTORS, Some quantities can be deseribed by single number. For e.g.:, Mass, time, distance, speed. One piece of infermation is, enough to describe them fully. These are called SCALAR, quantities., To tell someone how to get to Lakshya from some location,, one piece of information is not enough. To describe this fully,, both distance and displacement are required. Quantities which, require both magnitude and direction to describe a situation, fully are known as VECTOR. For e.g.: displacement, velocity, , G G, Then, r1 r2, , (a1 a 2 )iˆ (b1 b 2 )ˆj, , G G, r1 r2, , (a1 a 2 )iˆ (b1 b2 )ˆj, , Multiplication of a vector by scalar quantity., , G, cr1, , c(a1ˆi b1ˆj) ca1ˆi cb1ˆj, , G, Representation of r1 on the co–ordinate axis, , The vectors are denoted by putting an arrow over the, symbols representing them., JJJG, For e.g.: AB vector can be represented by AB, 10.1 Unit vector, A unit vector has a magnitude of one and so it really gives just, the direction of the vector., A unit vector can be found by dividing the original vector by, its magnitude, â, , G, a, a, , unit vectors along different co–ordinate axis, , G, magnitude and direction of r1, G G, Magnitude of r1( r1 ), , a12 b12, , G, direction of r1, , tan T=, , b1, a1, , component y axis, component along x axis, , §b ·, T tan 1 ¨ 1 ¸, © a1 ¹, , 10.3 Parallel vectors, Two vectors are parallel if and only if they have the same, 10.2 Addition, subtraction and scalar multiplication, of vectors, , direction. When any vector is multiplied by a scalar, a vector, parallel to the original vector is formed., , Suppose, we have two vectors, , G, If b, , G, r1, , a1ˆi b1ˆj, , if two vectors are parallel or not we must find their unit vectors., , G, r2, , a 2ˆi b 2ˆj, , G, G, G, ka then b and a are parallel vector. In general to find
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KINEMATICS, 10.4 Equality of vectors, Two vectors (representing two values of the same physical, quantity) are called equal if their magnitudes and directions, are same., , Another way is parallelogram rule of vector addition, G, G, on this we draw vectors a and b, with both the tails co–, inciding. Taking these two odjacent sides we complete the, parallelogram. the diagonal through the common tails gives, the sum of two vectors., , For e.g. (3iˆ 4ˆj)m and (3iˆ 4ˆj)m / s, Cannot be compared as they represent two different, physical quantities., 10.5 Addition of vectors, When two or more vectors are added, the answer is called, the resultant. The resultant of two vectors is equivalent to, the first vector followed immediately by the second vector., , G G, Finding magnitude of a b and its direction, |AD|2 = AE2 + ED2, AE = |a| + |b cos T|, ED = b sin T, AD2 = a2 + b2 cos2T + 2ab cos T + b2 sin2T, AD2 = a2 + b2 + 2ab cos T, AD =, , G, G, To find the resultant of vectors a and b, the tail of vector, , a 2 b 2 2ab cos T, , G, G, where, T is the angle contained between a and b, , G, G G, G, b must join to the head of vector a . The resultant a b, G, is the direct vector from the tail of vector a to the head of, G, vector b ., , tan D =, , ED, AE, , bsin T, a b cos T, , where D is the angle which the resultant makes with + x axis, Subtraction of vectors :, G, G, G G, Let a and b be two vectors. We define a b as sum of, G, G, vectors a and the vectos b ., G, G, or, a b, , This is known as triangle rule of vector addition
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KINEMATICS, Note : The direction of instantaneous velocity at any point, on the path of an object is tangent to the path at that point, and is in the direction of motion., 11.4 Averge acceleration, G, 'v, 't, , G, a avg, G, a avg, , 'v x ˆ 'v y ˆ, i, j, 't, 't, , a x ˆi a y ˆj, , 11.5 Instantaneous acceleration, , G, a, , G, dv, dt, , G, a, , G, a x ˆi a y ˆj, , dv x ˆ dv y ˆ, i, j, dt, dt, , 12. PROJECTILE MOTION, When a particle is projected obliquely near the earth, surface, it moves simultaneously in horizontal and vertical, directions. Motion of such a particle is called projectile, motion., , Horizontal axis, , Vertical axis, , ux = u cosT, , uy = u sin T, , ax = 0, , ay = – g, , (In the absence of any, , sy = uyt + 1/2 ay t2, , external force ax will be, , 0 – 0 = u sin T t – 1/2g t2, , assumed to be zero), , T=, , sx = ux t + 1/2ax t2, , vy = uy + ay t, , x – 0 = u cos T t, , vy = u sin T – gt, , x = u cos T× 2uy/g, , It depends on time ‘t’, , 2u sin T, g, , 2u y, g, , x=, , 2u 2 cos T sin T, g, , It in not constant, , R=, , u 2 sin 2T, g, , It’s magnitude first, , (' 2 cos T sin T = sin 2T), , decreases becomes zero, , horizontal distance covered, , and then increases., , is known as Range, vx = ux + ax t, , maximum height obtained, by the particle, , vx = u cos T, , Method 1 : using time of, , It is independent of t, , ascent, , It is constant, , sy = uyt1 +, , time of ascent and time of, , H=, , 1, ay t12, 2, , u 2 sin 2 T, 2g, , descent, , In this case a particle is projected at an angle T with an, initial velocity u. For this particular case we will calculate, , At top most point vy = 0, , Method 2 : using third, , vy = uy + ay t, , equation of motion, , 0 = u sin T – gt, , v2y u 2y, , the following :, (a) time taken to reach A from O, , t1 =, , u sin T, g, , (b) horizontal distance covered (OA), (c) maxm height reached during the motion, , t2 = T – t1 =, , 0 – u2 sin2T = – 2g sy, u sin T, g, , = u sin T ×, , (d) velocity at any time ‘t’ during the motion, t1 = t2 =, , T, 2, , 2a y s y, , u sin T, g, , H=, , u sin T 1 u 2 sin 2 T, g, g, 2, g2, , u 2 sin 2 T, 2g
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KINEMATICS, Maximum Range, R, , 12.2 Equation of trajectory, , u 2 sin 2T, and R max, g, , u2, g, , Range is maximum when sin 2Tis maximum, max (sin 2T) = 1 or, T = 45°, 12.1 Analysis of velocity in case of a projectile, , Trajectory is the path traced by the body. To find the, trajectory we must find relation between y and x by, eliminating time., [Ref. to the earlier diag], Horizontal Motion, , Vertical Motion, , ux = u cos T, , uy = u sin T, , ax = 0, , ay = – g, , sx = u cos T t = x, , sy = uyt +, , t=, , x, u cos T, , y = x tan T – , , 1, a t2, 2 y, , x2, § x · 1, y = u sin T ¨, ¸ g 2, © u cos T ¹ 2 u cos2 T, gx 2, y = bx – ax2, 2u 2 cos 2 T, , From the above equations;, (i) v1x = v2x = v3x = v4x = ux = u cos T, which means that the velocity along x axis remains constant, [as there is no external force acting along that direction], , (i) This is a equation of a parabola, (ii) Because the co–officient of x2 is negative, it is an inverted, parabola., , (ii) a) magnitude of velocity along y axis first decreases and, then it increases after the top most point, b) at top most point magnitude of velocity is zero., c) direction of velocity is in the upward direction while, ascending and is in the downward direction while, descending., d) magnitude of velocity at A is same as magnitude of, , Path of the projectile is a parabola, , velcoity at 0; but the direction is inverse, e) angle which the net velocity makes with the horizontal, can be calculated by, tan D, , vy, vx, , velocity along y axis, velocity along x axis, , net velocity is always along the tangent, , R, , 2u 2 sin T cos T, 2u 2, or,, g, g, , R, sin T cos T, , Substituting this value in the above equation we have,, , ª xº, y = x tan T «1- », ¬ R¼
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KINEMATICS, A boatman starts from point A on one bank of a river, G, with velocity vbr in the direction shown in figure. River, G, is flowing along positive x–direction with velocity v r ., , 13. RELATIVE MOTION, Relative is a very general term, In physics we use relative very oftenly., , Width of the river is d. Then, , For eg, , G, vb, , Case I : If you are observing a car moving on a straight, road then you say velocity of car is 20 m/s which means, velocity of car relative to you is 20 m/s or, velocity of, car relative to ground is 20 m/s (as you are standing, on the ground., , G G, v r v br, , Therefore,, , vbx = v rx + vbrx = v r – v br sin T, , and, , vby = vby + vbry = 0 + vbr cosT = vbr cosT, , Now, time taken by the boatman to cross the river is :, , Case II : If you go inside a car and observe you will find, that the car is at rest while the road is moving back, wards. you will say;, , t, , d, v by, , d, or t, v br cos T, , d, v br cos T, , ...(i), , velocity of car relative to the car is 0 m/s, Further, displacement along x–axis when he reaches, on the other bank (also called drift) is, , Mathematically, velocity of B relative to A is represented, as, G, G G, v BA = v B - v A, , vr v br sin T, , x, , v bx t, , x, , v r v br sin T, , d, v br cos T, , This being a vector quantity direction is very important, or, G, G, v BA ≠ v AB, , ?, , From eq. (i) we can see that time (t) will be minimum, when T = 0° i.e., the boatman should steer his boat, perpendicular to the river current., , In river–boat problems we come across the following, three terms :, G, v r absolute velocity of river.., , Condition when the boat wants to reach point B, i.e.,, at a point just opposite from where he started (shortest, distance), , G, v br, , velocity of boatman with respect to river or, G, velocity of boatman in still water and v b, absolute, G, Hence, it is important to note that vbr is the velocity, G, of boatman with which he steers and vb is the actual, , velocity of boatman relative to ground. Further, G, G, G, v b v br v r, Now, let us derive some standard results and their, special cases., , ...(ii), , Condition when the boatman crosses the river in, shortest interval of time, , 14. RIVER–BOAT PROBLEMS, , velocity of boatman., , d, v br cos T, , In this case, the drift (x) should be zero., ?, or, , or, , x=0, v r v br sin T, , sin T, , d, v br cos T, , vr, or T, v br, , 0 or v r, , v br sin T, , §v ·, sin 1 ¨ r ¸, © v br ¹, , Hence, to reach point B the boatman should row at an
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KINEMATICS, G, vm, , §v ·, angle T sin ¨ r ¸ upstream from AB., © v br ¹, 1, , t, , d, vb, , d, , JJJG, OC , which will be represented by diagonal, , JJJG, OD of rectangle OBDC., ?, , v rm, , v2r vm2 2v r v m cos 90q, , vr2 v 2m, , v 2br v 2r, , Since sin T ! 1 . So, if vr > vbr, the boatman can never, reach at point B. Because if vr = vbr, sinT = 1 or T = 90°, and it is just impossible to reach at B if T = 90°., Similarly, if vr > vbr, sinT > 1, i.e., no such angle exists., Practically it can be realized in this manner that it is, not possible to reach at B if river velocity (vr) is too, high., , 15. RELATIVE VELOCITY OF RAIN W.R.T, THE MOVING MAN, , G, If T is the angle which v rm makes with the vertical, , G, Consider a man walking west with velocity v m ,, JJJG, represented by OA . Let the rain be falling vertically, JJJG, G, downwards with velocity v r , represented by OB ., , direction then, , Figure. To find the relative velocity of rain with respect, G, to man ( i.e. v rm ) bring the man at rest by imposing a, G, velocity v m on man and apply this velocity on rain, , Here, angle T is from vertical towards west and is, written as T, west of vertical., , also. Now the relative velocity of rain with respect to, JJJG, G, man will be the resultant velocity of v r OB and, , tan T, , BD, OB, , vm, or T, vr, , §v ·, tan 1 ¨ m ¸, © vr ¹, , Note : In the above problem if the man wants to protect, himself from the rain, he should hold his umbrella in, the direction of relative velocity of rain with respect, to man i.e. the umbrella should be held making an angle, T (= tan–1 vm/vr) west of vertical.
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