VIEWS: 10 PAGES: 152 CATEGORY: High School POSTED ON: 9/14/2012
AIEEE IIT-JEE CBSE STUDY MATERIAL PHYSICS SAMPLE PAPERS TEST PAPER KEY SOLUTIONS ANSWERS QUESTIONS
collision 1 A process of strong and rapid interaction between two bodies which brings about a tremendous (quick) change in momenta of the interacting bodies is called collision. During collision, the colliding bodies may or may not touch physically. 2 2 For instance, in the collision between ball and bat physical contact takes place. On the other hand, in Rutherford’s gold foil experiment, the α-particles do not touch the gold nuclei. Explosion can also be called collision because it is a short timed interaction. 3 Collision •In a collision a relatively large force acts on each colliding particle for a relatively short time. •The basic idea of a ‘collision’ is that the motion of the colliding particles (or of at least one of them) changes rather abruptly . •We can make a relatively clean separation of times that are ‘before the collision’ and those that are ‘after the collision’. 4 4 Types Of Collision Collisions are of 3 types :- 1. ELASTIC (ALSO CALLED PERFECTLY ELASTIC) COLLISION Symptom: The colliding bodies separate out after collision. Properties (i) Total momentum of the system is conserved. (ii) Total K.E of the system is conserved. 5 5 2. PERFECTLY INELASTIC COLLISION) Symptom: Colliding bodies get embedded to each other and become one body after collision. In this case (i) Total momentum remains conserved, as in the case of Perfectly Elastic Collisions. (ii) Total K.E. of the system generally decreases. Sometimes, total K.E. of the gross body may increase also, if some internal energy gets released during collision. 6 6 3. INELASTIC COLLISION It is a case of lying between perfectly elastic and perfectly inelastic collisions. 7 7 8 GENERAL SOLUTON FOR HEAD- ON COLLISION BETWEEN TWO BODIES 9 Head On Collision If the initial and final velocities of colliding masses lie along the same line (along the common normal), then it is known as head-on collision. m1 m2 m1 m2 u1 u2 v1 v2 10 10 Head on Collision Let the smooth small balls of masses m1 and m2 moving in positive x-direction with speeds u1 and u2 (u1 > u2) collide. u1 u2 m1 m2 before collision 11 11 m1 m2 V During collision J J m1 m2 12 12 Head on Collision m1 m2 V v1 v2 m1 m2 After collision 13 13 Just after the collision let they move in positive x-direction with speeds v1 and v2 respectively. v1 v2 m1 m2 Just after collision 14 In case of collision linear momentum of system of particles is remains unchanged. Since no net impulse acts on the system (m1 + m2), its momentum remains constant Or Pf = Pi m1v1 + m2v2 = m1u1 + m2 u2 ...( i ) 15 15 Problem 33. Consider a head- on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 in the same direction. The collision starts at t = 0 and the particles interact for a time interval ∆t. During the collision, the speed of the first particle varies as t υ(t ) = u1 + (υ1 − u1 ) ∆t Find the speed of the second particle as a function of time during the collision. 16 16 Solution. m1 m2 Before collision : u1 u2 At t = 0 collision occure the particles interact for a time ∆t. m1 m2 v1(t) v2(t) As linear momentum of the system of particles remain constant. m1u1 + m2 u2 = m1v1 (t ) + m2v2 (t ) 17 ⎡ t ⎤ m1u1 + m2u2 = m1 ⎢ u1 + (ν1 − u1 )⎥ + m2v2 (t ) ⎣ ∆t ⎦ t m2v2 (t ) = m2u2 − m1 (v1 − u1 ) ∆t m1t v2 (t ) = u2 − (v1 − u1 ) m2 ∆t m1 m2 Before collision : u1 u2 m1 m2 During collision : v1(t) v2(t) 18 18 Problem 34. A bullet of mass m moving at a speed v hits a ball of mass M kept at rest. A small part having mass m’ breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed v1 in the direction of the bullet. Find the velocity of the bullet after the collision. 19 19 Solution. m v m M Just Before collision : bullet ball v’ v1 Just After collision : (m + m’) (m - m ’) (M-m) bullet ba ll Linear momentum in x-direction ‘bullet + ball’ system will be conserved. mv + 0 = (m + m ') v '+ (M − m ') v1 mv − (M − m ') v1 v' = (m + m ') 20 NEWTON’S EXPERIMENT LAW Newton’s Empirical Formula for Collision 21 NEWTON’S EXPERIMENT LAW When two bodies collide directly, the relative velocity after collision bears a constant ratio to their relative velocity before collision and is opposite in direction. 22 22 If two bodies collide their relative velocity along their common normal after collision bears a constant ratio to their relative velocity before collision. This constant ratio is called ‘e’ = coefficient of restitution. –(v1 – v2 ) Co-efficient of restitution,e = ; (u1 – u2 ) 23 23 where, v1 = velocity of particle (1) after collision v2 = velocity of particle (2) after collision u1 = velocity of particle (1) before collision u2 = velocity of particle (2) before collision (–) Relative velocity after collision e= Relative velocity before collision 24 24 [0 ≤ e ≤ 1] i.e., the quantity ε lies between 0 and 1. where e = 0, the collision is said to be perfectly inelastic (because in case of perfectly inelastic collision v1 = v2). e = 1, the collision is said to be perfectly elastic. 25 Velocity Of Approach Since u1 > u2, m1 approaches m2 with a velocity u12 = vapp = u1 – u2, just before the collision. 1 2 u1 - u2 m1 approaches m2 with vapp = u1 - u2 26 Velocity Of Separation Just after the collision m1 separates from m2 with a velocity V12 = vsep = v1 – v2 1 2 v1 − v 2 m1 separates from m2 with vsep = v1 - v2 27 → → Since v sep and vapp are oppositely directed and is generally less than vapp, we can write - e vapp = vsep 28 Where e is called the coefficient of restitution which tells us what fraction (coefficient) of the relative velocity between the colliding bodies (vapp) is restituted (restored). or e ( u1 − u2 ) = v2 − v1 ...( ii ) 29 General Equations for Direct Impact If u1, u2 are velocities before of the masses m1, m2, and v1,v2 are the velocities after impact, then we have m1u1 + m2u2 = m1v1 + m2v2 m1 m2 and v1 - v2 = -e(u1 - u2) u1 u2 Combining these equations, we get, before impact m1 − em2 (1 + e)m2 v1 = u1 + u2 m1 m2 m1 + m2 m1 + m2 (1 + e)m1 m2 − em1 v1 v2 v2 = u1 + u2 m2 + m1 m2 + m1 after impact 30 (b) For oblique impact: When two bodies collide obliquely, their relative velocity resolved along their common normal after impact is in a constant ratio to their relative velocity before impact (resolved along common normal) and is in the opposite direction. Common normal φ α β θ v2 u1 u2 v1 31 v1 cos θ − v2 cos φ = −e u1 cos α − u2 cos β v1 cos θ – v2 cos φ = – e (u1 cos α – u2 cos β) 32 Common Tangent (CT) & Common Normal (CN) ) CT on CT m t( c om en ng ta CN These CN & CT direction co rm have no nothing to do m N with the directions of m o on N velocity of the two al bodies. (C ) 33 Oblique collision in case of smooth surfaces (a) Common Normal (CN) Impule is exerted in CN–direction only. Both bodies exert equal and opposite forces (action and reaction) on each other. Hence, momentum and velocities change accordingly in CN–direction. Momentum changes in CN-direction only Apply ‘e’ (coefficient of restitution) in CN–direction only. 34 (b) Common Tangent (CT) No impulse in Common Tangen tdirection (in case of smooth surfaces) Neither momentum nor velocities change in CT- direction. 35 If a ball strikes with a velocity u1 with a velocity wall which itself approaching it with a velocity u2 u1 |Relation velocity of separation| = |Relative velocity of approach| v1 v1 – v2 = –e(u1 – u2 ) v1 – (–u2 ) = –e[u1 – (–u2 )] u2 v1 = –e(u1 + u2 ) – u2 v 2 = u2 36 In case of perfectly elastic collision, u1 v1 = –e(u1 + u2 ) – u2 v1 = –u1 – u2 – u2 = –(u1 + 2u2 ) v1 v1 = –(u1 + 2u2 ) u2 v 2 = u2 37 Illustration 32. A ball moving with a velocity v strikes a wall moving toward the wall with a velocity u. An elastic impact occurs. Determine the velocity of the ball after the impact. What is the cause of the change in the kinetic energy of the ball ? Consider the mass of the wall to be infinitely great. v u 38 Solution. (v2 – v1) Co-efficient of restitution,e = ; (u1 – u2 ) ((− u) – v1) e =1 = ; (v – (−u)) v + u = −(u + v1 ) v1 = −(v + u) − u = −(v + 2u) 39 1 2 Kinetic energy before impact = mv 2 1 Kinetic energy after impact = m (v + 2u)2 2 The change in kinetic energy is equal to 2mu(u + v). 40 Now we calculate the work of reaction forces acting on the ball during the impact . Let the collision continue for t seconds. Assume the reaction force to be constant (the result does not depend on this assumption). Since the impact changes the momentum by 2m(v + u), the force of reaction is 2m (v + u) F = t 41 The work of this force is 2m (v + u)ut W = Fs = F (ut ) = = 2m (v + u) u t So we can see that this work is equal to change in kinetic energy. 42 Problem. Two particle of masses m1 = 1 kg and m2 = 10 kg are moving in a straight line as shown in the figure. 10 kg 10 m/s 5 m/s 1 kg If particle collide elastically find the velocity of the particles after collision. 43 43 Solution. Linear momentum of colliding particle will be conserved. v1 v2 m1 m2 m1u1 + m2u2 = mv1 + mv 2 1× 10 + 10 × 5 = 1× v1 + 10v 2 60 = v1 + 10v 2 …(i) 44 44 Now applying equation of coefficient of restitution. ⎛ v 2 − v1 ⎞ v 2 − v1 e=⎜ ⇒ 1= ⎝ u1 − u2 ⎟⎠ u1 − u2 v 2 − v1 = u1 − u2 = 10 − 5 v 2 − v1 = 5 …(ii) Solving equation (i) and (ii) we can get the values of v1 and v2. 10 v1 = m/s 11 65 v2 = m/s 11 45 45 Problem. Two identical balls of mass m are moving on a horizontal surface with velocity u1 and u2 as shown in figure. The balls collide elastically find the velocities of the balls after collision. 46 46 Solution. * Both balls are moving on horizontal smooth surface with velocity u1 & u2 (given u1 > u2). * They collide elastically (e = 1) * Find the velocities of the balls after collision. 47 47 m m m1 m2 u1 u2 v1 v2 48 48 Conservation of linear moment mu1 + mu2 = mv1 + mv2 v1 + v2 = u1 + u2 Now using restitution equation v 2 − v1 e= =1 u1 − u2 v 2 − v1 = u1 − u2 Solving equation (i) and (ii) v1 = u2 v2 = u1 49 49 Inelastic Collision Let us consider a case of inelastic collision 9e > 1) in which a projectile of mass m1 collides with a stationary target of mass m2 with a velocity v1 as shown below . m1 m2 m1 m2 u1 Rest v1 v2 Before After A moving particle m1 hits a stationary particle m2 inelastically (e > 1). 50 Applying momentum conservation m1u1 = m1v1 + m2v 2 For inelastic collisions, e > 1, we have eu1 = v 2 − v1 After solving the above two equations, we get ⎡ m1 − em2 ⎤ v1 = ⎢ ⎥ u1 ⎣ m1 + m2 ⎦ ⎡ (1 + e )m1 ⎤ v2 = ⎢ ⎥ u1 ⎣ m1 + m2 ⎦ 51 Note :- • Velocity of the target is always more than that of the projectile for finite values of m1 and m2 and O < e < 1. * For completely inelastic collision (e = 0) both the particles move together. * When the target is infinitely massive, it remains stationary after the collision and the projectile rebounds with a speed eu1. 52 The loss in kinetic energy of the system is given by 1 ⎛ m1m2 ⎞ ∆K = ⎜ ⎟ (1 − e 2 ) u1 2 2 ⎝ m1 + m2 ⎠ 53 The fraction of kinetic energy loss is given by ∆K m2 (1 − e 2 ) = Ki m1 + m2 54 Note :- * for an elastic collision (e = 1), there is no kinetic energy loss. * for completely inelastic collision (e = 0), the kinetic energy loss is maximum and it is given by ⎛ ∆K ⎞ m2 ⎜ ⎟ = ⎝ K i ⎠max m1 + m2 * greater the mass of the target, larger the loss of kinetic energy. 55 Problem. A particle is dropped from a height h0 on a horizontal floor. The coefficient of restitution between the particle and the floor is e. (i) Determine the first height of rebound h1. h0 h1 56 56 Solution. (i) The ball strikes the floor with a speed v 0 = 2 gh0 And it rebounds from the ev0 floor with a speed v0 v1 = ev 0 = e 2 gh0 The approach velocity is Therefore, the first v0 and the separation height of rebound is velocity is ev0. v1 2 h1 = = e 2 h0 2g 57 57 Problem. A particle is dropped from a height h0 on a horizontal floor. The coefficient of restitution between the particle and the floor is e. (ii) Determine the height of the particle after nth rebound h0 h1 58 58 u=0 h e2 h e4 h e6 h u eu e2 u e3u 59 (ii) obviously, the particle’s velocity after nth rebound will be v n = e nv 0 Therefore, the height after nth rebound will be vn 2 hn = = e 2 n h0 2g 60 60 Problem. A particle is dropped from a height h0 on a horizontal floor. The coefficient of restitution between the particle and the floor is e. (iii) Determine the number of rebounds after which the rebound height h0 becomes half of the initial height h0. h1 61 61 vn 2 hn = = e 2 n h0 2g (iii) Taking log on both the sides, we get log hn = 2n log e + log h0 ⎛ h ⎞ log ⎜ n ⎟ n= ⎝ h0 ⎠ 2log e hn 1 Since = (given) h0 2 log(1/ 2) n= 2log e 62 62 Problem. A particle is dropped from a height h0 on a horizontal floor. The coefficient of restitution between the particle and the floor is e. (iv) Compute the total distance travelled by the particle before it stops bouncing. h0 h1 63 63 (iv) The total distance travelled by the particle is H = h0 + 2h1 + 2h2 + ..... H = h0 + 2e 2 h0 + 2e 4 h0 + ...... H = h0 [1 + 2e 2 (1 + e 2 + e 4 + .....)] ⎡ ⎛ 1 ⎞⎤ 1 H = h0 ⎢1 + 2e 2 ⎜ 2 ⎟ [∵ 1 + e 2 + e 4 = ....... = ⎝ 1− e ⎠⎥ ] ⎣ ⎦ 1− e 2 ⎡1 + e2 ⎤ H = h0 ⎢ 2⎥ ⎣1 − e ⎦ 64 64 Problem. A particle is dropped from a height h0 on a horizontal floor. The coefficient of restitution between the particle and the floor is e. (v) Compute the total time taken to stop bouncing. h0 h1 65 65 ⎡1 + e2 ⎤ H = h0 ⎢ 2⎥ ⎣1 − e ⎦ (v) The total time taken by the particle is T = t0 + 2t1 + 2t2 + ...... 2h0 2h1 2h2 T= +2 +2 + ..... g g g Since h1 = e2h0, h2 = e4h0 , therefore 2h0 T= ⎡1 + 2e + 2e 2 + .....⎤ g ⎣ ⎦ 2h0 T= ⎡1 + 2e(1 + e + e 2 + .....) ⎤ g ⎣ ⎦ ⎛ 1 + e ⎞ 2h0 ⎡ ∵ 1 = e + e 2 + ...... = 1 ⎤ T =⎜ ⎟ ⎢ ⎝ 1− e ⎠ g ⎣ 1− e ⎥ ⎦ 66 66 Problem. A particle of mass m moving with a speed v strikes a smooth horizontal surface at an angle α. The particle rebounds at an angle β with a speed v. Determine an expression for v and β if the coefficient of restitution is e. m m v u α β A ball strikes a smooth horizontal surface at an angle α and rebounds at an angle β. 67 67 Solution. Since no external impulse acts in the horizontal direction, therefore, momentum of the ball is conserved in the horizontal direction. That is, mu cos α = mv cos β or u cos α = v cos β …(i) Using the definition of the coefficient of restitution, we get e(u sin α) = v sin β …(ii) 68 68 Dividing (ii) by (i), we get tan β = e tan α or β = tan-1(e tan α)] On squaring and adding, we get v 2 = u 2 (cos2 α + e 2 sin2 α ) v = u cos2 α + e 2 sin2 α 69 69 Problem. Refer to fig. The pendulum on the left is pulled aside to the positions shown. It is then released and allowed to collide with the other pendulum, which h is at rest. (i) What is the speed of the ball on the left just before collision? 70 (ii) If collision is elastic calculate the maximum height reach by the right ball. (iii) How high, in terms of h, does the combination swing? Assume that the two balls have equal masses. h 71 Solution. Using conservation of mechanical energy 1 mu2 = mgh 2 1 u = 2 gh h u (i) If collision is elastic :- 1 2 The ball (1) will become at rest and ball (2) will start moving with velocity u and swing to the same height h same as ball (1). 72 (ii) If the balls stick together just after collision:- Using conservation of linear momentum just before and just after collision. mu + 0 = 2mv u ⇒ v= …(i) 2 For maximum height reached by h’ combined body using v conservation of mechanical energy. 73 1 2 mg h ' = (2m) v 2 2 2 ⎛u⎞ v '2 ⎜ 2 ⎟ ⎝ ⎠ =u 2 h' = = 2g 2g 8g ( 2 gh )2 h Hence, h ' = = 8g 4 h h' = 4 h’ v 74 Illustration 33. In figure, there are n identical spheres of mass m suspended with wires of equal length. The spheres are almost in contact with each other. The sphere 1 is pulled aside and released. If sphere 1 strikes sphere 2 with velocity v1,,find an expression for the velocity vn of the nth sphere immediately after being v1 vn struck by the one adjacent to ' ⎛1 + e ⎞ n −1 it. The coefficient of vn = ⎜ ⎟ v1 ⎝ 2 ⎠ restitution for all the impacts is e. 75 76 Solution. From the definition of coefficient of restitution, we have, ' ' v2 − v1 e= v1 − 0 or, ' ' ev1 = v2 − v1 … (i) From law of conservation of momentum, we have ' ' mv1 = mv2 + mv1 ' ' v1 = v2 + v1 77 ' ' or, v1 = v2 + v1 ' ' v1 = v1 − v2 … (ii) From equations (i) and (ii), on eliminating v’1, we get ' ⎛ e + 1⎞ … (iii) v2 = ⎜ ⎟ v1 ⎝ 2 ⎠ 78 Similarly for second and third balls, ' '' v 3 − v2 e= ' v2 − 0 ' ' '' ev2 = v 3 − v2 … (iv) From conservation of momentum, we get, ' ' '' mv2 = mv 3 + mv2 '' ' ' v2 = v2 − v 3 … (v) 79 From equations (iv) and (v), we can solve v’3. ' (e + 1)v2 ' v = 3 … (vi) 2 On substituting expression for v’2 from equation (iii) in equation (vi), we get, 2 ⎡ e + 1⎤ … (viii) ' v3 = ⎢ ⎥ v1 ⎣ 2 ⎦ In a similar way, we can generalize the result, n −1 ' ⎛1 + e ⎞ vn = ⎜ ⎟ v1 ⎝ 2 ⎠ 80 Problem. A heavy ball of mass 2M moving with a velocity v0 collides elastically head-on with a cradle of three identical balls each of mass M as shown in figure. Determine the velocity of each ball after collision. 2M v0 M M M A heavy ball collides a cradle of three identical balls. 81 81 Solution. Conserving linear momentum 2Mu0 = 2Mv1 + Mv2 2u0 = 2v1 + v2 v 2 − v1 v 2 − v1 Now using restitution equation e= = u1 − u2 v 0 − 0 =1 v 2 − v1 = v 0 Heavy ball 1, 2M v0 v0 M M M ⎛ 2M − M ⎞ v1 = ⎜ ⎟v = ⎝ 2M + M ⎠ 0 3 1 2 3 4 ⎛ 2(2M ) ⎞ 4 ball 2. v2 = ⎜ ⎟ v0 = v0 ⎝ 2M + M ⎠ 3 82 82 As the ball 2 gains momentum, immediately it transfers it to the ball 3, which in turn transfer to the ball 4. Thus, the ball 4 comes out with a velocity v4 = 4/3 v0 leaving the balls 1 and 2 stationary. 1 u0/3 2 4u0/3 3 4 83 83 Now, the heavy ball collides the ball 2 with a velocity u0/3 and their final velocities are: v0 4 v1 ' = v2 = v0 ' 9 9 The ball 2 transfer, its momentum to the ball 3, which starts moving with a velocity v2 = 4/9 u0, leaving the ball 2 stationary. 1 4v0/9 4v0/3 v0/9 2 3 4 84 84 Now, the heavy ball collides with the ball 2 with a velocity u0/9 and their final velocities are: u0 4 v'= ; v1 = ' u 27 27 0 2M u0/27 4u0/27 4u0/9 4u0/3 M M M 1 2 3 4 Final 85 85 ⎛u ⎞ ⎛ 4u ⎞ ⎛ 4u ⎞ pi = 2Mu0 Pf = 2M ⎜ 0 ⎟ + M ⎜ 0 ⎟ + M ⎜ 0 ⎟ ⎝ 27 ⎠ ⎝ 9 ⎠ ⎝ 3 ⎠ ⎡ 1 2 2 2⎤ Pf = 2Mu0 ⎢ + + + ⎥ ⎣ 27 27 9 3 ⎦ Pf = 2Mu0 Note: that the collision does not occur actually in the sequence described in the problem. In fact all the balls attain their final velocities simultaneously. 86 86 Problem 56. Two balls having masses m and 2m are fastened to two light strings of some length l. The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls rise after the collision ? (b) How high will the balls rise after the collision ? 1 0 1 m 2m 87 87 Solution. Applying m 2m conservation of O L linear momentum 1 2 just before and O just after collision mu – 2mu = mv1 + 2mv2 v1 + 2v2 = -u … (i) u = 2g u = 2g v1 v2 88 88 Applying restitution equation v2 – v1 = e (u1 - u2) (v2 – v1) = 1 (u - (-u)) v2 – v1 = 2u … (ii) Solving (i) & (ii), we get, 50 g v1 = toward left 3 2g v2 = toward right 3 89 Now applying conservation of mechanical energy for ball 1 & 2 2 1 ⎛ 50 g ⎞ For ball (1) : m⎜ ⎟ = mgh 2 ⎝ 3 ⎠ 1 50 g mgh = m 2 9 25 h= 9 90 But the ball can not go beyond 2 hence maximum height reached by ball 1 is 2 . 2 1 ⎛ 2g ⎞ ⇒ h' = m⎜ ⎟ = mgh ' 2 ⎝ 3 ⎠ 3 91 91 Particle & Wedge Collision 92 Problem. Determine the velocity of approach in the following situations. Rest m u1 u1 M u2 m M θ θ 93 Solution. Velocity of approach VOA = (u1 − u2 )n VOA = u1 sin θ − 0 VOA = u1 sin θ t n + Common tangent m u1 direction 2 θ 1 Common normal θ M direction 94 m 1 t u1 + 2 θ M θ u2 VOA = (u1 − u2 )n = u1 cos θ − (− u2 sin θ) VOA = u1 cos θ + u2 sin θ 95 Problem. Determine the velocity of separation in the following situations. m θ v1 α β m v M V θ v2 96 Solution. VOS = (v 2 − v1 )n n v1 = [(v 2 ) − (−v1 cos θ)] θ 1 VOS = v 2 + v1 cos θ + 2 v2 97 Solution. VOS = (v 2 sin θ) − (−v1 cos β ) VOS = v 2 sin θ + v1 cos β m n α + β m v1 M V2 θ 98 Illustration 28. A smooth ball is dropped from a height h on a smooth incline, as shown in figure. After collision the velocity of the ball is directed horizontally (a) Find the coefficient of restitution. (b) If the collision is elastic, what is the impulse on the ball? h v’ θ 99 Solution. (a) Normal axis and tangential axis are shown in the figure. Reaction of the incline is along n-axis and in the absence of friction there is no force along t-axis; therefore velocity along t-axis remains unchanged, n-axis υ cos θ = u sin θ …(i) j v sin θ v θ v cos θ i u cos θ u u sin θ θ t- ax (a) is 100 From the definition of coefficient of restitution, υ sin θ e=− …(ii) − u cos θ υ = eu cot θ …(iii) From equations (i) and (iii), (eu cot θ)cos θ = u sin θ e = tan θ 2 101 (b) When the collision is elastic, the component of velocity along n-axis is reversed in direction. Therefore the change in velocity n-axis ∆υ = 2υ cos θ θ There is no change in velocity θ v along the t-axis, therefore no impulse along t-axis. u θ = m ∆υ (b) = 2m υ cos θ 102 Velocity of the ball when it strikes the plane = 2gh Thus impulse = 2m 2gh cos θ 103 Problem. A ball of mass ‘m’ moving horizontally which velocity ‘u’ hits a wedge of mass ‘M’. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in v horizontal plane. Calculate (i) The velocity (v) at which the ball moves V u in vertical direction. (ii) The velocity of m θ wedge V. 104 Problem. A ball of mass ‘m’ moving horizontally which velocity ‘u’ hits a wedge of mass ‘M’. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in v horizontal plane. Calculate (iii) The impulse imparted by the ball on the wedge. V u m (iv) The impulse imparted θ by ground on the wedge. 105 Problem.31 We can solve this problem by one of two methods * Using conservation of linear momentum * Impulse approach. 1 Using conservation of θ V 2 vn linear momentum As no external force are M acting on the system in θ horizontal direction the t linear momentum should be constant conserved in horizontal direction. 106 Solution. For ball : J sin θ mu − J sin θ = 0 …(i) J cos θ J sin θ = mu …(ii) J cos θ = mv M J sin θ θ From (i) and (ii) J’ v = u cot θ 107 From (i) J = mu cos ec θ J sin θ For wedge: J ' = J cos θ J cos θ J ' = mu cot θ J sin θ M J sin θ = MV ⇒ V= J sin θ M θ mu V= M J’ 108 Problem. A ball of mass ‘m’ moving horizontally which velocity ‘u’ hits a wedge of mass ‘M’. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in v horizontal plane. Calculate (v) The coefficient of restitution e = ? u V (vi) The heat developed m θ during collision . 109 (v) The coefficient of restitution is given as (v 2 − v1 )n = e (u1 − u2 )n v n [V sin θ − (−v cos θ)] = e [ u sin θ − 0] V sin θ + v cos θ e= 2 θ u sin θ V u 1 V v m e = + cot θ = + (cot θ) cot θ u u M θ m e= + cot 2 θ M + 110 (vi) The heat developed 1 ⎛1 1 ⎞ H = mu2 − ⎜ mv 2 + MV 2 ⎟ 2 ⎝2 2 ⎠ 1 ⎡1 1 ⎛ mu ⎞ ⎤ 2 H = mu2 − ⎢ m (u cot θ)2 + M ⎜ ⎟ ⎥ 2 ⎢2 ⎣ 2 ⎝ M ⎠ ⎥ ⎦ 1 ⎡ m⎤ H = mu2 ⎢1 − cot 2 θ − ⎥ 2 ⎣ M⎦ 111 A Wedge of mass M Is kept at rest on sooth surface , a particle of mass m hits the wedge normally .Find the velocity of wedge and particle just after collision. Take coefficient of restitution as e. v0 vn V 112 mv 0 sin θ = MV + mv n sin θ …(i) Using Newton’s law of collision. (v 2 − v1 )n = e (u1 − u2 ) (v sin θ − v n ) = e (v 0 − 0) v0 …(ii) vn Solving (i) and (ii) we get V (1 + e ) mv 0 sin θ V= M + m sin θ 2 113 Using impulse method : The impulse acting on the ball and wedge can be represented as v v0 J vn J vn V J sin θ V θ 114 J v V J sin θ vn J θ J sin θ = MV …(i) mv − J = mv n …(ii) and Newton’s law of restitution (v sin θ − v n ) = e (v 0 − 0) …(iii) 115 Solving (i), (ii) and (iii) we get (1 + e ) mv 0 sin θ V= M + m sin θ 2 116 Collision of two Discs 117 Oblique Collision R2 U2 O2 d O1 U1 R1 t n O2 d θ sin θ = R1 + R2 d θ O1 A 118 At the time of collision let the velocity of Disc ‘1’ and Disc ‘2’ just after collision in the direction of normal and tangent are (v1,n, v1, t) and (v2,n, v2, t) respectively then t n O2 V2,n θ d θ O1 A v1,n v2t v1t 119 The velocity of Discs will be unchanged in the direction of tangent v1,t = u1 sin θ and v 2,t = u2 sin θ …(i) t n O2 V2,n θ d θ O1 A v1,n v2t v1t 120 Conservation of linear momentum is applicable in any direction as there is no external impulse on the system. We have already calculated the velocity in tangent direction. We need to calculate velocity in normal direction. Hence conserving linear momentum in normal direction. 121 …(ii) m1u1 cos θ + m2u2 cos θ = m1v1,n + m2v 2,n Now using Newton law of restitution we get (v 2,t − v1,n ) = e (u1 cos θ − u2 cos θ) …(iii) t n O2 V2,n θ d θ O1 A v1,n v2t v1t 122 From equation (ii) and (iii) we can calculate the value of v1,n and v2,n. t n O2 V2,n θ d θ O1 A v1,n v2t v1t 123 Illustration 35. In figure ball 1 with an initial speed of 10 m/s collides elastically with the stationary balls 2 and 3, whose centres are on a line perpendicular to the initial velocity of ball 1 and that are initially in contact with each other. The three balls are identical. Ball 1 is aimed directly towards contact point of 2 and 3. Assume all motion is frictionless. Determine the velocity of each ball after the collision. 2 v0 x 1 3 124 Solution. In the absence of friction each impulse is directed along the line joining centres of the colliding balls, normal to the colliding surfaces. After collision the target balls leave the collision along normals, while the incident ball leaves along x-axis. Let v0 be the velocity of the incident ball before the collision, V be its velocity afterwards and v be the speed of each of the target balls after the collision. Each ball has mass m. 125 From conservation of momentum along x-axis, we have, mv 0 = mV + 2mv cos θ 2 v R Sin θ = R/2R =1/2 v 1 R R x θ θ R R v 126 3 From conservation of momentum along x-axis, we have, mv 0 = mV + 2mv cos θ … (i) = −2.0 m/s Thus the incident ball rebounds in the negative x-direction. 127 Aditional Problems on Particle & Wedge Collision 128 Illustration 37. A right angled wedge ABC of mass M = 4 kg and base angle a = 53o is resting over a smooth horizontal plane. A shell of mass m = 0.5 kg moving horizontal with velocity v0 = 40 m/s collides with the wedge, just above point A. As a consequence, wedge starts to move towards left with velocity v = 5 m/s. Calculate C (a) maximum height reached by the shell. (b) heat generated during v0 collision, B 53° A 129 Illustration 31. Figure shows a smooth spherical ball of mass m striking two identical equilateral triangular m wedges of mass M. At the instant of impact velocity of the ball is v0. v0 Taking coefficient of M M restitution e, determine the velocities of the 60° 60° sphere and the wedges just after collision. 130 Solution. Let J be impulse between ball and wedges and v1 and v2 be the velocities of the ball and the wedge. From impulse-momentum equation on the ball, 2J sin 30° = mv1 – (-mv0) v1 J = mv1 + mv0 … (i) For the wedge, J J cos 30° = Mv2 … (ii) J v2 v2 M J 60° 60° 131 On eliminating J from equations (i) and (ii), we have, 2 Mv2 = mv1 + mv 0 … (iii) 3 From the definition of coefficient of restitution, v1 cos 60° + v2 cos 30° v1 + 3v2 e= = v 0 cos 60° v0 or, ev 0 = v1 + 3v2 … (iv) 132 On solving equations (iii) and (iv) for v1 and v2, we get, (2eM − 3m) v 0 v1 = 2M + 3m 3 (1 + e)mv 0 v2 = 2M + 3m 133 Illustration 36. A small particle of mass m is released from a height h on a large smooth sphere kept on a perfectly smooth surface, as h shown in figure. Collision between particle and sphere is perfectly inelastic. Determine O the velocities of particle and R/2 sphere after collision. 2 2 6m2 gh gh v = v n + vt = 2 + 4 (4M + m) 2 134 Solution. Collision takes place between smooth bodies; therefore tangential velocity of particle is unchanged. v t = u sin θ R /2 1 Where sin θ = = and u = 2 gh R 2 or, θ = 30° y θ u n-axis n si si n vn u x θ u R vn V θ θ vt R/2 t-axis vt 135 u vt = … (i) 2 From conservation of momentum along x-axis, Pi = Pf 0 + 0 = − MV + mv n sin θ + mv t cos θ … (ii) y θ u n-axis n si si n vn u x θ u R vn V θ θ vt R/2 t-axis vt 136 From the definition of coefficient of restitution, V sin θ − (−v n ) e= =0 u cos θ − 0 [e = 0, for a perfectly inelastic collision] or, V sin θ = −v n V vn = − … (iii) 2 137 On substituting expressions for vn and vt in equation (ii), we obtain ⎛ V ⎞1 ⎛u⎞ 3 MV = m ⎜ − ⎟ + m ⎜ ⎟ ⎝ 2 ⎠2 ⎝2⎠ 2 ⎛ m⎞ u 3 where u = 2 gh ⎜ M+ ⎟ V =m ⎝ 4⎠ 4 or, m 6 gh V = (4M + m) 2 2 6m2 gh gh v = v n + vt = 2 + 4 (4M + m) 2 138 Illustration 34. A small particle of mass m, moving with velocity u horizontally, strikes a triangular wedge of mass M. After collision the particle starts moving along the incline. Determine the velocity of the wedge immediately after collision and the maximum height to which it can ascend on the wedge. Assume that friction is negligible at all the surfaces. M m θ 139 u2 M 2 cos2 α h= 2 g (M + m)(M + m cos2 α) mu sin2 α Mu cos α V = 2 and v rel = M + m sin α M + m sin2 α 140 Solution. Let vrel be the velocity of the particle relative to the wedge and V the velocity of the wedge after collision. After collision x- and y-components of resultant velocity of particle are v x = v rel cos θ + V v y = v rel sin θ Vrel sin θ vrel N vrel v N θ α Vrel cos θ V 141 Momentum of system is conserved along x-direction only. mu = m (v rel cos θ + V ) + MV … (i) At the instant of collision the force between the particles is normal to the surface, hence the velocity of the particle along the wedge remains constant. u cos α = v rel + V cos α … (ii) 142 On solving equations (i) and (ii), we get, mu sin2 α Mu cos α V = 2 and v rel = 2 … (iii) M + m sin α M + m sin α The particle and wedge moves on a single body till the block reaches maximum height. At this position the particle has only x-component of velocity, let it be u. 143 From law of conservation of energy, we have, mu mu = (m + M)v ' or, v ' = … (iv) (m + M) From law of conservation of energy, we have, 1 1 1 … (v) 2 MV + m 2 (v rel + V 2 ) = (M + m)v '2 2 2 2 On substituting V, vrel and v’ in equation (v) and solving for h, we get, u2 M 2 cos2 α h= 2 g (M + m)(M + m cos2 α) 144 Problem. A wedge of mass M = 3.6 kg and having base angle q= 370 is resting over a smooth horizontal surface. A ball of mass m = 1 kg is thrown vertically downwards such that it strikes the wedge with velocity v0 = 11 ms-1 at height h = 48 cm from the base of the wedge as shown in figure. m v0 M h = 48cm θ 145 Coefficient of restitution between ball and wedge is e = 0.5. Assuming all the surfaces to be smooth. calculate 146 (i) Velocity of wedge just after collision. (ii) Vertical component of velocity of ball just after collision. (iii) Time of flight of ball from the instant of collision with wedge to the instant when ball strikes the floor. (iv) Distance between ball and right edge of the wedge when ball strikes the floor. m v0 M h = 48cm θ 147 Solution. When the ball with strike the wedge the impulsive reaction will develop normal to the inclined n surface of the wedge. The horizontal component of this θ impulse will give motion to the V J wedge in horizontal direction. Which vertical component of θ the impulsive reaction will be J’ t supported by impulse given by ground in vertical direction, as shown in the diagram given below, 148 vy n θ y vx θ x J vS If we consider (wedge + ball) a system there exits a external impulse in vertical direction. Hence we can conserve the momentum of the system only in horizontal direction as there will be no component of external impulse (from ground) in the direction. The momentum of the ball can be conserved in the direction of common tangent as there will be no impulse on the ball in the common tangent direction. 149 Conserving momentum of the system in horizontal direction. ˆ ˆ = − MV i + mv x i …(i) Where v = velocity of the wedge just after collision Vx = horizontal component of velocity of ball just after collision. For the ball velocity of it should be same as tangent direction just before and just after collision. v 0 sin θ = v x cos θ − v y sin θ …(ii) 150 Now we are left with three unknowns v, vx and vy hence we need another equation which can be achieved by Newton’s restitution equation. (v 2 − v1 )n v sin θ + (v x sin θ + v y cos θ) e= = (u1 − u2 )n v 0 cos θ From equations (i), (ii) and (iii) we get V =2ms-1, vx = 7.2 ms-1 and vy = 1.4 ms-1 (downward) Let the time when ball strikes the ground be T 1 h = v tT = gT 2 2 151 Substituting the values we get T = 0.2 sec Distance between ball and right edge of wedge when ball strikes the ground l = R – x + s V = vx, T – h/tan θ = V.T = 1.20 m V h θ x x R S l 152