VIEWS: 3 PAGES: 151 CATEGORY: High School POSTED ON: 9/14/2012
AIEEE IIT-JEE CBSE STUDY MATERIAL PHYSICS SAMPLE PAPERS TEST PAPER KEY SOLUTIONS ANSWERS QUESTIONS
collision 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 For instance, in the collision between ball and bat physical contact takes place. On the other hand, in Rutherford’s gold foil experiment, the -particles do not touch the gold nuclei. Explosion can also be called collision because it is a short timed interaction. 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 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 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 3. INELASTIC COLLISION It is a case of lying between perfectly elastic and perfectly inelastic collisions. 7 GENERAL SOLUTON FOR HEAD- ON COLLISION BETWEEN TWO BODIES 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 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 m1 m2 V During collision J J m1 m2 12 Head on Collision m1 m2 V v1 v2 m1 m2 After collision 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 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 m2u2 ...(i ) 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 (t ) u1 (1 u1) t Find the speed of the second particle as a function of time during the collision. 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 m2u2 m1v1(t ) m2v2 (t ) t m1u1 m2u2 m1 u1 (1 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 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 Solution. m v m M Just Before collision : bullet ball v’ v1 Just After collision : (m + m’) (m - m ’) (M-m) bullet ball 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 ') NEWTON’S EXPERIMENT LAW Newton’s Empirical Formula for Collision 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 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 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 [0 e 1] i.e., the quantity 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. 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 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 Since vsep and vapp are oppositely directed and is generally less than vapp, we can write - e vapp = vsep 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 ) (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 f b q v2 u1 u2 v1 v1 cos q v2 cos f e u1 cos u2 cos b v1 cos q – v2 cos f = – e (u1 cos – u2 cos b) Common Tangent (CT) & Common Normal (CN) CT CN These CN & CT direction have no nothing to do with the directions of velocity of the two bodies. 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. (b) Common Tangent (CT) No impulse in Common Tangen tdirection (in case of smooth surfaces) Neither momentum nor velocities change in CT- direction. 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 v2 = u2 In case of perfectly elastic collision, u1 v1 –e(u1 u2 ) – u2 v1 –u1 – u2 – u2 –(u1 2u2 ) v1 v1 –(u1 2u2 ) u2 v2 = u2 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 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) 1 Kinetic energy before impact mv 2 2 1 Kinetic energy after impact m (v 2u)2 2 The change in kinetic energy is equal to 2mu(u + v). 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 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. 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 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 v 1 10v 2 …(i) 44 Now applying equation of coefficient of restitution. v v1 v 2 v1 e 2 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 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 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 m m m1 m2 u1 u2 v1 v2 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 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). 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 m em2 v1 1 u1 m1 m2 (1 e )m1 v2 u1 m1 m2 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. The loss in kinetic energy of the system is given by 1 m1m2 K (1 e 2 ) u1 2 2 m1 m2 The fraction of kinetic energy loss is given by K m2 (1 e 2 ) Ki m1 m2 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. 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 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 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 u=0 h e2 h e4 h e6 h u eu e2 u e3 u (ii) obviously, the particle’s velocity after nth rebound will be v n e nv0 Therefore, the height after nth rebound will be vn 2 hn e 2 n h0 2g 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 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 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 (iv) The total distance travelled by the particle is H h0 2h1 2h2 ..... H h0 2e2 h0 2e 4 h0 ...... H h0 [1 2e2 (1 e2 e 4 .....)] 1 1 H h0 1 2e 2 2 [ 1 e2 e 4 ....... 1 e ] 1 e 2 1 e2 H h0 1 e2 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 1 e2 H h0 1 e2 (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 Problem. A particle of mass m moving with a speed v strikes a smooth horizontal surface at an angle . The particle rebounds at an angle b with a speed v. Determine an expression for v and b if the coefficient of restitution is e. m m v u b A ball strikes a smooth horizontal surface at an angle and rebounds at an angle b. 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 = mv cos b or u cos = v cos b …(i) Using the definition of the coefficient of restitution, we get e(u sin ) = v sin b …(ii) 68 Dividing (ii) by (i), we get tan b = e tan or b = tan-1(e tan )] On squaring and adding, we get v 2 u2 (cos2 e2 sin2 ) v u cos2 e 2 sin2 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? (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 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). (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 combined body using h’ v conservation of mechanical energy. 1 2 mg h ' (2m) v 2 2 2 u v '2 2 u2 h' 2g 2g 8g Hence, h ' ( 2 gh )2 h 8g 4 h h' 4 h’ v 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 vn v1 it. The coefficient of 2 restitution for all the impacts is e. 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 ' ' 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 Similarly for second and third balls, ' '' v 3 v2 e ' v2 0 ' ' '' ev2 v3 v2 … (iv) From conservation of momentum, we get, ' ' '' mv2 mv3 mv2 '' ' ' v2 v2 v 3 … (v) From equations (iv) and (v), we can solve v’3. ' (e 1)v2 ' v3 … (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 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 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 v M M M 2M M v1 v0 0 2M M 3 1 2 3 4 2(2M ) 4 ball 2. v2 v0 v0 2M M 3 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 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 Now, the heavy ball collides with the ball 2 with a velocity u0/9 and their final velocities are: u0 4 v' ; v1 ' u0 27 27 2M u0/27 4u0/27 4u0/9 4u0/3 M M M 1 2 3 4 Final 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 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 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 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 Now applying conservation of mechanical energy for ball 1 & 2 2 1 50 g For ball (1) : m mgh 2 3 1 50g mgh m 2 9 25 h 9 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 Particle & Wedge Collision Problem. Determine the velocity of approach in the following situations. Rest m u1 u1 M u2 m M q q Solution. Velocity of approach VOA (u1 u2 )n VOA u1 sin q 0 VOA u1 sin q t n + Common tangent m u1 direction 2 q 1 Common normal q M direction m 1 t u1 + 2 q M q u2 VOA (u1 u2 )n u1 cos q ( u2 sin q) VOA u1 cos q u2 sin q Problem. Determine the velocity of separation in the following situations. m q v1 b m v M V q v2 Solution. VOS (v 2 v 1 )n n v1 [(v 2 ) ( v 1 cos q)] q 1 VOS v 2 v 1 cos q + 2 v2 Solution. VOS (v 2 sin q) ( v 1 cos b ) VOS v 2 sin q v1 cos b m n + b m v1 M V2 q 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’ q 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 cos q u sin q …(i) j v sin q v q v cos q i u cos q u u sin q q (a) From the definition of coefficient of restitution, sin q e …(ii) u cos q eu cot q …(iii) From equations (i) and (iii), (eu cot q)cos q u sin q e tan q 2 (b) When the collision is elastic, the component of velocity along n-axis is reversed in direction. Therefore the change in velocity n-axis 2 cos q q There is no change in velocity q v along the t-axis, therefore no impulse along t-axis. u q m (b) 2m cos q Velocity of the ball when it strikes the plane 2gh Thus impulse 2m 2gh cos q 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 in vertical direction. V u (ii) The velocity of m q wedge V. 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 q by ground on the wedge. Problem.31 We can solve this problem by one of two methods * Using conservation of linear momentum * Impulse approach. 1 Using conservation of q 2 vn linear momentum V As no external force are M acting on the system in q horizontal direction the t linear momentum should be constant conserved in horizontal direction. Solution. For ball : J sin q mu J sin q 0 …(i) J cos q J sin q mu …(ii) J cos q mv M J sin q q From (i) and (ii) J’ v u cot q From (i) J mu cos ec q J sin q For wedge: J ' J cos q J cos q J ' mu cot q M J sin q MV J sin q J sin q V M q mu V M J’ 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 during collision . q (v) The coefficient of restitution is given as (v 2 v1 )n e (u1 u2 )n v n V sin q (v cos q) e u sin q 0 V sin q v cos q e 2 q u sin q V u 1 V v m e cot q (cot q) cot q u u M q m e cot 2 q M + (vi) The heat developed 1 1 1 2 H mu mv MV 2 2 2 2 2 1 1 1 mu 2 H mu2 m (u cot q)2 M 2 2 2 M 1 m H mu 2 1 cot 2 q 2 M 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 mv 0 sin q MV mv n sin q …(i) Using Newton’s law of collision. (v 2 v1 )n e (u1 u2 ) (v sin q v n ) e (v 0 0) v0 …(ii) vn Solving (i) and (ii) we get V (1 e ) mv 0 sin q V M m sin q 2 Using impulse method : The impulse acting on the ball and wedge can be represented as v v0 J vn J vn V J sin q V q J v V J sin q vn J q J sin q MV …(i) mv J mv n …(ii) and Newton’s law of restitution (v sin q v n ) e (v 0 0) …(iii) Solving (i), (ii) and (iii) we get (1 e ) mv 0 sin q V M m sin q 2 Collision of two Discs Oblique Collision R2 U2 O2 d O1 U1 R1 t n O2 d q sin q R1 R2 d q O1 A 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 q d q O1 A v1,n v2t v1t The velocity of Discs will be unchanged in the direction of tangent v1,t u1 sin q and v2,t u2 sin q …(i) t n O2 V2,n q d q O1 A v1,n v2t v1t 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. …(ii) m1u1 cos q m2u2 cos q m1v1,n m2v2,n Now using Newton law of restitution we get (v2,t v1,n ) e (u1 cos q u2 cos q) …(iii) t n O2 V2,n q d q O1 A v1,n v2t v1t From equation (ii) and (iii) we can calculate the value of v1,n and v2,n. t n O2 V2,n q d q O1 A v1,n v2t v1t 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 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. From conservation of momentum along x-axis, we have, mv0 mV 2mv cos q 2 v R Sin q = R/2R =1/2 v 1 R R x q q R R v 3 From conservation of momentum along x-axis, we have, mv0 mV 2mv cos q … (i) 2.0 m/s Thus the incident ball rebounds in the negative x-direction. Aditional Problems on Particle & Wedge Collision 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 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 v0 velocity of the ball is v0. Taking coefficient of M M restitution e, determine the velocities of the 60° 60° sphere and the wedges just after collision. 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) v J 2 v2 M J 60° 60° 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) On solving equations (iii) and (iv) for v1 and v2, we get, (2eM 3m) v0 v1 2M 3m 3 (1 e)mv0 v2 2M 3m 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 Solution. Collision takes place between smooth bodies; therefore tangential velocity of particle is unchanged. vt u sin q R /2 1 Where sin q and u 2gh R 2 or, q 30 y n-axis vn x u R vn V q q vt R/2 t-axis vt u vt … (i) 2 From conservation of momentum along x-axis, Pi Pf 0 0 MV mv n sin q mvt cos q … (ii) y n-axis vn x u R vn V q q vt R/2 t-axis vt From the definition of coefficient of restitution, V sin q (v n ) e 0 u cos q 0 [e = 0, for a perfectly inelastic collision] or, V sin q v n V vn … (iii) 2 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 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 q u2 M2 cos2 h 2g (M m)(M m cos2 ) mu sin2 Mu cos V 2 and v rel M m sin M m sin2 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 q V v y v rel sin q Vrel sin q vrel N vrel v N q Vrel cos q V Momentum of system is conserved along x-direction only. mu m (v rel cos q 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 v rel V cos … (ii) On solving equations (i) and (ii), we get, mu sin2 Mu cos V 2 and v rel 2 … (iii) M m sin M m sin 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. 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 MV m (v rel V ) (M m)v '2 … (v) 2 2 2 2 2 2 On substituting V, vrel and v’ in equation (v) and solving for h, we get, u2 M2 cos2 h 2g (M m)(M m cos2 ) 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 q Coefficient of restitution between ball and wedge is e = 0.5. Assuming all the surfaces to be smooth. calculate (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 q 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 q impulse will give motion to the V J wedge in horizontal direction. Which vertical component of q the impulsive reaction will be J’ t supported by impulse given by ground in vertical direction, as shown in the diagram given below, vy n q y vx q 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. 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. v0 sin q v x cos q v y sin q …(ii) 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 q (v x sin q v y cos q) e (u1 u2 )n v 0 cos q 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 vtT gT 2 2 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 q = V.T = 1.20 m V h q x x R S l