VIEWS: 3 PAGES: 91 CATEGORY: High School POSTED ON: 9/14/2012
AIEEE IIT-JEE CBSE STUDY MATERIAL PHYSICS SAMPLE PAPERS TEST PAPER KEY SOLUTIONS ANSWERS QUESTIONS
Conservation of Linear Momentum and Mechanical Energy In absence of external forces the linear momentum of the system is conserved. Conservation Of Linear Momentum If no external force acts on a system of masses, then its linear momentum remains constant. Consider a system of masses m1, m2, m3, ....... moving with velocities v1, v2, v3,... respectively. The net linear momentum of the system is p m1v1 m2v2 m3v3 ... 2 If Fext = O, then In terms of X & Y components px initial px final py initial py final Conservation Of Linear Momentum As linear momentum depends on frame of reference, observers in different frames would find different values of linear momentum of a given system but each would agree that his own value of linear momentum does not change with time provided the system is isolated and closed, i.e., law of conservation of linear momentum is independent of the frame of reference though linear momentum depends on frame of reference. 4 Conservation of linear momentum is equivalent to Newton’s III law of motion. This law is universal, i.e., it applies to both macroscopic as well as microscopic systems. It holds good even in atomic and nuclear physics where classical mechanics fails. 5 Further it is more generally applicable than the law of ‘conservation of mechanical energy’ because ‘internal forces’ are often non- conservative and so mechanical energy is not conserved but momentum is (if Fext = 0). Principal applications of conservation of linear momentum are in the field of collisions. 6 Problem 6 . An object of mass m is projected with a velocity u at angle q with horizontal. At the top of the trajectory it explode into two pieces in the ratio of mass 1 : 2. If heavier piece starts falling in vertical direction find Velocity of lighter piece just after explosion. y 2m 1 3 3m u q M A O N Problem 6 . y 2m 1 3 3m u q M A O N Problem 38. Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball toward A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he rolls the ball for the first time. (b) Find the speed of A after he catches the ball for the first time. (c) Find the speed of A and B after the ball has made 5 round trips and is held by A. (d) How many times can A roll the ball ? 9 A B ball A B ball A B ball A B ball Solution. After every throw ‘A’ or ‘B’ gain the momentum 20 kgm/s in the direction opposite to the motion of ball. Ist round A B pA= 20kg m/s p = 20 kg m/s ball A B pA= 20kg m/s 20kg m/s 20 kg m/s = pA,1 20kg m/s 40kg m/s 40kg m/s 14 Second round A B pA = 60kg m/s 40kg m/s 20kg m/s 20 kg m/s pA =60kg m/s 20kg m/s 80kg m/s 20kg m/s pA,2 = 80kg m/s At the end of each round both the man receive a momentum 40 kg m/s. (a) pA = 20 = 40 vA 1 vA m/ s 2 15 (b) pA,1 = 40 = (40+4) vA,1 40 10 vA,1 m/ s 44 11 (c) At the end of 5 round trip the linear momentum of (ball + ‘A) system pA,5 40 5 200 kg m / s 200 (40 4) vA,5 50 vA,5 m/ s 11 16 (d) ‘A’ can catch the ball untill the velocity of ‘A’ is less than 5 m/s. After 5 round when ‘A’ through the ball i.e., 6th time the velocity of ‘A’ become greater than 5 m/s hence A can through the ball only 6 times. (e) As no external force is acting on ‘A’, ‘B’ and ‘ball’ system hence the position of centre of mass of system should remain constant always. Let the centre of mass is at a distance x cm from the initial position of A. (40 40 4) x cm 40 0 4 0 40 d 10 xcm m From initial position of A. 21 18 CONSIDER THE SITUATION A smooth wedge of mass M is placed on horizontal surface and a block of mass m is kept at the highest position of wedge. Initially, the system is held. m h M v 19 CONSIDER THE SITUATION A smooth wedge of mass M is placed on horizontal surface and a block of mass m is kept at the highest position of wedge. Initially, the system is held. m h M 20 If the wedge is held and block is released from rests. m h M v The block starts sliding on the curved path of the wedge. 21 If we take the wedge and block as system Can we apply conservation of linear momentum for this system? Can we apply conservation of mechanical energy for this system ?. m h M v We can not apply conservation of linear momentum as we have external forces acting on the system in horizontal as well as vertical direction. External force in vertical direction is gravity as well as normal reaction of ground, m h M v We can use conservation of mechanical energy for this system. We have external force in horizontal direction but it is not doing any work. While applying conservation of mechanical energy earth is the part of the system hence no work done by external force in vertical direction. Calculate the velocity of the block when it reaches at ground. Apply C.O.M.E. K + U = 0 1 mv 2 0 0 mgh 0 v 2 gh 2 m h M v m h M What will happen if we remove stopping force on the wedge and system is released. h M h M h M V h M v When system is released both the block and wedge will starts moving. The wedge starts moving backward in horizontal direction while the block starts sliding over the wedge. m h M m Can we apply C.O.L.M in h horizontal direction for M the system (block + wedge) ? V Can we apply C.O.M.E for h m this system ?. M v We can apply C.O.L.M. in horizontal direction as there are not any external forces in horizontal direction. We can also apply C.O.M.E. for the system as no external forces are doing work on the system What is the velocity of the block and wedge, when the block reaches at the base of the wedge ? We can apply C.O.L.M in horizontal direction As no force is acting on the system in horizontal direction linear momentum during entire process will be conserved m Initial linear momentum h Linitial = 0 M Final linear momentum Lfinal = mv - MV V h m v As Linitial = Lfinal M 0 = mv - MV MV v ...( i ) m m Now applying C.O.M.E for this system. h M K + U = 0 1 1 MV 2 mv 2 0 0 mgh 0 2 2 2 V 1 1 MV MV m 2 mgh h m v 2 2 m M 2mgh V 2 M2 M m The velocity of the block and wedge, when the block reaches at the base of the wedge The velocity of wedge V 2m 2 gh M (m M ) 2 Mgh The velocity of the block v (m M ) Now an another identical wedge is placed in front of wedge (I). The block will slide over wedge I and reaches at its base with velocity v. Then the block slide over smooth horizontal surface and interact with wedge II. m h M M m h M M h M M h M M h M M V h M v M V h M v M h v M M II V M M v II M M v II M m h M M m h M M V h m v M M Vh mv M M v m h M M 2 Mgh The velocity of the block v …(i) (m M ) Now considering the ‘m’ and wedge ‘2’ as system. Using conservation of linear momentum m mv (m M )V ' …(ii) h’ V’ m v M Conservation of mechanical energy 1 1 mv 2 (M m)V '2 mg h ' …(iii) 2 2 From equations(i) (ii) and (iii) we get 2 M h' h M m Illustration 88. A smooth wedge of mass M rests on a smooth horizontal surface. A block of mass m is projected from its lowermost point with velocity v0 . What is the maximum height reached by the block? m v0 M 51 m v0 M 52 m M 53 m M 54 m M 55 At the instant the block breaks contact with the wedge, they have common x-component of velocity. In addition, the block has a vertical component of velocity. The trajectory of the block will be parabola. m Due to this vertical component the block rises upwards till the vertical component of velocity vanishes. M 56 m M 57 m M 58 m M 59 m M 60 m M 61 m M 62 From momentum conservation along x-axis, m0 (m M )V …(i) m0 V …(ii) (m M ) m m M M 63 From energy conservation between initial and final positions of block, 1 1 m2 (m M )V 2 mgh …(iii) 0 2 2 1 1 m2 2 m2 0 mgh 2 0 2 m M M 2 h 0 2g m M 64 Illustration Calculate the velocity of the wedge when the ball reaches at the bottom of the groove. r O m r M r O m r M Two identical buggies each of mass ‘M’ moves one after due to inertia (without friction) with the some velocity V0 . A man of mass m rides the rear buggy. At a certain moment, the man jumps into the front buggy with velocity u relative to this buggy . Knowing that the mass of each buggy is equal to M. Find the velocity with which the buggies will move after that . V0 m V0 M M Solution. Considering the rear buggy and man ‘m’ as system. (M m)V0 MVr m (Vr u) mu Vr V0 ( M m) Vr (Vr + u) Rear buggy Considering front buggy and mass as system M0 V0 m (u Vr ) (m M )Vf mu MV0 m u V0 (m M )Vf (M m) u(M m) mu (M m)V0 m (M m)Vf (M m) m Mmu Vf V0 Vf (M m)2 M Illustration 85. Two identical buggies 1 and 2 with one man in each move along parallel rails. When the buggies are opposite to each other, the men jump in a direction perpendicular to the direction of motion of buggies, so as to exchange their places. Buggy 1 v1 Buggy 2 v2 Top view (a) 70 Illustration 85. As a consequence buggy 1 stop and buggy 2 keeps moving in the same direction with its final velocity v. Find the initial velocities v1 and v2 of buggies. Mass of each buggy (without man) equals M, mass of each man is m; ignore frictional effects anywhere and the buggies are constrained to move along the rails only. Buggy 1 v1 Buggy 2 v2 Top view (a) 71 Solution. In the problems involving mass ejection and mass addition, the ejected as well as added mass both are part of a system. When the men jump they push on the floor of the buggy (action); the same force acts on them as well (reaction). Buggy 1 v1 Buggy 2 v2 Top view (a) 72 This action- reaction pair is internal force for the system of man and buggy. Similarly, when men jump in, there is an action-reaction force between them and the buggy. Buggy 1 v1 Buggy 2 v2 Top view (a) 73 Let the buggy “1” move to the right and buggy “2” to the left. When the men jump. Man in “1” brings momentum mv1 directed towards right in the buggy “2”. Similarly the man in “2” brings momentum mv2 in “1” directed towards left. Buggy 1 v1 Buggy 2 v2 Top view (a) For system 1: Pi (M m)1 m1 m2 M 1 m2 Pf 0 System 1 From law of conservation of momentum, Pi Pf v1 M 1 m2 0 …(i) Moving in v1 v2 Going out 75 For system 2: Pi (M m)2 m1 m2 Going out Pf (M m) v1 v2 From law of conservation Moving in of momentum, v2 Pi Pf System 2 (M m)2 m1 m2 (M m) (b) …(ii) 76 On solving equations (i) and (ii) simultaneously, we obtain m M 1 2 (M m ) (M m ) Problem 19. A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically. When at a height h from the ground, he notices that the H h ground below him is pretty hard, but there is a pond at a horizontal distance x from the x line of fall. hard ground pond 78 Problem 19. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum H horizontal velocity imparted to h the bag so that the man lands in the water x hard ground pond 79 Solution. Considering (man + bag) as a system; No external forces in horizontal direction is acting on man. To avoid the man falling on hard floor the man H h should be displaced by a horizontal displacement x during which he fall a x height h. hard ground pond 80 Solution. The time taken by man to fall a height h T = time to fall H – time to fall (H – h) 2H 2(H h) t g g H Hence horizontal velocity of man h required …(i) Vman x x t hard ground pond 81 Solution. As no external force in horizontal direction hence, the linear momentum of ‘man + bag’ system should be conserved (i.e. equal to zero). Mman .V man Mbag V bag 0 H h x M mv b 0 t M x x v bag . m t Mx g hard ground v bag pond m[ 2H 2(H 2)] 82 Illustration 89. A wedge of mass m1 with its upper surface hemispherical in shape, as shown in figure (a), rests on a smooth horizontal surface near the wall. A small block of mass m2 slides without friction on the hemispherical surface of the m2 wedge. r What is the maximum velocity attained by the wedge? m1 (a) 83 Solution. As long as the block moves from A to B, the reaction on the wedge presses it to the wall. When the block reaches the lowermost positions, its velocity from energy conservation is 2gr A N N’ m2 C N 2gr N’ r B m1 (b) (a) 84 When the block moves along the right half of the wedge, during its upward journey as well as downward journey the reaction of the block on the wedge is towards right as shown in figure (b). Therefore during the entire motion of the block from B to C and C to B, the wedge is accelerated towards right. A N N’ m2 C N 2gr N’ r B m1 (b) (a) 85 Thus maximum velocity is attained by the wedge at the instant when the block passes point B during the return journey. From conservation of momentum, at the instant when the wedge is separated from the wall, Pi Pf m2 2gr m11 m2 2 …(i) A N N’ m2 C N 2gr N’ r B m1 (b) (a) 86 From M.E. energy conservation, E i Ef m11 m2 2 2 2 m2gr …(ii) 2 2 On solving equations (i) and (ii) simultaneously, we obtain two solutions 1 0, 2 2gr 1 2m2 m2 m1 and 2gr , 2 2gr m1 m2 m1 m2 The first solution corresponds to the instant when the block reaches for the first time at point B. At this instant the block moves with velocity v2 and the wedge is at rest. 87 The second solution corresponds to the instant when the block has the maximum velocity 2m2 2gr (1 )max m1 m2 Problem 27. A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 m where m is the mass of one shell. If the muzzle velocity of the shells is 200 m/s, what is the recoil speed of the car after the second shot ? Neglect friction. gun vs,g =200 m/s v1 89 Solution. Taking Gun and car as a system no external force are acting on the system in horizontal direction at any time. After Ist shot 49 mv1 m .200 200 v1 m/ s 49 gun vs,g =200 m/s v1 After IInd shot 200 200 49 m 48mv 2 m 200 49 49 1 1 v 2 200 49 48 gun vs,g =200 m/s v1 91