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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 towards 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 five round trips and is held by A. d. How many times can A roll the ball? e. Where is the centre of mass of the system A + B + ball at. the end of the nth trip? |
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Answer» Solution :After Every throw `A` or `B` gains a momentum of `20 kg m//s` in the direction opposite to the motion of the ball. First round  Second Round  At the end of each round, both men receive a momentum `40 kg m//s` a. `pA=20=40 v_(A)` `v_(A)=1/2m//s` `b. p_(A,1)=40=(40+4)v_(A,1)` `v_(A,1)=40/44=10/11m//s` C. At the end of five round trips, the linear momentum of (ball `+ A` ) system, `p_(A,S)=40xx5=200kgm/s` `200=(40+4)v_(A,5)` `V_(A,5)=50/11m/s` Momenum transfered to `B` in five trips `=40xx5=200impliesv_(B,5)=5m//s` d. A can catch the ball until the velocity of `A` is less than `5 m/s`. after fie ROUNDS when `A` THROUWS the balls, i.e., `6 th` time, the velocity of `A` becomes greater than `5m//s` hence `A` can throw the ball only six times. e. `A` no external force is ACTION on As, `B` and ball system hence the postion of centre of mass of the system should remain constant always. Let the centre of mass be at a distance `x` cm from the position of `A`. `(40+4+10)x_(CM)=40xx0+4xx0+40d` `x_(CM)=(10d)/21m` from initial position of `A`. |
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