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A ball of mass m is distributed from the top of a fixed smooth circular tube in a verticalp lane and falls impinging on a ball of mass 2 m at the bottom. The coefficient of restitution is 1/2. Find the heights to which the balls rise after a second impact. |
Answer» Solution :The situation is SHOWN in the figure. If `u` be the velocity of ball `A` just before the impact. Then  `u=sqrt(2g(2a))` consider the first impact between `A` and `B` if velocities of the balls become `v_(1)` and `v_(2)`. respectively, according to linear momentum conservation. `mu+0=mv_(1)+2mv_(2)` ..........i coefficient of restitution is given as `e=(v_(2)-v_(1))/u=1/2`........ii on solving eqn i and ii we GET `v_(1)=0` and `v_(2)=u/2` ball `A` comes to rest, second impact between `A` and `B` occurs, when ball `B` RETURNS to its initial position owht the sam SPEED `u/2`. if after the second impact the velocities ofthe two balls becomes `v_(3)` and `v_(4)` again from linear momentum conservation and coefficient of restitution. `2m(u/2)+0=mv_(3)+2mv_(4)`..........III `e=(v_(3)-v_(4))/(u/2)=1/2`.........iv on solving eqn iii and iv we get `v_(3)=u/2` and `v_(4)=u/4` both the velocities are positive, it implies that both masses will move in the same direction after the second impact. if `h_(1)` and `h_(2)` are the heights to which the masses `m` and `2m` will rise after the second impact according to energy conservation we have for ball of mass `m`, `1/2m(u/2)^(2)=mgh_(1)` or `h_(1)=u^(2)/(8g)=a/2` for ball of mass `2m` `1/2(2m)(u/4)^(2)=2mgh_(2)` or `h_(2)=u^(2)/(32g)=a/8` |
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