A ball P of mass m kg is dropped from a point A, which is 2 m vertically above a point B on a horizontal floor. After P hits the floor at B, it rebounds and hits another ball Q, of the same mass, which has also been dropped from A. The impact between the two balls is direct and takes place at the mid-point of AB. The coefficient of restitution in each impact is 5/7 . Neglecting air resistance, find the speed of P
(i) immediately after it hits the floor,
(ii) immediately after it collides with Q.
Ans: (i)4.52 m/s (ii) 3.74 m/s
e=(velocities after collision)/(velocities before collision)
Law of conservation of momentum
The Attempt at a Solution
Please see attachment.
I tried to calculate this problem but some of them said my calculation is wrong because they mentioned that I can't use law of conservation of momentum for last part because of gravity. So how to calculate to get correct answer?Thanks a lot.