An elevator platform is going up at a speed of `20 m//s` and during its upward motion a small ball of `50 g` mass, falling in downward direction, strikes the platform at speed `5 m//s`. Find the speed with which the ball rebounds.

Answer

4,5

Explanation

The situation is analysed in the figure. We casn consider mass of platform to be very large compared to that of ball, so we have

`vecv_(1)=vecu_(1)e=((v_(2)-v_(1))/(u_(2)-u_(1)))=((v_(2)-u_(1))/(u_(1)-(-u_(2))))impliesvecv_(2)=2vecu_(1)-vecu_(2)`
Thus rebound velocity of ball is
`vecv_(2)=2xx20-(-5)=45ms^(-1)` upward