A small ball is projected horizontally between two large blocks. The ball is given a velocity `V ms^(-1)` and each of the large blocks move uniformly with a velocity of `2V ms^(-1)`. The ball collides elastically with the blocks. If the velocity of the blocks do not change due to the collision, then find out the velocity of the ball after the `2nd` collision.

Options

A. `5v`
B. `7V`
C. `9V`
D. none of these

Answer

C

Explanation

Use the following result

`e=(v_(2)-v_(1))/(u_(1)-u_(2))implies1=(-v_(0)(-v_(1)))/(u_(1)-(-v_(0)))impliesv_(1)u_(1)+2v_(0)`
using the above result
Now velocity after first collision
`V'=V+2(2V)=5V`
Velocity after second collision
`V''=V'+2(2V)=5V+4V=9V`