A smooth sphere is moving on a horizontal surface with velocity vector `2hati+2hatj` immediately before it hits a vertical wall. The wall is parallel to `hatj` vector and the coefficient of restitution between the sphere and the wall is `e = 1//2`. The velocity vector of the sphere after it hits the wall is

Options

A. `hati-hatj`
B. `-hati+2hatj`
C. `-hati-hatj`
D. `2hati-hatj`

Answer

B

Explanation

`hatj` component i.e., component of velocity parallel to wall remains unchanged while `hatj` component will become `(-1/2)(2hati)or -hati`. Therefore velocity of the sphere after it hits the wall `-hati+2hatj`.