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

A smooth sphere is moving on a horizontal surface with velocity vector

1.	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
Answer» `HATI-HATJ` `-hati+2hatj` `-hati-hatj` `2hati-hatj` Solution :`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`.