A smooth sphere of mass m is moving on a horizontal plane with a velocity 3hati+hatj when it collides with a vertical wall which is parallel to the hatj vector. If the coefficient of restitution between the sphere and the wall is 1//2. find a. the velocity of the sphere after impact, b. the loss in kinetic energy caused by the impact. c. the impulse that acts on the sphere.

A smooth sphere of mass m is moving on a horizontal plane with a veloc

1.	A smooth sphere of mass m is moving on a horizontal plane with a velocity 3hati+hatj when it collides with a vertical wall which is parallel to the hatj vector. If the coefficient of restitution between the sphere and the wall is 1//2. find a. the velocity of the sphere after impact, b. the loss in kinetic energy caused by the impact. c. the impulse that acts on the sphere.
Answer» Solution :Let `v` be the velocity of the sphere after the impact. Now we separate the velocity components PARALLEL and PERPENDICULAR tothe wall. Using the law of restitution, the COMPONENT of velocity paralel to the wall remains unchanged while the component perpendicular to the wall becomes e times in opposite direction. Thus `vecv=-3/2hati+hatj` a. THEREFORE, the velocity of the sphere after the impact. `=-3/2hati+hatj` b. the loss in `KE =1/2mu^(2)-1/2mv^(2) ` `=1/2m(3^(2)+1^(2))-1/2m({3/2}^(2)+1^(2))=27/8m` C. `vecJ=/_\vecP=vecP_(f)-m(vecv)-m(vecu)` `=m(-3/2hati+hatj)-m(3hati+hatj)=-9/2mhati`