Two identical uniform solid spherical balls `A` and `B` of mass `m` each are placed on a the fixed wedge as shown in figure. Ball `B` is kept at rest and it is released just before two balls collides. Ball `A` rolls down without slipping on inclined plane and collide elastically with ball `B`. The kinetic energy of ball `A` just after the collision with ball `B` is:

Options

A. `(mgh)/7`
B. `(mgh)/2`
C. `(2mgh)/5`
D. `(7mgh)/5`

Answer

A

Explanation

Just before collision between two balls
Potential energy lost by bal `A=` kinetic energy gained by ball `A`.
`mgh/2=1/2I_(cm)omega^(2)+1/2mv_(cm)^(2)`
`=1/2xx2/5 mR^(2)xx((v_(cm))/R)^(2)+1/2mv_(cm)^(2)=1/5mv_(cm)^(2)+1/2mv_(cm)^(2)`
`implies5/7mgh=mv_(cm)^(2)implies(mgh)/7=1/5mv_(cm)^(2)`
After collision only translattion kinetic energy is transferred to ball `B`.
So just ater collision, rotational kinetic energy of ball
`A=1/5mv_(cm)^(2)=(mgh)/7`