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

Two identical uniform solid spherical balls `A` and `B` of mass `m` ea

1.	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: A. `(mgh)/7`B. `(mgh)/2`C. `(2mgh)/5`D. `(7mgh)/5`
Answer» Correct Answer - A 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`

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