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Two identical balls A and B of mass m each are placed on a 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?

Two identical balls A and B of mass m each are placed on a 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?

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Grade:12

1 Answers

Vikas TU
14149 Points
3 years ago
A
Solution :
Just before collision between two balls
Potential energy lost by bal A= kinetic energy gained by ball A.
mgh/2=1/2Icmω^2+1/2mv^2cm
=1/2×25/mR^2×(vcm/R)2+1/2mv^2cm=1/5mv2cm+1/2mv2cm
⇒5/7mgh=mv^2cm⇒mgh/7=1/5mv^2cm
After collision only translattion kinetic energy is transferred to ball B.
So just ater collision, rotational kinetic energy of ball
A=1/5mv^2cm=mgh/7

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