Question

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. $\frac{mgh}{7}$
• B. $\frac{mgh}{2}$
• C. $\frac{2mgh}{5}$
• D. $\frac{7mgh}{5}$

Answer 1

The correct option is A $\frac{mgh}{7}$

Just before collision between two balls,
Potential energy lost by ball A = kinetic energy gained by ball A.
$mg\frac{h}{2}=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}m{v}_{cm}^2$
$=\frac{1}{2}\times \frac{2}{5}m{R}^2\times {\left(\frac{v_{cm}}{R}\right)}^2+\frac{1}{2}m{v}_{cm}^2=\frac{1}{5}m{v}_{cm}^2+\frac{1}{2}m{v}_{cm}^2$
$\Rightarrow \frac{5}{7}mgh=m{v}_{cm}^2\Rightarrow \frac{mgh}{7}=\frac{1}{5}m{v}_{cm}^2$
After collision, only translational kinetic energy is transferred to ball B.
So just after collision, rotational kinetic energy of ball $A=\frac{1}{5}m{v}_{cm}^2=\frac{mgh}{7}$