Question

Two identical uniform solid spherical ball $A$ 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 ball 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 $B$ is:

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

Answer 1

Answer: (A)

Just before collision between two balls
Potential energy lost by $A=$ kinetic energy gained by ball A.
$mg\frac{h}{2}=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}m{v}_{cm}^2$
$=\frac{1}{2}\times 2\frac{2}{5}m{R}^2\times {\left(\frac{v_{cm}}{R}\right)}^2+\frac{1}{2}m{v}_{cm}^2+\frac{1}{2}m{v}_{cm}^2$
$\Rightarrow \frac{4}{7}mgh=m{v}_{cm}^2\Rightarrow \frac{mgh}{7}=\frac{1}{5}m{v}_{cm}^2$
After collision, only translational kinetic energy is transferred to ball B.
So just after collision, rotational kinetic energy of ball
$A=\frac{1}{5}m{v}_{cm}^2=\frac{mgh}{7}$