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. $\frac{m g h}{7}$
• B. $\frac{\operatorname{mgh}}{2}$
• C. $\frac{2 mgh }{5}$
• D. $\frac{7 mgh }{5}$

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} mv _{ cm }^2$
$=\frac{1}{2} \times 2 \frac{2}{5} m R^2 \times\left(\frac{ v _{ cm }}{ R }\right)^2+\frac{1}{2} mv _{ cm }^2+\frac{1}{2} mv _{ cm }^2$
$\Rightarrow \frac{4}{7} mgh = mv _{ cm }^2 \Rightarrow \frac{ mgh }{7}=\frac{1}{5} mv _{ cm }^2$
After collision, only translational kinetic energy is transferred to ball B.
So just after collision, rotational kinetic energy of ball
$A =\frac{1}{5} mv _{ cm }^2=\frac{ mgh }{7}$