Question

A ball of mass $m$ is projected with a speed $v$ into the barrel of a spring gun of mass $M$ initially at rest lying on a frictionless surface. The mass sticks in the barrel at the point of maximum compression in the spring. The fraction of kinetic energy of the ball stored in the spring is :

• A. $\frac{m}{M}$
• B. $\frac{M}{m+M}$
• C. $\frac{m}{M+M}$
• D. None of these

Answer 1

The correct option is B $\frac{M}{m+M}$

Initial energy $=K_1=\frac{1}{2}m{v}^2$

Initial momentum$=p_1=mv$

From conservation of momentum $=p_2=mv$

Final kinetic energy $=K_2=\frac{p_2^2}{2(m+M)}=\frac{m^2{v}^2}{2(m+M)}$

This loss in kinetic energy =$K_1-K_2$ gets stored in the spring on compression.

$E=K_1-K_2=\frac{mM{v}^2}{2(m+M)}$

Hence, required ratio$=\frac{E}{K_1}=\frac{M}{m+M}$

Answer-(B)