Question

A rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits at one end of the rod with a velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After collision, the particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of ${}^{\prime }{x}^{\prime }$ will be ______.

• A. 4.00
• B. 4
• C. 4.0

Answer 1

Answer: (A), (B), (C)

Just before collision :


Just after collision :


From linear momentum conservation,

$P_i=P_f$

$\Rightarrow mu=Mv$

$\Rightarrow v=\frac{mu}{M}\dots .\left(i\right)$

From angular momentum conservation about center of rod $O$,

$L_i=L_f$

$\Rightarrow mu\times \frac{L}{2}=\frac{M{L}^2}{12}\times \omega$

$\Rightarrow \omega =\frac{6mu}{ML}\dots .\left(ii\right)$

Also, coefficient of restitution for the particle and the point of rod where it collided,

$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}$

$\Rightarrow 1=\frac{v+\frac{\omega L}{2}}{u}$

$\Rightarrow v+\frac{\omega L}{2}=u$

$\Rightarrow v+\frac{3mu}{M}=u\left[\text{From}\right(ii\left)\right]$

$\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\left[\text{From}\right(i\left)\right]$

$\Rightarrow \frac{4mu}{M}=u$

$\Rightarrow \frac{m}{M}=\frac{1}{4}=\frac{1}{x}$

$\therefore x=4$