Question

A rod AB of mass M and length 4L rests on a smooth horizontal plane. A small particle P moving with a velocity V strikes the rod elastically in a direction normal to the length, at a point distant L from the centre C of the rod. Immediately after the collision, if P comes to rest, then its mass should be

• A. $\frac{3M}{4}$
• B. $\frac{4M}{3}$
• C. $\frac{4M}{7}$
• D. $\frac{3M}{7}$

Answer 1

The correct option is C

$\frac{4M}{7}$

Let m be the mass of that particle from conservation of linear momentum, $mv=0+M{V}_{c}where{V}_{c}is$

linear velocity of cm of the rod.

From conservation of angular momentum, $mVL=I\omega$

As collision is elastic,

$V=V_c+wL\Rightarrow V=\frac{mV}{M}+\frac{mV{L}^2}{I}\Rightarrow (\frac{m}{M}+\frac{m{L}^2}{I})=1ButI=\frac{M(4L{)}^2}{12}Then\frac{m}{M}(1+\frac{3}{4})=1\Rightarrow m=\frac{4M}{7}$