Question

A homogenous rod of length ℓ = ηx and mass, M is lying on a smooth horizontal floor. A bullet of mass 'm' hits the rod at a distance 'x' from the middle of the rod at a velocity $v_0$ perpendicular to the rod and comes to rest after collision. If the velocity of the farther end of the rod just after the impact is in the opposite direction of $v_0$, then:

• A. $\eta >3$
• B. $\eta <3$
• C. $\eta >6$
• D. $\eta <6$

Answer 1

The correct option is D

$\eta <6$

Let the centre of mass of rod moves with velocity 'v' in forward direction.

Considering the (Bullet + Rod) as one system and applying law of conservation of linear momentum before and after the impact.

$m{V}_0=MV\Rightarrow V=\frac{m}{M}{V}_0$. {given that bullet comes to rest}

let the rod acquire an angular velocity '$\omega$'.

Now conserving angular momentum about point 0 as shown,

$m{V}_{0}x=I\omega$

$m{V}_{0}x=\frac{m{l}^2}{12}\omega \Rightarrow m{V}_{0}x=\left(\frac{M}{12}\right)\times \left({\eta }^2{x}^2\right)\times \left(\omega \right)$

$\Rightarrow \omega =\frac{12m{V}_0}{M{\eta }^{2}x}$

Now velocity of p with respect to ground$=v-\frac{l}{2}\omega$ for it to be opposite to $v_0$

$v-\frac{l}{2}\omega <0$

substituting the corresponding values and solving,

$\frac{m}{M}{v}_0<\frac{\eta x}{2}\times \frac{12m{V}_0}{M{\eta }^{2}x}$

$\Rightarrow \eta <6$

Hence option D is correct