1. Set up the situation

A rod of mass $M$ and length $L$ lies on a frictionless horizontal surface. A particle of mass $m$ strikes one end of the rod with speed $u$, moving perpendicular to the rod.

After collision, the particle comes to rest.

Let after collision:

• center of mass speed of rod $= V$
• angular speed of rod about its center $= \omega$

Since the surface is frictionless, linear momentum and angular momentum are conserved. Also, collision is elastic.

2. Conservation of linear momentum

Initially, only the particle has momentum: $p_i = mu$

Finally, particle stops, so only the rod has translational momentum: $p_f = MV$

Thus, $mu = MV$ $V = \frac{mu}{M}$

3. Conservation of angular momentum about the center of the rod

Take moments about the rod's center at the instant of collision.

Initially, the particle strikes at the end of the rod, so perpendicular distance from rod's center is $L/2$. Hence initial angular momentum of particle about rod's center is $L_i = mu\left(\frac{L}{2}\right)$

Finally, particle is at rest, so rod alone has rotational angular momentum: $L_f = I\omega$ where for a uniform rod about its center, $I = \frac{1}{12}ML^2$

Therefore, $mu\left(\frac{L}{2}\right) = \frac{1}{12}ML^2\omega$

Solve for $\omega$: $\omega = \frac{6mu}{ML}$

4. Use conservation of kinetic energy

Collision is completely elastic, so total kinetic energy is conserved.

Initial kinetic energy: $K_i = \frac{1}{2}mu^2$

Final kinetic energy:

• particle is at rest, so zero
• rod has translational and rotational kinetic energy

Thus, $\frac{1}{2}mu^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$

Substitute $V = \frac{mu}{M}$ and $I=\frac{1}{12}ML^2$, $\omega=\frac{6mu}{ML}$:

$\frac{1}{2}mu^2 = \frac{1}{2}M\left(\frac{mu}{M}\right)^2 + \frac{1}{2}\cdot \frac{1}{12}ML^2 \left(\frac{6mu}{ML}\right)^2$

Now simplify each term.

First term: $\frac{1}{2}M\cdot \frac{m^2u^2}{M^2} = \frac{1}{2}\frac{m^2u^2}{M}$

Second term: $\frac{1}{2}\cdot \frac{1}{12}ML^2 \cdot \frac{36m^2u^2}{M^2L^2}$ $= \frac{1}{24}\cdot 36 \cdot \frac{m^2u^2}{M}$ $= \frac{3}{2}\frac{m^2u^2}{M}$

So, $\frac{1}{2}mu^2 = \left(\frac{1}{2} + \frac{3}{2}\right)\frac{m^2u^2}{M}$ $\frac{1}{2}mu^2 = 2\frac{m^2u^2}{M}$

Cancel $u^2$ and multiply by $2$: $m = 4\frac{m^2}{M}$

Assuming $m\neq 0$, $1 = 4\frac{m}{M}$ $\frac{m}{M} = \frac{1}{4}$

Given $\frac{m}{M} = \frac{1}{x}$ so, $\frac{1}{x} = \frac{1}{4} \implies x=4$

5. Final answer

$\boxed{4}$