1. Set up the collision equations

Let the mass $M$ initially move along the $x$-axis with speed $V_0$. Mass $m$ is initially at rest.

After elastic collision:

• mass $M$ moves with speed $v_1$ at angle $\theta_1$
• mass $m$ moves with speed $v_2$ at angle $\theta_2$

We are given the special case: $\theta_1=\theta_2=\theta$

Since the two bodies go on opposite sides of the initial direction for transverse momentum to cancel, take their $y$-components with opposite signs.

So momentum conservation gives:

$MV_0 = Mv_1\cos\theta + mv_2\cos\theta$ $0 = Mv_1\sin\theta - mv_2\sin\theta$

From the second equation, $Mv_1 = mv_2$ $v_2 = \frac{M}{m}v_1$

2. Use the $x$-momentum equation

Substitute $mv_2 = Mv_1$ into the $x$-momentum equation:

$MV_0 = (Mv_1 + mv_2)\cos\theta = (Mv_1 + Mv_1)\cos\theta$ $MV_0 = 2Mv_1\cos\theta$

Hence, $v_1 = \frac{V_0}{2\cos\theta}$

And therefore, $v_2 = \frac{M}{m}\cdot \frac{V_0}{2\cos\theta}$

3. Apply conservation of kinetic energy

Since collision is elastic, $\frac12 MV_0^2 = \frac12 Mv_1^2 + \frac12 mv_2^2$

Substitute $v_2 = \frac{M}{m}v_1$:

$MV_0^2 = Mv_1^2 + m\left(\frac{M}{m}v_1\right)^2$ $MV_0^2 = Mv_1^2 + \frac{M^2}{m}v_1^2$

Divide by $M$: $V_0^2 = v_1^2\left(1+\frac{M}{m}\right)$

Now use $v_1 = \frac{V_0}{2\cos\theta}$

So, $V_0^2 = \frac{V_0^2}{4\cos^2\theta}\left(1+\frac{M}{m}\right)$

Cancel $V_0^2$: $1 = \frac{1}{4\cos^2\theta}\left(1+\frac{M}{m}\right)$

Thus, $4\cos^2\theta = 1+\frac{M}{m}$

So, $\frac{M}{m} = 4\cos^2\theta - 1$

4. Find the largest possible value

Since $0\le \cos^2\theta \le 1$ we get $\frac{M}{m} \le 4(1)-1 = 3$

Hence the largest possible value is $\boxed{3}$

5. Check options

• A: $4$ ❌
• B: $1$ ❌
• C: $3$ ✅
• D: $2$ ❌

Therefore, the correct option is: $\boxed{\text{C}}$