A body of mass $$M$$ moving at speed $$V_0$$ collides elastically with a mass $$m$$ at rest. After the collision, the two masses move at angles $$\theta_1$$ and $$\theta_2$$ with respect to the initial direction of motion of the body of mass $$M$$. The largest possible value of the ratio $$\frac{M}{m}$$, for which the angles $$\theta_1$$ and $$\theta_2$$ will be equal, is:

We have a body of mass $$M$$ moving initially with speed $$V_0$$ along the positive $$x$$-axis. It collides elastically with a second body of mass $$m$$ that is at rest. After the collision the two bodies move with speeds $$u$$ (for mass $$M$$) and $$v$$ (for mass $$m$$) making angles $$\theta_1$$ and $$\theta_2$$ respectively with the original direction. The question states that $$\theta_1$$ and $$\theta_2$$ are equal, so we write $$\theta_1=\theta_2=\theta$$.

Because the collision is perfectly elastic, both linear momentum and kinetic energy are conserved. We now translate those words into equations.

Conservation of linear momentum in component form:

Along $$x$$:

$$M V_0 \;=\; M u\cos\theta + m v\cos\theta$$

Along $$y$$ (taking the upward direction for the angle of mass $$M$$ and the downward direction for mass $$m$$ so that the net initial $$y$$-momentum is zero):

$$0 \;=\; M u\sin\theta \;-\; m v\sin\theta$$

From the second equation we solve for $$u$$:

$$M u\sin\theta = m v\sin\theta \;\Longrightarrow\; u = \dfrac{m}{M}\,v.$$

Substituting this value of $$u$$ into the first momentum equation gives

$$M V_0 = M\bigl(\tfrac{m}{M}v\bigr)\cos\theta + m v\cos\theta = m v\cos\theta + m v\cos\theta = 2 m v\cos\theta.$$

Hence the speed of mass $$m$$ after collision is

$$v = \dfrac{M V_0}{2 m\cos\theta}.$$

Conservation of kinetic energy now supplies a second relation. The usual formula is $$\tfrac12 M V_0^2 = \tfrac12 M u^2 + \tfrac12 m v^2.$$ Removing the factor $$\tfrac12$$ gives

$$M V_0^2 = M u^2 + m v^2.$$

We already have $$u = \dfrac{m}{M}v,$$ so we substitute:

$$M V_0^2 = M\!\left(\dfrac{m}{M}v\right)^{\!2} + m v^2 = M\!\left(\dfrac{m^2}{M^2}v^2\right) + m v^2 = \dfrac{m^2}{M}\,v^2 + m v^2 = v^2\!\left(\dfrac{m^2}{M} + m\right).$$

Factor out $$m$$ to simplify the right-hand side:

$$M V_0^2 = v^2 \, m \!\left(\dfrac{m}{M} + 1\right) = v^2 \, m \!\left(\dfrac{m + M}{M}\right).$$

Multiplying both sides by $$M$$ removes the denominator:

$$M^2 V_0^2 = v^2\,m\,(m+M).$$

Now we substitute for $$v$$ from the momentum result:

$$v^2 = \left(\dfrac{M V_0}{2 m\cos\theta}\right)^{\!2} = \dfrac{M^2 V_0^2}{4 m^{\,2}\cos^{2}\theta}.$$

Putting this into the energy equation yields

$$M^2 V_0^2 \;=\; \dfrac{M^2 V_0^2}{4 m^{\,2}\cos^{2}\theta}\;\,m\,(m+M).$$

The factor $$M^2 V_0^2$$ appears on both sides, so it cancels out completely:

$$1 = \dfrac{m(m+M)}{4 m^{\,2}\cos^{2}\theta} = \dfrac{m+M}{4 m\cos^{2}\theta}.$$

Re-arranging this gives the squared cosine of the common angle:

$$\cos^{2}\theta = \dfrac{m+M}{4 m}.$$

We know from trigonometry that $$\cos^{2}\theta \le 1$$ for all real angles. Therefore

$$\dfrac{m+M}{4 m} \;\le\; 1.$$

Multiplying both sides by $$4m$$ we get

$$m + M \;\le\; 4 m.$$

Isolating $$M$$ leads directly to

$$M \;\le\; 3 m.$$

Thus the largest possible value of the ratio $$\dfrac{M}{m}$$ that still allows the two deflection angles to be equal is

$$\dfrac{M}{m} = 3.$$

Hence, the correct answer is Option A.