Animated Solution for Physics - System of Particles: 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 $M/m$, for which the angles ${\theta }_1$ and ${\theta }_2$ will be equal, is

Imagine you are observing a game of cosmic billiards. A heavy body of mass $M$ is cruising along the X-axis with an initial speed $v_0$. Waiting patiently at the origin is a lighter, stationary body of mass $m$.

When they collide, it isn't a simple head-on bump. They scatter obliquely! The problem tells us that they scatter symmetrically, meaning both masses veer off at the exact same angle $\theta$ relative to the original path. Our mission? To find the absolute maximum possible value for the mass ratio $M/m$ that allows this symmetric scattering to happen in a perfectly elastic collision.

Momentum Conservation

In any collision where no external forces are acting, the total momentum of the system is strictly conserved. Because this is a 2D collision, we must conserve momentum along both the X and Y axes independently.

Let's start with the Y-axis. Before the collision, everything was moving horizontally, so the initial vertical momentum is zero. After the collision, the vertical momenta of the two masses must perfectly cancel each other out to maintain that zero total:

Because the angles are equal, the $\sin \theta$ terms cancel out beautifully, leaving us with a direct relationship between their final speeds:

Now, let's look at the X-axis. The initial momentum is entirely carried by the heavy mass $M$. After the collision, both masses contribute to the forward momentum via the cosine of their scattering angles:

We can substitute our expression for $v_2$ into this equation:

Notice how the small $m$ cancels out in the second term! This simplifies to:

Dividing both sides by $M$, we get a clean expression for $v_1$:

The problem explicitly states that the collision is elastically perfect. This is a massive hint! It means that not a single joule of kinetic energy is lost to heat or sound. The total kinetic energy before the collision must equal the total kinetic energy after:

Let's multiply the entire equation by 2 to clear those fractions, and then substitute our $v_2$ expression once again:

We can factor out $M{v}_1^2$ on the right side:

We are in the endgame now. Let's take the expression for $v_1$ we found from the X-axis momentum and plug it into our energy equation:

Look at the magic of algebra! The $M$ and the $v_0^2$ terms exist on both sides and completely cancel out, leaving us with a pure relationship between the mass ratio and the scattering angle:

Rearranging this to isolate the mass ratio, we get:

The question asks for the largest possible value of this ratio. To maximize the right side of the equation, we need ${\cos }^2\theta$ to be as large as possible. The maximum value of any cosine function squared is exactly $1$, which occurs when $\theta =0^{\circ }$.

Plugging in ${\cos }^2\theta =1$:

Physically, an angle of $0^{\circ }$ means the collision wasn't oblique at all—it was a perfectly head-on collision where both masses continued moving straight forward!