A particle of mass $M$ collides elastically with a particle of mass $m$ at rest. At what maximum angle can the first particle be deflected after the collision, if $M>m$?
I. solution. It is advisable to examine the elastic collision of two particles in the center of mass system, as in this coordinate system, the absolute value of the particles' velocity remains unchanged, only their direction changes. Let the velocity of the particle with mass $M$ before the collision in the laboratory-fixed coordinate system be $v_{0}$. Then the velocity of the center of mass is $\mathbf{v}_{C M}=\mathbf{v}_{0} M /(M+m)$. In the center of mass system, the velocity of the particle with mass $M$ before the collision is $\mathbf{v}_{0}-\mathbf{v}_{C M}$, and its velocity after the collision is $\left|\mathbf{v}_{0}-\mathbf{v}_{C M}\right| \cdot \mathbf{e}$, where $\mathbf{e}$ is a unit vector whose possible directions are determined by the conservation of momentum. Transforming back to the laboratory coordinate system, the velocity after the collision is:

$$
\mathbf{v}=\mathbf{v}_{C M}+\left|\mathbf{v}_{0}-\mathbf{v}_{C M}\right| \mathbf{e}=\mathbf{v}_{0} \frac{M}{M+m}+v_{0} \frac{m}{M+m} \mathbf{e}
$$

Let's geometrically represent the vector addition given by equation (1) (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_7a1cf3e9661d44101f22g-1.jpg?height=212&width=374&top_left_y=653&top_left_x=862)

Figure 1

The vector $\mathbf{v}$ starts at point $A$ and its endpoint lies on a sphere with center $O$ and radius $r=v_{0} m /(M+m)$, where $\overline{A O}=v_{0} M /(M+m)$. Since $M>m$, point $A$ lies outside the sphere. The direction of $M$'s velocity changes the most when the line of $\mathbf{v}$ is tangent to the sphere, in which case

$$
\sin \alpha=r / \overline{A O}=m / M
$$

We can show that this deflection angle is indeed realizable.

Peter Sáfár (Debrecen, Tóth Á. Gymnasium III. grade)

II. solution. Suppose that after the collision, the velocity of $M$ is $\alpha$ and the velocity of $m$ is $\beta$ with respect to the initial velocity $\mathbf{v}_{0}$ of $M$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_7a1cf3e9661d44101f22g-1.jpg?height=265&width=436&top_left_y=1472&top_left_x=823)

Figure 2

We can write the equations expressing the conservation of momentum and energy:

$$
\begin{gathered}
M v_{M} \sin \alpha=m v_{m} \sin \beta \\
M v_{0}=M v_{M} \cos \alpha+m v_{m} \cos \beta \\
(1 / 2) M v_{0}^{2}=(1 / 2) M v_{M}^{2}+(1 / 2) m v_{m}^{2}
\end{gathered}
$$

From the system of equations, we can express $\alpha$ as a function of $\beta$:

$$
\operatorname{tg} \alpha=\frac{\sin 2 \beta}{(M / m)-\cos 2 \beta}
$$

By differentiating, we can find the angle $\beta$ at which $\operatorname{tg} \alpha$ is maximal, for which $\cos 2 \beta=(m / M)$ is obtained. Substituting this back into (6), we get the maximum deflection angle of the particle:

$$
\operatorname{tg} \alpha=\frac{m}{\sqrt{M^{2}-m^{2}}}, \quad \text { i.e., } \sin \alpha=m / M
$$