 Solution

Let the projectile's initial laboratory velocity be $u\hat{\mathbf{x}}$. The centre-of-momentum frame moves atIn that frame, the projectile's initial speed is $w=u-V=mu/(M+m)$. An elastic collision preserves that speed, but can change its direction. Transforming the outgoing projectile velocity back by Galilean velocity addition givesThis is the velocity-circle construction for elastic scattering: the outgoing tip lies on a circle of radius $w$, centred a distance $V$ from the origin in the velocity plane. Since $M>m$, $V>w$ and the origin lies outside this circle. The largest scattering angle occurs when the ray from the origin is tangent to it. The resulting right triangle givesEquality is kinematically possible; in the centre-of-momentum frame it corresponds to $\cos \chi =-w/V=-m/M$, where $\chi$ is the outgoing projectile angle to its initial direction. In particular, a heavier projectile cannot reverse its laboratory direction in an isolated elastic collision with an initially stationary lighter target.