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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.

Choose the cylinder's velocity to be $v\hat{\mathbf{x}}$, and the ball's laboratory velocity to be $u(\cos \theta ,\sin \theta )$. By Galilean velocity addition, the ball's velocity relative to the cylinder isFor a suitably centred entry without hitting a wall, the cylinder's axis must be parallel to this relative velocity. Its oriented angle $\alpha$ to the cylinder's direction of travel is thereforeEquivalently, $\tan \alpha =u\sin \theta /(u\cos \theta -v)$, with the quadrant chosen from the two components. For $v/u\ll 1$,Thus the extra tilt relative to the ball's original direction is approximately $(v/u)\sin \theta$. The use of $atan2$ also covers nearly transverse incidence, where taking an unqualified inverse tangent could select the wrong axis direction.

When $0\le v,v^{\prime }\ll c$, expand the denominator:The leading term is ordinary Galilean velocity addition. When both $v,v^{\prime }\to c$ from below, the exact formula instead gives $\boxed{v^{\prime \prime }\to c}$, rather than $2c$. Indeed, writing $a=v/c$, $b=v^{\prime }/c$, we have $1-v^{\prime \prime }/c=(1-a)(1-b)/(1+ab)>0$ whenever $a,b<1$.

In a nonrelativistic elastic collision of incident mass $M$ at speed $u$ with initially stationary mass $m$, the centre-of-momentum frame moves at $V=Mu/(M+m)$. The projectile's outgoing velocity in that frame has fixed magnitude $w=mu/(M+m)$. Its laboratory velocity tips lie on a circle of radius $w$ centred at $V\hat{\mathbf{x}}$. For $M>m$, a tangent from the origin gives the largest laboratory scattering angle, $\sin {\theta }_{\max }=w/V=m/M$. This geometrical result follows from Galilean velocity addition and conservation of energy.