Solution to Question 26
Frame of Reference
Since the cylinder is massive, its velocity $u_1$ remains unchanged. Switch to the Cylinder’s Frame:
- Velocity of Cylinder = 0
- Velocity of Ball = $u_1 + u_2$ (towards cylinder)
Collision Geometry
Let $R$ be the radius. Impact parameter is $b$. Angle of incidence $\alpha$: $$ \sin \alpha = \frac{b}{R} $$ In the cylinder’s frame, the ball reflects elastically. The speed remains $v_{rel} = u_1 + u_2$. The angle of deflection is $2\alpha$ relative to the incoming line (or reflection angle relative to normal is $\alpha$).
Back to Ground Frame
Final velocity vector $\vec{v}$ is the vector sum of the cylinder’s velocity $\vec{u}_1$ and the ball’s reflected relative velocity $\vec{v}_{rel}$. $$ \vec{v} = \vec{u}_1 + \vec{v}_{rel}’ $$ Using the Law of Cosines (since we want speed): $$ v^2 = u_1^2 + (u_1+u_2)^2 + 2u_1(u_1+u_2)\cos(2\alpha) $$ Wait, geometry check. Reflected velocity points away from cylinder center. Using simple coordinate transformation: $$ v_{x} = u_1 + (u_1+u_2)\cos(\pi – 2\alpha) $$ Actually, simpler formula using geometry: $$ v = \sqrt{ u_2^2 + 4u_1(u_1+u_2)\left(1 – \frac{b^2}{R^2}\right) } $$