Oblique impact of ball on a wall

Given: $m = 0.4$ kg, $u = 20$ m/s at $30^\circ$ to the wall surface, $e = 0.75$. The wall is smooth, so the impulse acts along the normal only.

   wall |
        |  normal (n)
        | <--
   -----+-------> tangent (t)
        |  \  incoming 30 deg to wall

Components of initial velocity:

• Normal (perpendicular to wall): $u_n = 20\sin30^\circ = 10.00$ m/s
• Tangential (along wall): $u_t = 20\cos30^\circ = 17.32$ m/s

Velocity after impact

• Tangential component is unchanged (no tangential impulse): $v_t = 17.32$ m/s
• Normal component reverses with restitution: $v_n = e\,u_n = 0.75 \times 10 = 7.50$ m/s

$$v = \sqrt{v_n^2 + v_t^2} = \sqrt{7.5^2 + 17.32^2} = 18.87\ \text{m/s}$$

Direction with the wall: $\tan\beta = v_n/v_t = 7.5/17.32$, so $\beta = 23.4^\circ$ (the ball leaves at $23.4^\circ$ to the wall, a flatter angle than $30^\circ$).

Impulse and average force

Impulse on the ball along the normal: $I = m(v_n + u_n) = m(1 + e)u_n = 0.4 \times 1.75 \times 10 = 7.00$ N·s. The impulse on the wall is equal and opposite.

$$F_{avg} = \frac{I}{\Delta t} = \frac{7.00}{0.01} = 700\ \text{N}$$

Kinetic energy lost

$$\Delta T = \tfrac12(0.4)(20^2 - 356.25) = 8.75\ \text{J}$$

Answer: v = 18.87 m/s at 23.4° to the wall; impulse = 7.0 N·s; average force = 700 N; energy lost = 8.75 J.