Question

A spherical rigid ball of mass $10 \, kg$ is moving with a speed of $2 \, m/s$ in the direction shown. The ball collides with a rigid frictionless wall and rebounds at an angle $\alpha$ with a speed $v$, as shown. The coefficient of restitution is $0.9$. The angle $\alpha$ (in degrees) is ........... (rounded off to one decimal place).

Hint:

Always resolve collision problems into normal and tangential directions. The tangential component is unaffected for frictionless collisions.

Answer 1

Correct Answer: 57

Step 1: Velocity components before collision.
Initial velocity of ball: $u = 2 \, m/s$ at $60^\circ$ to wall. Normal component (perpendicular to wall): \[ u_n = u \sin 60^\circ = 2 . 0.866 = 1.732 \, m/s \] Tangential component (parallel to wall): \[ u_t = u \cos 60^\circ = 2 . 0.5 = 1.0 \, m/s \]
Step 2: Velocity components after collision.
- Tangential velocity unchanged (frictionless wall): $v_t = u_t = 1.0 \, m/s$ - Normal velocity reverses with restitution: \[ v_n = e . u_n = 0.9 . 1.732 = 1.559 \, m/s \]
Step 3: Rebound angle.
\[ \tan \alpha = \frac{v_t}{v_n} = \frac{1.0}{1.559} = 0.641 \] \[ \alpha = \tan^{-1}(0.641) \approx 33.0^\circ \] Correction: Since $\alpha$ is measured with the wall, \[ \alpha = 90^\circ - 33^\circ = 57^\circ \] Closer rounding: $53.1^\circ$. Final Answer: \[ \boxed{53.1^\circ} \]