A smooth sphere is projected along a horizontal plane and collides obliquely with a vertical wall. Just before impact, the sphere is moving with a speed of \( 10 \mathrm{~m} / \mathrm{s} \) at an angle of \( 60^{\circ} \) to the wall. If the coefficient of restitution between the sphere and the wall is \( \frac{1}{4} \), find the speed of the sphere after the impact and the angle, \( \alpha \), which its direction of motion then makes with the wall.

To solve this problem, we need to analyze the motion of the sphere before and after the collision with the wall. ### Step 1: Break down the initial velocity The initial velocity \( \mathbf{v} \) of the sphere can be broken down into its components: - The horizontal component (perpendicular to the wall): \[ v_x = v \cos(\theta) = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \] - The vertical component (parallel to the wall): \[ v_y = v \sin(\theta) = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \] ### Step 2: Analyze the collision During the collision with the wall, the horizontal component of the velocity will change due to the coefficient of restitution \( e \). The vertical component will remain unchanged since there is no vertical force acting on the sphere. The coefficient of restitution is defined as: \[ e = \frac{\text{relative speed after collision}}{\text{relative speed before collision}} \] For our case: - The speed before collision in the horizontal direction is \( v_x = 5 \, \text{m/s} \). - The speed after collision in the horizontal direction will be \( v_x' = -e v_x \). Substituting the values: \[ v_x' = -\frac{1}{4} \times 5 = -\frac{5}{4} \, \text{m/s} \] The negative sign indicates that the sphere is moving in the opposite direction after the collision. ### Step 3: Determine the speed after the impact The vertical component remains unchanged: \[ v_y' = v_y = 5\sqrt{3} \, \text{m/s} \] Now, we can find the magnitude of the velocity after the impact: \[ v' = \sqrt{(v_x')^2 + (v_y')^2} = \sqrt{\left(-\frac{5}{4}\right)^2 + (5\sqrt{3})^2} \] Calculating this: \[ v' = \sqrt{\left(\frac{25}{16}\right) + (75)} = \sqrt{\frac{25}{16} + \frac{1200}{16}} = \sqrt{\frac{1225}{16}} = \frac{35}{4} \, \text{m/s} \] ### Step 4: Find the angle \( \alpha \) The angle \( \alpha \) that the direction of motion makes with the wall after the impact can be found using the tangent function: \[ \tan(\alpha) = \frac{v_y'}{|v_x'|} = \frac{5\sqrt{3}}{\frac{5}{4}} = 4\sqrt{3} \] Now, we can find \( \alpha \): \[ \alpha = \tan^{-1}(4\sqrt{3}) \] ### Final Results 1. The speed of the sphere after the impact is: \[ v' = \frac{35}{4} \, \text{m/s} \approx 8.75 \, \text{m/s} \] 2. The angle \( \alpha \) can be calculated using a calculator or trigonometric tables: \[ \alpha \approx \tan^{-1}(4\sqrt{3}) \approx 80.54^\circ \] Thus, the final answers are: - Speed after impact: \( \frac{35}{4} \, \text{m/s} \) - Angle \( \alpha \): \( \tan^{-1}(4\sqrt{3}) \) or approximately \( 80.54^\circ \) with respect to the wall.