Question

A racquetball of mass \(41.05 \mathrm{~g}\) has a speed of \(15.49 \mathrm{~m} / \mathrm{s}\) and collides with the wall of the court at an angle of \(43.53^{\circ}\) relative to the normal to the wall. The coefficient of restitution of the racquetball is \(0.8199 .\) What is the angle relative to the normal at which the ball leaves the wall?

Short answer

Answer: The racquetball leaves the wall at an angle of \(38.14^{\circ}\) below the horizontal normal.

Step-by-step solution

Understand the information given and find velocities parallel and perpendicular to the wall

We know the racquetball's speed before the collision (\(15.49 m/s\)), the angle it makes with the normal to the wall (\(43.53^\circ\)), and the coefficient of restitution (\(0.8199\)). We can use this information to find the components of the velocity parallel (\(V_{1x}\)) and perpendicular (\(V_{1y}\)) to the wall before the collision. $$V_{1x} = V_{1} \cos(43.53^\circ)$$ $$V_{1y} = V_{1} \sin(43.53^\circ)$$

Calculate \(V_{1x}\) and \(V_{1y}\)

Plug the values of \(V_1\) and the angle in the equations and calculate \(V_{1x}\) and \(V_{1y}\). $$V_{1x} = 15.49 \cos(43.53^\circ) \approx 11.22~m/s$$ $$V_{1y} = 15.49 \sin(43.53^\circ) \approx 10.61~m/s$$

Determine the components of the final velocity of the racquetball (\(V_{2x}\) and \(V_{2y}\))

After the collision with the wall, the perpendicular component of the velocity changes, but the parallel component remains the same due to the conservation of momentum. To find \(V_{2y}\), we will use the coefficient of restitution \(e\): $$e = \frac{|V_{2y} - V_{1y}|}{|V_{1y}|}$$ We know that after the collision, the velocity components change their directions so \(V_{2y} = -e V_{1y}\). Now we have both the final velocity components: $$V_{2x} = 11.22~m/s$$ $$V_{2y} = -0.8199 \times 10.61 \approx -8.68~m/s$$

Calculate the final angle (\(\theta\)) with the normal to the wall

Now that we have the components of the final velocity, we can use the tangent function to find the new angle \(\theta\) with respect to the normal to the wall: $$\tan(\theta) = \frac{V_{2y}}{V_{2x}}$$ Solve for \(\theta\): $$\theta = \arctan{\left(-\frac{8.68 \mathrm{~m/s}}{11.22 \mathrm{~m/s}}\right)} \approx -38.14^\circ$$ Since the angle is negative, it means that after the collision, the racquetball leaves the wall at an angle of \(38.14^{\circ}\) below the horizontal normal (and not above as before the collision).