Question

A racquetball of mass \(41.97 \mathrm{~g}\) has a speed of \(15.69 \mathrm{~m} / \mathrm{s}\) and collides with the wall of the court at an angle of \(48.67^{\circ}\) relative to the normal to the wall. The racquetball leaves the wall at an angle of \(55.75^{\circ}\) relative to the normal to the wall. What is the coefficient of restitution of the ball?

Short answer

The coefficient of restitution, \(e\), is given by the expression: $$ e = \frac{\sqrt{v_{2n}^2 + 11.43^2} \cos{55.75^{\circ}} - 10.47}{-10.47} $$

Step-by-step solution

Identify the known values

The mass of the ball is \(41.97 \,\text{g}\), the initial speed of the ball is \(15.69 \, \tfrac{\text{m}}{\text{s}}\), the angle of incidence is \(48.67^{\circ}\), and the angle of reflection is \(55.75^{\circ}\).

Calculate the velocity components along the normal and parallel to the wall before the collision

The velocity along the normal before the collision is given by \(v_{1n} = v_1 \cos{48.67^{\circ}}\) and the velocity parallel to the wall before the collision is given by \(v_{1p} = v_1 \sin{48.67^{\circ}}\). Using the values of the angle and initial speed, we find $$ v_{1n} = 15.69 \cos{48.67^{\circ}} \approx 10.47 \, \tfrac{\text{m}}{\text{s}} \\ v_{1p} = 15.69 \sin{48.67^{\circ}} \approx 11.43 \, \tfrac{\text{m}}{\text{s}} $$

Calculate the velocity components along the normal and parallel to the wall after the collision

Using the conservation of momentum, we can deduce that the velocity parallel to the wall does not change after the collision. Therefore, the velocity parallel to the wall after the collision is given by \(v_{2p} = v_{1p} = 11.43 \, \tfrac{\text{m}}{\text{s}}\). Now, we need to find the magnitude of the velocity of the ball after the collision, knowing the angle of reflection, \(55.75^{\circ}\). Let's denote the magnitude of the velocity of the ball after the collision as \(v_2\). The velocity along the normal after the collision is given by \(v_{2n} = v_2 \cos{55.75^{\circ}}\). For now, we'll leave \(v_{2n}\) in terms of \(v_2\).

Calculate the coefficient of restitution

The coefficient of restitution, denoted as \(e\), is defined as the ratio of the relative speeds of the object after and before the collision along the line of collision: $$ e = \frac{v_{2n} - v_{1n}}{-v_{1n}} = \frac{v_2 \cos{55.75^{\circ}} - 10.47}{-10.47} $$ The only unknown quantity in this formula is \(v_2\), the magnitude of the velocity of the ball after the collision. We have two equations relating \(v_2\) and \(v_{2p}\) (conservation of momentum parallel to the wall), and \(v_2\) and \(v_{2n}\): $$ v_{2p} = v_{1p} = 11.43 \\ v_{2n} = v_2 \cos{55.75^{\circ}} $$ From the first equation, we have \(v_2 \sin{55.75^{\circ}} = 11.43\). Now square both sides of the two equations and then add them together to get: $$ v_2^2 \cos^2{55.75^{\circ}} + v_2^2 \sin^2{55.75^{\circ}} = v_{2n}^2 + 11.43^2 $$ Now using the fact that \(\cos^2{\alpha} + \sin^2{\alpha} = 1\) (a trigonometric identity), we get \(v_2^2 = v_{2n}^2 + 11.43^2\). Hence, we get: \(v_2 = \sqrt{v_{2n}^2 + 11.43^2}\) Now, we can plug this back into the formula for \(e\): $$ e = \frac{\sqrt{v_{2n}^2 + 11.43^2} \cos{55.75^{\circ}} - 10.47}{-10.47} $$ Finally, to find the numerical value of \(e\), we need to solve for \(v_{2n}\) in the expression of \(v_2\). As we have only one equation for \(v_{2n}\), we don't have a unique solution, and hence the coefficient of restitution depends upon the final velocity along the normal (\(v_{2n}\)).