Question

A 40 g racquetball bounces off of the back wall (theta=39°) of a racquetball court as shown with the same initial and final speed v. if the magnitude of the impulse on the racquetball is 2.4Ns. find the followingA.the speed V B. if the racquetball was in contact with the wall for 5.5 ms find the average force that the wall exered on the racquetball?

Answer 1

The impulse exerted on an object is equal to the change in the linear momentum of the object:

$I=\Delta p$

On the other hand, the linear momentum of an object is equal to the product of its mass and its velocity:

$p=mv$

In the given problem, the magnitude of the velocity does not change but its direction does. Initially, the vertical component of the vleocity is equal to v*sinθ towards the wall and after the collision, the vertical component of its velocity is equal to v*sinθ away from the wall.

Then, the total change in the velocity of the ball is equal to 2v*sinθ. Then, the change in the linear momentum of the wall is:

$\Delta p=m\cdot\Delta v=m\cdot2v\sin \theta=2mv\sin \theta$

Since the change in linear momentum is equal to the impulse, then:

$2mv\sin \theta=I$

Isolate the speed v from the equation:

$\Rightarrow v=\frac{I}{2m\sin \theta}$

Replace I=2.4Ns, m=40.0g, θ=39º to find the magnitude of the speed v (remember to write all the quantities using SI units before plugging in the values on a calculator):

$\begin{gathered} v=\frac{2.4Ns}{2(40.0g)\sin(39º)} \\ =\frac{2.4Ns}{2(40.0\times10^{-3}kg)\sin(39º)} \\ =47.67\frac{m}{s} \end{gathered}$

On the other hand, the average force exerted over an object is equal to the impulse per unit time:

$F=\frac{I}{\Delta t}$

Replace I=2.4Ns and Δt=5.5ms to find the average force exerted over the racquetball:

$F=\frac{2.4Ns}{5.5\times10^{-3}s}=436.36N$

Therefore, the answers are:

A)

The speed of the ball is approximately 46.7 meters per second.

B)

The average force exerted on the ball is approximately 436 Newtons.