Answer :
The impulse exerted on an object is equal to the change in the linear momentum of the object:
[tex]I=\Delta p[/tex]On the other hand, the linear momentum of an object is equal to the product of its mass and its velocity:
[tex]p=mv[/tex]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:
[tex]\Delta p=m\cdot\Delta v=m\cdot2v\sin \theta=2mv\sin \theta[/tex]Since the change in linear momentum is equal to the impulse, then:
[tex]2mv\sin \theta=I[/tex]Isolate the speed v from the equation:
[tex]\Rightarrow v=\frac{I}{2m\sin \theta}[/tex]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):
[tex]\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}[/tex]On the other hand, the average force exerted over an object is equal to the impulse per unit time:
[tex]F=\frac{I}{\Delta t}[/tex]Replace I=2.4Ns and Δt=5.5ms to find the average force exerted over the racquetball:
[tex]F=\frac{2.4Ns}{5.5\times10^{-3}s}=436.36N[/tex]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.