Question

A 0.0950-kg ball is thrown at 28.0 m/s toward a brick wall. 1) Determine the impulse that the wall imparts to the ball when it hits and rebounds at 28.0 m/s in the opposite direction. The positive direction is toward the wall. (Express your answer to three significant figures). kg⋅mskg⋅ms Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 2) Determine the impulse that the wall imparts to the ball when it hits and rebounds at an angle of 45.0∘∘. The positive direction is toward the wall. (Express your answer to three significant figures). kg⋅mskg⋅ms Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 3) If the ball thrown in part (b) contacts the wall for 0.0100 s, determine the magnitude and direction of the force that the wall exerts on the ball. The positive direction is toward the wall. (Express your answer to three significant figures).

Answer 1

1) $-5.32 kg m/s$

The impulse exerted by the wall on the ball is equal to the change in momentum of the ball:

$I=\Delta p = m (v-u)$

where

m = 0.0950 kg is the mass of the ball

v = -28.0 m/s is the final velocity of the ball (negative because it is away from the wall)

u = +28.0 m/s is the initial velocity of the ball (positive because it is towards the wall)

Substituting into the equation, we find

$I=(0.0950 kg)(-28.0 m/s-(+28.0 m/s))=-5.32 kg m/s$

2) $-3.76 kg m/s$

The problem is similar to before, but this time we must consider only the component of the initial and final velocities that are perpendicular to the wall. So we have:

$u_x = u sin 45^{\circ}=(+28.0 m/s)(sin 45^{\circ} C)=+19.8 m/s$ is the component of the initial velocity perpendicular to the wall

$v_x = v sin 45^{\circ}=(-28.0 m/s)(sin 45^{\circ} C)=-19.8 m/s$ is the component of the final velocity perpendicular to the wall

Using again the formula for the impulse ,we find

$I=m(v_x-u_x)=(0.0950 kg)(-19.8 m/s-(+19.8 m/s))=-3.76 kg m/s$

3) -376 N

We know that the impulse is equal to the product between the average force exerted on the ball and the contact time:

$I=F \Delta t$

and in this case we have

$I=-3.76 kg m/s$ is the impulse

$\Delta t = 0.0100 s$ is the contact time

So we can solve the formula for F, and we find

$F=\frac{I}{\Delta t}=\frac{-3.76 kg m/s}{0.0100 s}=-376 N$

And the negative sign means the direction of the force is away from the wall.