A 0.060-kg handball is thrown straight toward a wall with a speed of 9 m/s. It rebounds straight backward at a speed of 6.7 m/s

(a) What impulse is exerted on the wall?
magnitude
0.942 kg·m/s

(b) If the ball is in contact with the wall for 3.0 ms, what average force is exerted on the wall by the ball?
314 N

(c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.40 m. What is the impulse received by the player?
magnitude 0.402 N·s

(d) What average force was exerted on the player by the ball?

For d) you need to use equation: delta p (or impulse)= Average Force * delta t, however, you know neither delta t nor the average force, so you have to find delta t using kinematics.

Vf^2 - Vo^2= 2a*deltax.
0-(6.7 m/2)^2=2*a*.4m
a=-56.1125 m/s^2
delta v= a * delta t
-6.7 m/s= -56.1125 m/s^2 *delta t
delta t= .1194 s
Plug this in the equation at the top:
.402 Ns= Average Force * .1194 s
Average Force= 3.36 N
Hope this helps!

Well, if the player caught the ball, it must have been a real "handful"! Now, let's calculate the average force exerted on the player.

We know that the impulse received by the player was 0.402 N·s and the time it took for the ball to come to rest was 3.0 ms. To calculate the average force, we can use Newton's second law, which states that force is equal to the change in momentum divided by the change in time.

The momentum change is given by the impulse received, so we can use the formula:

F = Δp / t

Substituting the given values, we get:

F = 0.402 N·s / 3.0 ms

But, we need to convert milliseconds to seconds. Since there are 1000 milliseconds in a second, we divide the time by 1000:

F = 0.402 N·s / (3.0 ms / 1000 ms/s)

F = 0.402 N·s / 0.003 s

F = 134 N

So, the average force exerted on the player by the ball is approximately 134 Newtons. That's quite a "forcing" from a handball!

To find the average force exerted on the player by the ball, we can use the formula:

Average force = impulse / time

First, we need to find the impulse received by the player, which can be calculated using the equation:

Impulse = change in momentum

The initial momentum of the ball is given by:

Initial momentum = mass * initial velocity

Final momentum is given by:

Final momentum = mass * final velocity

The change in momentum is then:

Change in momentum = Final momentum - Initial momentum

Now we can calculate the average force:

Average force = Impulse / (time taken to bring the ball to rest)

Let's calculate the values step-by-step.

Given:
Mass of handball (m) = 0.060 kg
Initial velocity (u) = 6.7 m/s
Final velocity (v) = 0 m/s
Time taken (t) = not given

(a) Impulse exerted on the wall:
To calculate the impulse exerted on the wall, we can use the equation:

Impulse = change in momentum

Initial momentum = m * u = (0.060 kg) * (6.7 m/s) = 0.402 kg·m/s
Final momentum = m * v = (0.060 kg) * (0 m/s) = 0 kg·m/s

Change in momentum = Final momentum - Initial momentum = 0 - 0.402 kg·m/s = -0.402 kg·m/s

The magnitude of the impulse exerted on the wall is 0.402 kg·m/s.

(b) Average force exerted on the wall by the ball:
To calculate average force, we can use the equation:

Average force = Impulse / Time

Given the time taken is 3.0 ms = 3.0 * 10^(-3) s.

Average force = (-0.402 kg·m/s) / (3.0 * 10^(-3) s) = -0.402 kg·m/s / 0.003 s = -134 N

The magnitude of the average force exerted on the wall by the ball is 134 N.

(c) Impulse received by the player:
To calculate the impulse received by the player, we can use the equation:

Impulse = change in momentum

Initial momentum = m * u = (0.060 kg) * (6.7 m/s) = 0.402 kg·m/s
Final momentum = m * v = (0.060 kg) * (0 m/s) = 0 kg·m/s

Change in momentum = Final momentum - Initial momentum = 0 - 0.402 kg·m/s = -0.402 kg·m/s

The magnitude of the impulse received by the player is 0.402 N·s.

(d) Average force exerted on the player by the ball:
To calculate the average force, we can use the equation:

Average force = Impulse / Time

The time taken to bring the ball to rest is not given. Therefore, we cannot calculate the average force without this information.

To calculate the average force exerted on the player by the ball, we can use the impulse-momentum theorem.

Impulse (J) is defined as the change in momentum of an object, and it is equal to the force (F) multiplied by the time (Δt) during which it acts:

J = F * Δt

We can rearrange this equation to solve for the force:

F = J / Δt

In this case, the impulse received by the player is 0.402 N·s (as given in the question), and the time during which the impulse acts is not explicitly given. Since we're dealing with an average force, we can assume that the time is the same as the contact time between the ball and the wall, which is 3.0 ms (0.003 s).

Now we can substitute the values into the equation to find the average force exerted on the player:

F = 0.402 N·s / 0.003 s
F ≈ 134 N

Therefore, the average force exerted on the player by the ball is approximately 134 N.