How Do You Calculate the Impulse Exerted by the Wall on a Ball?

[v0rtexza]

I am having trouble on the fourth part of a question that states calculate the impulse exerted by the wall on the ball. The original question was:

A ball of mass 300 g is thrown against a wall with a speed of 15 m/s. The ball rebounds elastically off the wall. The original questions were:

1. Calculate the momentum of the ball before it strikes the wall
2. Calculate the momentum of the ball after it strikes the wall
3. Calculate the change in the momentum of the ball
4. Calculate the impulse exerted by the wall on the ball

Homework Equations

FΔt = mΔv
Δp = mΔv
mΔv = m(v-u) = mv - mu = momentum now - momentum previous (change in momentum)

Note:
F = force
Δt = time
m = mass (in kilograms)
Δv = velocity
Δp = impulse
v = Vf or final velocity
u = Vi or initial velocity

The Attempt at a Solution

1.1) m = 0,3 kg
vi = 15 m/s
vf = -15 m/s (elastic collision)
Δp = mΔv
therefore Δp = 0,3 x 15 = 4,5 kgm/s

1.2) Δp = mΔv
Δp = 0,3 x -15
Δp = -4,5 kgm/s or 4,5 kgm/s away from the wall

1.3) Δp = m(v-u)
= 0,3(-15-15)
= -9 kgm/s or 9 kgm/s away from the wall

I think my current solutions for the above are correct but I am unsure of how to calculate 1.4

1.4) I think that Δp = mΔv and we know that the change in momentum of the ball is -9 kgm/s from 1.3)

so then do we assume that the impulse is 9 kgm/s away from the wall too as Δp = mΔv? Or would this be an incorrect way of thinking?

This is my first post here on PF so I hope I have laid out my question correctly.

Thanks, any help is appreciated!

 

[v0rtexza]

If there is any more information required, Please don't hesitate to ask...
I think I am right but I do not have the answers to 1.4) and would rather be safe than sorry!

 

[cepheid]

I am having trouble on the fourth part of a question that states calculate the impulse exerted by the wall on the ball. The original question was:

A ball of mass 300 g is thrown against a wall with a speed of 15 m/s. The ball rebounds elastically off the wall. The original questions were:

1. Calculate the momentum of the ball before it strikes the wall
2. Calculate the momentum of the ball after it strikes the wall
3. Calculate the change in the momentum of the ball
4. Calculate the impulse exerted by the wall on the ball

Homework Equations

FΔt = mΔv
Δp = mΔv
mΔv = m(v-u) = mv - mu = momentum now - momentum previous (change in momentum)

Note:
F = force
Δt = time
m = mass (in kilograms)
Δv = velocity
Δp = impulse
v = Vf or final velocity
u = Vi or initial velocity

The Attempt at a Solution

1.1) m = 0,3 kg
vi = 15 m/s
vf = -15 m/s (elastic collision)
Δp = mΔv
therefore Δp = 0,3 x 15 = 4,5 kgm/s

1.2) Δp = mΔv
Δp = 0,3 x -15
Δp = -4,5 kgm/s or 4,5 kgm/s away from the wall

Your answers and method are correct above, but your notation is wrong. The Δ (delta) symbol means "change", so you shouldn't include it in 1.1 and 1.2, because you're not looking for the change in momentum or velocity. You should just write p and v. I mention this because it was confusing at first what you were doing.

1.3) Δp = m(v-u)
= 0,3(-15-15)
= -9 kgm/s or 9 kgm/s away from the wall

I think my current solutions for the above are correct but I am unsure of how to calculate 1.4

1.4) I think that Δp = mΔv and we know that the change in momentum of the ball is -9 kgm/s from 1.3)

so then do we assume that the impulse is 9 kgm/s away from the wall too as Δp = mΔv? Or would this be an incorrect way of thinking?

Yes, of course. The impulse is equal to the change in momentum, which you computed correctly in 1.3

 

[v0rtexza]

Thanks!

 

[Nickie]

Hello,

Thank you for reaching out with your question. It seems like you are on the right track with your solutions for the first three parts of the question. To calculate the impulse exerted by the wall on the ball, you can use the formula FΔt = mΔv, where F is the force exerted by the wall on the ball, Δt is the time interval during which the force is applied, m is the mass of the ball, and Δv is the change in velocity of the ball.

In this case, we know that the ball has a mass of 300 g (0.3 kg) and that its velocity changes from 15 m/s before the collision to -15 m/s after the collision. This means that the change in velocity is 30 m/s (since the direction of velocity changes from positive to negative). Now, we need to determine the time interval during which the force is applied.

We can use the fact that the ball rebounds elastically off the wall to determine the time interval. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. This means that the ball has the same kinetic energy before and after the collision. We can use the formula for kinetic energy, KE = 1/2 mv^2, to calculate the kinetic energy of the ball before the collision and after the collision.

KE before = 1/2 (0.3 kg) (15 m/s)^2 = 33.75 J
KE after = 1/2 (0.3 kg) (-15 m/s)^2 = 33.75 J

Since the kinetic energy is the same before and after the collision, we can set these two equations equal to each other and solve for the time interval, Δt.

1/2 mv^2 = 1/2 mv^2
⇒ 1/2 (0.3 kg) (15 m/s)^2 = 1/2 (0.3 kg) (-15 m/s)^2
⇒ 33.75 J = 33.75 J
⇒ Δt = 0.001 s

Now, we can use this value for Δt in the equation FΔt = mΔv to calculate the impulse.

FΔt = mΔv
⇒ F (0.001 s) = (0