# Impulse and momentum question

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!

Last edited:

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I think I am right but I do not have the answers to 1.4) and would rather be safe than sorry!

cepheid
Staff Emeritus
Gold Member
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 shouldnt 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

Thanks!