- #1

v0rtexza

- 13

- 0

**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:

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:

- Calculate the momentum of the ball before it strikes the wall
- Calculate the momentum of the ball after it strikes the wall
- Calculate the change in the momentum of the ball
- 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: