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

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To calculate the impulse exerted by the wall on a ball, the mass of the ball (0.3 kg) and its initial and final velocities (15 m/s and -15 m/s, respectively) are used. The change in momentum is found to be -9 kgm/s, indicating the direction away from the wall. The impulse is equal to this change in momentum, confirming it is also 9 kgm/s away from the wall. The calculations and reasoning presented are correct, affirming the understanding of impulse and momentum in elastic collisions.
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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!
 
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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!
 
v0rtexza said:
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.

v0rtexza said:
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!
 
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