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

  • Thread starter Thread starter v0rtexza
  • Start date Start date
  • Tags Tags
    Impulse Momentum
Click For Summary
SUMMARY

The discussion focuses on calculating the impulse exerted by a wall on a ball during an elastic collision. A ball with a mass of 300 g (0.3 kg) is thrown at a speed of 15 m/s and rebounds elastically, resulting in a change in momentum of -9 kgm/s. The impulse exerted by the wall on the ball is confirmed to be equal to this change in momentum, thus 9 kgm/s away from the wall. The calculations utilize the equations FΔt = mΔv and Δp = mΔv, ensuring clarity in notation for momentum and velocity.

PREREQUISITES
  • Understanding of elastic collisions
  • Familiarity with momentum and impulse concepts
  • Proficiency in using the equations FΔt = mΔv and Δp = mΔv
  • Basic knowledge of physics units (kg, m/s)
NEXT STEPS
  • Study the principles of elastic and inelastic collisions in physics
  • Learn about the conservation of momentum in closed systems
  • Explore advanced applications of impulse in real-world scenarios
  • Review the mathematical derivation of impulse-momentum theorem
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to momentum and impulse in elastic collisions.

v0rtexza
Messages
13
Reaction score
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:

  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:
Physics news on Phys.org
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!
 
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!
 

Similar threads

Replies
10
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
1
Views
2K
Replies
7
Views
2K