Impulse/Momentum ball into wall Problem

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In summary: and so the impulse the wall applies is equal in magnitude and opposite in direction to the impulse the ball applies.
  • #1
adorrkable
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Homework Statement


1) A 0.1 kg ball is rolling across the floor at 0.5 m/s when it hits a wall and reverses direction without any change in speed.

A) Can you determine the impulse delivered to the ball by the wall? If so, what is it? If not, explain why not.

B) Can you determine the impulse delivered by the ball to the wall? If so, what is it? If not, explain why not.

C) Can you determine the force exerted by the wall on the ball? If so, what is it? If not, explain why not.


Homework Equations


Jnet = [tex]\Delta[/tex]p
p = m*v
J = Fnet * [tex]\Delta[/tex]t


The Attempt at a Solution


My problem - I managed to calculate that the change in momentum for the ball is -0.1 m/s, but if this is the amount of impulse it exerts on the wall (since Jnet = Δp), wouldn't the wall exert an equal and opposite impulse? But wouldn't the wall's momentum and therefore change in momentum be zero, because Δp = m * Δv, and the wall never has any velocity.
 
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  • #2
Hi adorrkable,

adorrkable said:

Homework Statement


1) A 0.1 kg ball is rolling across the floor at 0.5 m/s when it hits a wall and reverses direction without any change in speed.

A) Can you determine the impulse delivered to the ball by the wall? If so, what is it? If not, explain why not.

B) Can you determine the impulse delivered by the ball to the wall? If so, what is it? If not, explain why not.

C) Can you determine the force exerted by the wall on the ball? If so, what is it? If not, explain why not.


Homework Equations


Jnet = [tex]\Delta[/tex]p
p = m*v
J = Fnet * [tex]\Delta[/tex]t


The Attempt at a Solution


My problem - I managed to calculate that the change in momentum for the ball is -0.1 m/s

Not to be picky, but remember the units are kg m/s.

, but if this is the amount of impulse it exerts on the wall (since Jnet = Δp),

Actually that's the impulse the wall exerts on the ball. When you apply:

[tex]
\mbox{impulse} = \Delta p
[/tex]

the two sides of the equation have to be applied to the same system. But yes, the impulse from the wall on the ball is equal in magnitude and opposite in direction from the impulse of the ball on the wall.

wouldn't the wall exert an equal and opposite impulse? But wouldn't the wall's momentum and therefore change in momentum be zero, because Δp = m * Δv, and the wall never has any velocity.

We wouldn't say that the no part of the wall never has any velocity, it's just that it's small enough to be ignored here. The wall will react in some way to the impact, but it's not important to the problem.
 
  • #3
So the ball exerts an impulse of -0.1 kg m/s, and the wall exerts an impulse of 0.1 kg m/s, am I correct?

Thank you very much!
 
  • #4
adorrkable said:
So the ball exerts an impulse of -0.1 kg m/s, and the wall exerts an impulse of 0.1 kg m/s, am I correct?

If the ball's change in momentum is -0.1 kg m/s, then that is what happened to it; and so the impulse of -0.1 kg m/s is what was applied to the ball (and the negative of that is what the ball applied to the wall).

That's what I meant about the equation; if you're looking at the ball as your system, then:

(impulse applied to ball)=(change in ball's momentum)
 

FAQ: Impulse/Momentum ball into wall Problem

1. What is the equation for calculating momentum?

The equation for calculating momentum is p = mv, where p is momentum, m is mass, and v is velocity.

2. How is impulse related to momentum?

Impulse is the change in momentum, or J = Δp. This means that the greater the impulse, the greater the change in momentum.

3. What happens to the momentum of a ball when it hits a wall?

When a ball hits a wall, its momentum changes. The ball's velocity decreases to zero, resulting in a change in momentum. The wall exerts an equal and opposite force on the ball, causing the change in momentum.

4. How does the mass of the ball affect its momentum?

The mass of an object directly affects its momentum. The greater the mass, the greater the momentum at a given velocity. This means that a heavier ball will have more momentum than a lighter ball at the same speed.

5. Can you use the impulse/momentum equation to solve for the force of impact in a ball into wall problem?

Yes, the impulse/momentum equation can be rearranged to solve for force, F = Δp/Δt. This means that by knowing the change in momentum and the time of impact, you can calculate the force of impact between the ball and the wall.

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