Linear momentum and impulse problem

[BrainMan]

Homework Statement

A 0.3-kg ball falls from a height of 30 m from rest and rebounds with half its velocity upon impact with a sidewalk. Find (a) the momentum delivered to the Earth and (b) the impulse delivered to the earth.

Homework Equations

FΔt= mv_f-mv_i

The Attempt at a Solution

What I tried to do was I found the velocity using the potential energy formula to find the energy and then the kinetic energy formula to find the velocity right before the ball hit the ground. I then multiplied the velocity by the mass to find the initial momentum. After that I divided the velocity by two because it said that when it rebounded it had half the velocity. Then I multiplied half the initial velocity by the mass to get the final momentum. Then I used the equation FΔt= mv_f-mv_i to find the impulse. I got -3.7125 and the answer was 11 for both (a) and (b)

 

[Simon Bridge]

What I tried to do was I found the velocity using the potential energy formula to find the energy and then the kinetic energy formula to find the velocity right before the ball hit the ground. I then multiplied the velocity by the mass to find the initial momentum. After that I divided the velocity by two because it said that when it rebounded it had half the velocity. Then I multiplied half the initial velocity by the mass to get the final momentum. Then I used the equation FΔt= mv_f-mv_i to find the impulse. I got -3.7125 and the answer was 11 for both (a) and (b)

Not too bad, but you missed out a bit.
1. Momentum is a vector (and it has units)
2. The collision involves the Earth - so you need to consider it in the calculation.
Law of conservation of momentum applies here.Note: It is easier to just use momentum directly, so you avoid awkward numbers.
i.e. if velocity halves and mass stays the same then momentum halves.
if it ends up going in the opposite direction, and the initial momentum was positive, then the final momentum is negative.
$$K=\frac{mv^2}{2}=\frac{p^2}{2m}$$... since $v=p/m$
You want the gravitational PE lost falling turning into kinetic energy at the moment of contact so:$$\frac{p^2}{2m}=mgh \implies p=m\sqrt{2gh}$$... see?

 

[BrainMan]

Not too bad, but you missed out a bit.
1. Momentum is a vector (and it has units)
2. The collision involves the Earth - so you need to consider it in the calculation.
Law of conservation of momentum applies here.


Note: It is easier to just use momentum directly, so you avoid awkward numbers.
i.e. if velocity halves and mass stays the same then momentum halves.
if it ends up going in the opposite direction, and the initial momentum was positive, then the final momentum is negative.
$$K=\frac{mv^2}{2}=\frac{p^2}{2m}$$... since $v=p/m$
You want the gravitational PE lost falling turning into kinetic energy at the moment of contact so:$$\frac{p^2}{2m}=mgh \implies p=m\sqrt{2gh}$$... see?

OK I see what I did wrong. I didn't make the velocity after the ball had bounced negative. I got the right numbers with my method but the negative threw off my answer. Thanks!

 

[Simon Bridge]

Well done :)

 

[HalfLostAndSearching]

respectively.Your approach is correct. However, there is a slight error in your final calculation. The impulse delivered to the Earth should be a positive value, since the force and change in time are both positive quantities. Therefore, the correct answer for (b) would be 3.7125 Ns. Other than that, your solution is correct. Keep up the good work!