Calculating Momentum Change in Elastic Collisions

[superaznnerd]

Homework Statement

a ball of 2kg falls from rest froma 5m height and collides elastically with the floor below. what is the magnitude change in the object's momentum?

Homework Equations

The Attempt at a Solution

I did S=So+.5aT^2
I solved for the time it takes to reach the ground, which is 1 second.
Next, I figured out that the velocity on impact with ground is 10m/s
I did 10m/s *2kg = 20kgm/s

however, the answer is 40kgm/s. Can someone explain? thanks

 

[rock.freak667]

You have the initial momentum, you need to get the velocity after impact to get the final momentum.

 

[inky]

You have the initial momentum, you need to get the velocity after impact to get the final momentum.

But problem asked the change in momentum.

Impulse=change in momentum=Ft=mg*∆t

 

[Redbelly98]

They want the change in momentum during the impact with the floor, not during the initial drop.

 

[Borek]

Think what "elastically" means.

 

[inky]

Think what "elastically" means.

I think elastically means after collision the ball moves up. If we consider this, we need coefficient of restitution.

 

[Redbelly98]

The coefficient of restitution = 1 for an elastic collision.

Please let the OP solve the problem now.

 

[inky]

The coefficient of restitution = 1 for an elastic collision.

Please let the OP solve the problem now.

e=1 for perfectly elastic collision. I mean before collisin and after collision speed are the same for e=1. Actual case is not perfectly elastic.

 

[Redbelly98]

Disagreed. From Giancoli's book Physics (5th ed.): "a collision, in which total kinetic energy is conserved, is called an elastic collision". And reading further: "collisions in which kinetic energy is not conserved are said to be inelastic collisions", and "if two objects stick together as a result of a collision, the collision is said to be completely inelastic."

 

[superaznnerd]

now I get the answer is 0. right before the collision it is 20kgm/s down. right after it is 20kgm/s up. Thus, there is no change in magnitude of momentum...

rebelly: can you explain how the stuff you wrote is relevant??

 

[Borek]

now I get the answer is 0. right before the collision it is 20kgm/s down. right after it is 20kgm/s up. Thus, there is no change in magnitude of momentum...

Hm, could be my English fails me here, but looks like you are right. Momentum magnitude doesn't change, even if momentum changes by 40 kgm/s. But I can be wrong.

What you posted is a direct quote of the question wording, or was it translated?

rebelly: can you explain how the stuff you wrote is relevant??

You can safely ignore discussion with inky.

 

[Redbelly98]

now I get the answer is 0. right before the collision it is 20kgm/s down. right after it is 20kgm/s up. Thus, there is no change in magnitude of momentum...

But there is a change in the direction of momentum. Momentum is a vector, so there is a change in the momentum. Hint: express the upward momentum as positive, and downward as negative.

rebelly: can you explain how the stuff you wrote is relevant??

It was addressed to stuff inky posted, don't worry about it.

 

[Redbelly98]

what is the magnitude change in the object's momentum?

Okay, I think I see the source of confusion here.

They are asking you to first find the change in momentum, and then get the magnitude of that change. It is not asking for the change in the magnitude of momentum.

 

[superaznnerd]

that sounds right ill ask my teacher 2moro