# Homework Help: Momentum versus kinetic energy

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1. Oct 19, 2014

### smcclenaghan

1. The problem statement, all variables and given/known data
A 2000kg car is moving at 50km/h. What is the force required to stop it?

2. Relevant equations
KE=1/2m(v^2)
p=m*v

3. The attempt at a solution

1/2(2000kg)*(50km/h)^2 =
1000kg*2500km^2/h^2 =
2,500,000 km^2/h^2
converting from hours to seconds and km to m (to get joules)
2,500,000 km^2/h^2 = 2,500,000,000 m^2/h^2
2,500,000,000 m^2/h^2 / 60 / 60 = 694444.4... J

My answer may or may not be right. (I'm actually not so concerned with the math). Rather, I have a fundamental problem understanding the difference between KE and momentum. Why am I using the formula for KE to compute the energy to stop a moving mass and not the formula for momentum?

If not here, when is the formula for momentum used?

I am extremely confused on whether to use KE or momentum to handle collisions. I've come across an analogy that KE is the work, and momentum is the transportation for that work.

The equal and opposite reaction (Newton's third law), is this based off of momentum or KE?

I really would love a simple place to start for a comparison of these two terms. Thanks for any help.

2. Oct 19, 2014

### NTW

The question that you state should be completed as follows:

(...) What is the force required to stop it in x seconds?

or:

(...) What is the force required to stop it in y meters?

Since no mass in motion can be stopped instantaneously, or in no distance...

3. Oct 19, 2014

### smcclenaghan

Ok, I think I follow.

If it is in seconds, we're looking at momentum. If it is in meters, we're looking at KE.

4. Oct 19, 2014

### Staff: Mentor

No. NTW was pointing out that your question cannot be answered, there is insufficient information.

We need to be given information on how quickly the vehicle is to be stopped.

5. Oct 19, 2014

### smcclenaghan

Right. So if the question were asking how much force in X seconds, we'd be looking at the formula for momentum.
And if the question were asking how much force in X meters, we'd be looking at the formula for KE.

Am I on the right track?

(I understand the original question is not answerable as posted).

6. Oct 20, 2014

### Staff: Mentor

Yes, you would be able to solve for the force that way.

7. Oct 20, 2014

### smcclenaghan

I appreciate the response, NascentOxygen (and to you NTW).