# Homework Help: Momentum Questions

1. Oct 10, 2005

### shannons

Hello,

Due to a teacher strike where I live, I've been trying to teach myself the momentum unit . Could anyone go through these questions with me? Or tell me what formulas I should use to solve these problems?

1. A mass of 6.3 kg traveling at 6m/s is acted on by a force giving an impulse of -31.5 N/s. What is the velocity of the mass after the impulse?

I know that I have to do something with the final momentum and the intial momentum, but I don't know what formula to use.

2. What force must be imparted to a 100g baseball to change the velocity from 40.0 m/s to -50.0 m/s in 1.2 ms?
b) what is the impulse of this force?

thanks

2. Oct 10, 2005

### arildno

Derive it then, from Newton's second law and the definition of impulse.

3. Oct 10, 2005

### shannons

When I tried that I got:

delta p = pf-pi (change in momentum= momentum final - momentum initial)
deltap=m(vf)-m(vi)
37.8=6.3(vf)-6.3(6)

The answer for number 1 is supposed to be 1.0 m/s. Where am I going wrong?

4. Oct 10, 2005

### arildno

Well, putting in the wrong number, for startes.
Where did you get 37.8 from, it is supposed to be -31.5

5. Oct 10, 2005

### shannons

I thought I was supposed to calculate the momentum and put that in...

-31.5= 6.3(Vf) - 6.3(6)
-31.5/6.3= Vf - 37.8
-42.8= Vf

Is that better?

6. Oct 10, 2005

### arildno

Impulse equals change in momentum, okay?
So, you have been given the impulse as -31.5, agreed?
You also know that initial momentum is 6.3*6=37.8, agreed?
Thus, you have:
-31.5=6.3*Vf-37.8
That is, adding 37.8 to both sides:
6.3=6.3*Vf

Understood?

7. Oct 10, 2005

### shannons

I see, I did it in the wrong order. Thank you so much!

8. Oct 10, 2005

### arildno

No problem.

9. Oct 10, 2005

### shannons

Would I use the same formula for the next problem too?

10. Oct 10, 2005

### arildno

It is simplest to answer b) first, and then determine the force afterwards.

11. Oct 10, 2005

### shannons

2. What force must be imparted to a 100g baseball to change the velocity from 40.0 m/s to -50.0 m/s in 1.2 ms?
b) What is the impulse of this force?

delta p= pf-pi
delta p= m(vf)-m(vi)
delta p= 100(-50)-100(40)
delta p= -5000 -4000
= -9000

is that how I set that up?

If i divide the impulse by the time I get the force: -9000/ 1.2 = -7500. Which gives me the right answer for a.

Last edited: Oct 10, 2005
12. Oct 10, 2005

### arildno

Rather, you should say:
a) By DEFINITION of the symbol delta(p), delta(p) = mvf-mvi.
b)By Newton's 2.law, we have
I=delta(p), where I is the impulse.

Thus, you have:
I=mVf-mvi

You have derived the right answer for I.

13. Oct 10, 2005

### shannons

Thats an easier way of thinking of it :), thank you.

How do I get the correct impulse? In the answer key, it says that the answer for b. is -9 N/s

edit: I think the answer key might be wrong:( thank you for your help, I'm starting to understand this now :D.

Last edited: Oct 10, 2005
14. Oct 10, 2005

### arildno

Oh, I see, your baseball weigh 100g, not 100kg as you used..