# Conservation of momentum

1. Jul 16, 2014

### Mltn12

1. The problem statement, all variables and given/known data

A 55-kg girl is standing near and to the left of a 75-kg boy on the frictionless surface of a frozen pond. The girl tosses a 1.8-kg ice ball to the girl with a horizontal speed of 7.5 m/s and he catches it. What are the velocities of the boy and the girl immediately after the boy catches the ball?

2. Relevant equations

MgVog+MbVb=0
Vfg=MballVb-Mb / Mg

3. The attempt at a solution

1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1

Vfb=1/2 MbVb/ ( -1.1Mg-Mb)
= -10x10^-2

I don't think I did this correctly, can someone walk me through this?

Last edited: Jul 16, 2014
2. Jul 16, 2014

### Nathanael

Try solving it for only the girl. If a 55 kg girl throws a 1.8kg ball at 7.5 m/s what will be her final speed? (When she starts at rest)

3. Jul 16, 2014

### SammyS

Staff Emeritus
Hello Mltn12. Welcome to PF !

4. Jul 16, 2014

### Mltn12

V=Vo+at?
A=t/v, but I don't have t.

5. Jul 16, 2014

### Mltn12

V=Vo+at?
A=t/v, but I don't have t.

6. Jul 16, 2014

### Nathanael

What about this equation? That is conservation of momentum, right? Shouldn't it also apply to the girl and the ball?

7. Jul 16, 2014

### Mltn12

Vfg=MballVb-Mb / Mg

1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1

8. Jul 16, 2014

### Nathanael

Where did you get $V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}$?

Use the other equation $M_{girl}V_{girl}+M_{ball}V_{ball}=0$

9. Jul 16, 2014

### Mltn12

(1.8 kg)( 7.5 m/s) / (55kg)
.25 m/s

10. Jul 16, 2014

### Nathanael

Yes correct.

Can you do something similar for the other person?

11. Jul 16, 2014

### Mltn12

MbVb+MballVball=0
(1.8kg)(7.5m/s) / 75
=.18

12. Jul 16, 2014

### Nathanael

Yes, that is right.

The reason is that the person slows down the momentum of the ball (by 1.8 kg times 7.5 m/s) and so the persons momentum must increase by the same amount

13. Jul 16, 2014

### Mltn12

So the final velocity for the girl would be .25m/s while the boys would .18 m/s?

14. Jul 16, 2014

### Nathanael

Yep!

15. Jul 16, 2014

### Mltn12

Wow really? It was that simple? I thought there was more to that problem! Thank you so much!!

16. Jul 16, 2014

### Nathanael

No problem!

Conservation of momentum is a very useful tool