# Homework Help: Conservation of Momentum and energy problem help

1. Jan 25, 2006

### kgbwolf

SOmeone explain to me this because im lost. i got 2 masses one velecity. the equation is like is mass times volocity + mass times velocity=0

122kg*22m/s+14kg*v=0 I made these numbers up what i do is multiply and add then move my anwser to the other side becoming negative but that doesnt seem to be the right anwser.

2. Jan 25, 2006

### kgbwolf

ki even tried multipling then dividing by the remaining number still gives me the wrong anwser

3. Jan 25, 2006

### Hootenanny

Staff Emeritus
Generally;
$$M_{1}V_{1} + M_{2}V_{2} = 0$$
Then Rearranging
$$V_{2} = - \frac{M_{1}V_{1}}{M_{2}}$$

4. Jan 25, 2006

### kgbwolf

Ok i worked the first problem out now i have a nother problem

A 125kg pile driver falls from a hieght of 10m to hit a piling.

From this i see mass is 125kg and distance is 10m and it says find the speed it hits the piling? and the momentum how can i do this with just mass and distance or is distance really velocity?

5. Jan 25, 2006

### Hootenanny

Staff Emeritus
Use conservation of energy. At 10m up the piledriver will have a potential energy of mgh. At the instant it hits the pile all the potential energy will be converted to kinetic $\frac{1}{2}mv^2$. You should be able to work out the velocity by rearranging and hence the momentum.

6. Jan 25, 2006

### kgbwolf

i did 1/2*m moved it to the other side negative and ttok the square root and 7.91m/s

7. Jan 25, 2006

### kgbwolf

and the right anwser is 14m/s

8. Jan 25, 2006

### Hootenanny

Staff Emeritus
I got a different answer. What value did you get for the potential energy?

9. Jan 25, 2006

### Hootenanny

Staff Emeritus
Yes, I got 14m/s.

10. Jan 25, 2006

### kgbwolf

12,250Newtons

11. Jan 25, 2006

### kgbwolf

how?................

12. Jan 25, 2006

### Hootenanny

Staff Emeritus
Check you units. Newtons for energy? The value is correct. Now show me how you re-arrange the kinetic energy equation.

13. Jan 25, 2006

### kgbwolf

do u have msn IM??

14. Jan 25, 2006

### Hootenanny

Staff Emeritus
No..............

15. Jan 25, 2006

### kgbwolf

1/2 times 125 = 62/1/2 and i took the square root gives me 7.9

16. Jan 25, 2006

### Hootenanny

Staff Emeritus
$$E_{k} = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{\frac{2E_{k}}{m}}$$

17. Jan 25, 2006

### kgbwolf

Whats Ek........

18. Jan 25, 2006

kinetic energy

19. Jan 25, 2006

### kgbwolf

potential got it

20. Jan 25, 2006

### Hootenanny

Staff Emeritus
The potential energy you calculated it would all be converted into kinetic energy. The same equation could be written
$$v = \sqrt{\frac{2mgh}{m}} \Rightarrow v = \sqrt{2gh}$$

21. Jan 25, 2006

### kgbwolf

ok.......... got it