Finding final velocity with conservation of momentum

[kgbwolf]

SOmeone explain to me this because I am 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 doesn't seem to be the right anwser.

 

[kgbwolf]

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

 

[Hootenanny]

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

 

[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?

 

[Hootenanny]

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.

 

[kgbwolf]

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

 

[kgbwolf]

and the right anwser is 14m/s

 

[Hootenanny]

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

 

[Hootenanny]

Yes, I got 14m/s.

 

[kgbwolf]

12,250Newtons

 

[kgbwolf]

how?...

 

[Hootenanny]

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

 

[kgbwolf]

do u have msn IM??

 

[Hootenanny]

No.....

 

[kgbwolf]

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

 

[Hootenanny]

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

 

[kgbwolf]

Whats Ek...

 

[courtrigrad]

kinetic energy

 

[kgbwolf]

potential got it

 

[Hootenanny]

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

 

[kgbwolf]

ok... got it