1. Oct 11, 2008

### ~*Noor*~

My Question:
A 1500 kg car moving east at 38 m/s collides with a 1780 kg car moving south at 15 m/s and the two cars stick together.
(a) What is the velocity of the cars right after the collision? _______m/s ______o

(b) How much kinetic energy was converted to another form during the collision? ______kJ

My attempt:

So i used the momentum formula and got the total momentum before collision and made it equal to the momentum after the collsion but didn't get the right answer.

Before collision: car1:
p=m1v1=(1500kg)(38m/s)=57000

car2:
p=m2v2=(1780kg)(-15m/s)=-26700

p1+p2=P total
57000+(-26700)=30300

After collision: car1:
p=1500v1

car2:
p=1780v2

v1=v2 because of collision therefore

30300=(1500+1780)v
v=9.24m/s

when i took out the negative i got a different answer (25.5m/s) which was also wrong.

Part B i've no idea how to do:uhh:

Please tell me what i'm doing wrong!!!!!

2. Oct 11, 2008

### Galileo's Ghost

Re: Momentum.....Please urgent help!!!

Remember that velocity, and hence momentum, are vector quantities and need to be treated as such when adding them together. I think you attempted to do this with the minus sign, but these vectors are not opposite in direction. They are at 90 degrees to one another. Draw a diagram of the momenta vectors. It may help you see how to combine them for the total.

Also, for part b, remember Kinetic energy is not a vector quantity. determine the total kinetic energy before the collision and compare it to thee total kinetic energy after the collision and determine the difference.

3. Oct 12, 2008

### ~*Noor*~

Re: Momentum.....Please urgent help!!!

First of all........thanks Galileo's Ghost for all your help

Ok.....part a: Yes i understand what your saying....my first try was without the negative sign and i still got that wrong.....so it must be my method that's messed up

And how am i supposed to find the angle?????

Lastly as long as i know how to get part a then i can get part b.

4. Oct 13, 2008