# Momentum/Kinetic Energy

1. Apr 1, 2006

### Drevin

Two objects with a mass of 4kg are sliding across ice at a constant velocity of 12 m/s east. After an explosive charge seperates the two masses, they are found to be travelling at 50 degrees north of east and 50 degrees south of east respectively. What is the change in kinetic energy of the system?

I got as far as to find out the inital kinetic energy from Ek = 1/2mv^2, but I'm not sure what to do next? Thanks.

(BTW, I hope this is in the right section...)

2. Apr 1, 2006

### Pengwuino

Well, if we take the assumption that the x-direction velocity doesn’t change, we only need to find the velocity in the y direction. If we know what the x-direction velocity is and we know the angle, you can determine what the velocity in the y-direction will be right?

From that, you can determine the new kinetic energy for both blocks. Compare that to the original kinetic energy and you’ll determine the added kinetic energy.

3. Apr 1, 2006

### Drevin

Ah, yes! That makes sense. After I find out the velocity in the y direction, should I use that velocity for kinetic energy, or should I use a^2 + b^2 = c^2 to find out the hypotenuse and use that velocity? Thanks!

4. Apr 1, 2006

### Pengwuino

You need to use the resultant velocity (hypotenus), you can do that or you can just multiply 1/cos(50) * x direction to give you the hypotenus

5. Apr 1, 2006

### Pengwuino

There is no change though since they both go off at equal angles and are of equal mass and are initially going at equal velocities. I know you would normally have to check conservation of momentum but since all things are equal, the vertical momentums cancel out.

Last edited: Apr 1, 2006
6. Apr 1, 2006

### nrqed

Yes, I agree. Sorry, Ideleted my message. I had misread the angles (for some reason I thought that one was 30 degrees and the other one 50 degrees.

7. Apr 1, 2006

### Pengwuino

Ohh i see.

As it pertains to the original poster, you always want to make sure that you check conservation of momentum like nrqed did. If they were different masses or went off at different angles and you weren't explicitely told the x-velocity changed, they are probably trying to catch you to see if you do verify conservation laws. It's a weakness many people have (like me haha) where they just do not make sure momentum and energy are conserved and the problem is trying to catch you on it.

8. Apr 1, 2006

### nrqed

Indeed...That's a good point.

A reason that I posted so quickly is that you had started your post by saying ''if we make the assumption that the x component of the velocity is the same after..'' which made react by wanting to warn you and the OP (because I still had the 30 and 50 degrees figures in my head).
However if you had said ''here the x component of the velocity does not change because the masses are equal and the angles are the same afterward'' then that would have prompted me to reread the original post.

Given that the fact that the x component of the velocity is unchanged here is a very special case, I think that it is indeed important to warn the OP why it is valid in this example and how to do it in general (as it is likely that in a test the masses and/or the final angles will not be the same).

I apologize again for not reading carefully but maybe something good came out of my blunder:shy:

Regards

Patrick