# Elastic Collison Problem

1. Mar 9, 2008

### the7joker7

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

A 10-kg mass moving at a speed of V$$_{1}$$ in a positive direction collides elastically with a 2-kg mass which is initially at rest. The magnitude of the net force that acts on each object during the collision is shown in the figure. Find the initial velocity, V$$_{1}$$, of the 10-kg mass prior to collision, the final velocity, v$$_{3}$$, of the 10-kg mass, and the final velocity, v$$_{4}$$, of the 2-kg mass.

I can't post the figure up, but it's a force with respect to time chart. It starts with nothing, then starts to spike upwards at a constant rate from time 1 millisecond to time 3 milliseconds, where it peaks at 10000N, then goes back down at the same rate, hitting 0N at 5 milliseconds. I've already figured the Impulse is 20 N*s from this.

2. Relevant equations

p = mv

I = F$$/delta$$T

$$/delta$$p = mv$$_{final}$$ - mv$$_{initial}$$

mv$$_{before collision}$$ = the some of mv$$_{after collision}$$

Possibly assorted others.

3. The attempt at a solution

So far, I know the Impulse is 20 NS.

Impulse = Momentum, I do believe.

So

20 = (10)V

Making Velocity of the first object initially = 2 m/s, correct?

That would answer the first question, but I'm not totally sure where to go from there.

2. Mar 9, 2008

### the7joker7

Here's a quick idea of what I tried to do...lemme know if it looks right.

20 = (10)V$$_{3}$$ + 2(V$$_{4}$$)
20 = .5(10)V$$_{3}$$$$^{2}$$ + .5(2)V$$_{4}$$$$^{2}$$

Those are the conservation of momentum and energy equations.

I solved the momentum one for V$$_{3}$$ and got .55279 m/s. I can post steps if anyone wants to see.

Then I plugged that into the energy one and got 1.7061 m/s. Sound about right?

It looks like it MIGHT be accurate, but the sums of mv before and the sums of mv after aren't equal.

I got 2(10) = 20 for before and 10(.55279) + 2(1.7061) = 8.94 for after. =/

Last edited: Mar 9, 2008
3. Mar 10, 2008

### physixguru

v1 + V3 = 10
2V1 = 12
V1 = 6 m/s
V3 = 4 m/s

4. Mar 10, 2008

### the7joker7

How is V1 6m/s?

Am I doing the impulse equation wrong?