# Implse and collisions

1. May 16, 2004

### loatisaf

impulse and collisions

I have one problem I've been trying to do, but I'm stuck on it. In this one, two ice skaters collide into each other. The specifics are as follows:
Skater 1 (mass 58.96 kg) is travelling at 7.15 m/s. She then collides with skater 2 (mass 49.89 kg) who is at rest. During the collision, skater 1 pushes skater 2 as hard as she can, imparting an average force of 1300 N over a period of .75 seconds. The questions are how fast are skater 1 and skater 2 moving after the impact.

I 've got the whole force over time thing is impulse, so I've got that
(m1v1f + m2v2f) - (m1v1i + m2v2i) = 975
and breaking it down, collecting like terms, and inserting values I already know, it works out to
5896 v1f-49.89 v2f = 1396.564
Unfortunately, this formula still has 2 unknown values, and I can't think of a second formula to use to do substitution.

Last edited: May 16, 2004
2. May 16, 2004

### arildno

The force imparted by skater 1 one skater 2 is an INTERNAL FORCE, hence, it does NOT contribute to momentum change of the center of mass of the 2 skaters.
The correct set of equations are therefore:
$$m_{1}v_{1,f}+m_{2}v_{2,f}-(m_{1}v_{1,i}+m_{2}v_{2,i})=0$$
$$m_{2}v_{2,f}=975$$

3. May 16, 2004

### Staff: Mentor

arildno's solution is, of course, completely correct. But a slightly simpler set of equations can be obtained by realizing that the skaters exert equal and opposite forces on each other:
$$m_{1}v_{1,f} -m_{1}v_{1,i} =-975$$
$$m_{2}v_{2,f}=975$$

But most important is to understand that both methods are valid and equivalent.

4. May 16, 2004

### arildno

I agree; using Newton's 3.law in an explicit manner is more straightforward