# Particles and Momentum

1. Nov 8, 2005

### DStan27

A 1550 kg weather rocket accelerates upward at 20 m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

is there a way to use momentum and distance to find the speed. i know the second particle goes downward, but i can't figure out how to do this. i've tried everything i can think of.

thanks

2. Nov 8, 2005

### DStan27

should i be using mv = m1v1 + m2v2??

3. Nov 8, 2005

### cepheid

Staff Emeritus
Is momentum conserved in this situation? If so, then you should be. Don't ask, "should I be using such and such equation?" Ask, "which physical principle applies to this situation?"

Anyway, have you figured out how to calculate v, the final velocity of the rocket? What about v1, the velocity of the less massive fragment? That should be simple kinematics in both cases. Once you have calculated those velocities, do you have the information you need to solve the problem?