# Two-ended rocket - Conservation of Momentum

1. Jul 28, 2017

### Ignitia

1. The problem statement, all variables and given/known data
A two-ended "rocket" is initially stationary on a frictionless floor, with its center at the origin of an axis. The rocket consists of a central block C (of mass M=6 kg) and blocks L and R (each of mass m = 2 kg) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence; (1) At time t = 0, block L is shot to the left with a speed of 3 m/s "relative" to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.8 sec, block R is shot to the right with a speed of 3 m/s "relative" to the velocity that block C then has. At t = 2.8 sec, what is the velocity of block C?

2. Relevant equations
L = left, R = right, C = center, CR = center and right combined.

So,
MCR = MC + MR
VCR = VC + VR
and so on.

Since it's in a closed system, and p=mv
P(initial)=P(final)
M(initial)V(initial) = M(final)V(final)

3. The attempt at a solution

Since it starts at stationary, for the first step I went:

0=MLVL + MCRVCR
0=2kg(-3m/s) + 8VCR
6=8VCR
3/4 m/s = VCR

2nd step has block CR moving at 3/4 m/s. I want to find VC so:

MCRVCR = MCVC + MRVR
8kg*(3/4 m/s) = 6kg*VC + 2kg*3m/s

6 = 6*VC + 6
0 = 6*VC
0 = VC

This cannot be the correct answer. Where did I go wrong?

2. Jul 28, 2017

### TSny

Is the second equation written correctly?

In the first equation, are VL and VCR velocities that are measured relative to the floor?
Note that the problem states that after the first explosion, L has a speed of 3 m/s relative to CR (not relative to the floor).