Two-ended rocket - Conservation of Momentum

[Ignitia]

Homework Statement

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?

Homework Equations

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

So,
M_CR = M_C + M_R
V_CR = V_C + V_R
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)

The Attempt at a Solution

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

0=M_LV_L + M_CRV_CR
0=2kg(-3m/s) + 8V_CR
6=8V_CR
3/4 m/s = V_CR

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

M_CRV_CR = M_CV_C + M_RV_R
8kg*(3/4 m/s) = 6kg*V_C + 2kg*3m/s

6 = 6*V_C + 6
0 = 6*V_C
0 = V_C

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

 

[TSny]

M_CR = M_C + M_R
V_CR = V_C + V_R

Is the second equation written correctly?

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

0=M_LV_L + M_CRV_CR
0=2kg(-3m/s) + 8V_CR

In the first equation, are V_L and V_CR 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).