# Homework Help: Finding final velocity and position of a two ended rocket

1. Oct 25, 2016

### Jrlinton

1. The problem statement, all variables and given/known data
The figure shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6.40 kg) and blocks L and R (each of massm = 1.90 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 2.80 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.90 s, block R is shot to the right with a speed of 2.80 m/s relative to the velocity that block C then has (after the second explosion). At t = 2.80 s, what are (a) the velocity of block C (including sign) and (b) the position of its center?

2. Relevant equations

3. The attempt at a solution
Okay so first I need to find the velocity of the center and right blocks after the explosion at t=0
Mass of right block(relative velocity- velocity of center and right blocks)=combined mass of right and left blocks*velocity of right and left blocks
1.90kg(2.80m/s-v)=8.30kg*
v=0.523 m/s
Taking that velocity and multiplying by 0.9 seconds gives the position of the center of the center block at the time that the second explosion occurs
0.523m/s*0.9s= 0.4707m

Next is to use the same method to find the velocity of the center block after the second explosion, I will skip the equation modeling as it is redundant and crude as I am unable to use notation.
1.90kg(2.80m/s-v)=6.40kg*v
v=0.6410 m/s but in the negative x direction so v=-0.6410 m/s
Now this is where I became uncertain. Is this velocity of -0.6410 m/s the new velocity of the center block and as it is a frictonless surface the answer to part a or should it be added the the original velocity of 0.523 m/s?
v=-0.6410m/s
or
v=-0.6410m/s+0.523m/s=-0.1703m/s
That would also affect part b as it would be
0.4707m+-0.6410(2.8-0.9)=-0.7472 m
or
0.4707m+-0.703(2.8-0.9)=0.14713m

The problem being that I entered both sets of solutions and both proved to be incorrect. What is wrong with my approach of using the conservation of momentum?

2. Oct 25, 2016

### haruspex

I take it the 8.3 is a typo. I agree with the 0.52 (but 0.522, not 0.523).
The idea is right, but check the arithmetic.

3. Oct 25, 2016

### Jrlinton

Okay so my problem was that instead of adding the velocity from the first explosion to the velocity of the second, I added the position of the first explosion to the second. This was just me quickly glancing at the wrong number. I got the final velocity to be -0.119 m/s and the final position to be 0.244 m

4. Oct 25, 2016

### haruspex

Looks right.

5. Oct 25, 2016

### Jrlinton

Haruspex comes to my rescue once again

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