# What is the speed of a piece of a spaceship that blew up?

oneamp

## Homework Statement

A spaceship of mass 2.30×106kg is cruising at a speed of 5.50×106m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.20×105kg , is blown straight backward with a speed of 2.30×106m/s . A second piece, with mass 8.40×105kg , continues forward at 1.30×106m/s .

## The Attempt at a Solution

Using mv=p, I calculated p of each piece like this:
p1 = 1.196*10^12
p2 = 1.092*10^12
p_total = 1.265*10^13

Then I found p3 by subtracting p2+p1 from p_total.

I divided p3 by the mass of the mystery piece: p/940,000 = 1.0302*10^13

But... the answer was wrong. I also tried subtracting: p1-p2 =pt and working from there. What am I doing wrong?

Thanks

nasu
Momentum is a vector. You need to add the three momenta (after explosion) as vectors. Here, in 1D, be careful with the signs. A diagram may help.

oneamp
But it's not 1D, right? Because one goes forward, one goes back, and the mystery piece I am solving for goes off in an unspecified direction...

oneamp
I got it, thanks for that tip.

Rawrr!
I believe you are right in using the conservation of momentum to tackle this problem. The initial momentum should equal the final momentum of the combined three pieces, such that

pinitial = p1 + p2 + p3

m(total)v = m1v1 + m2v2 + m3v3

p(total) = (2.3x106kg)(5.50 x 106m/s) = (1.265x1013)

p1 = (5.20×105kg)(-2.30×106m/s) = (-1.20x1012)

p2 = (8.40×105kg)(1.30×106m/s) = (1.092x1012)

p3 = (9.4x105kg) v3

Because the first piece was blown backwards, it becomes important to apply the negative to its velocity component, and should look like,

(1.265x1013) = (-1.20x1012) + (1.092x1012) + (9.4x105kg)(v3)

It becomes simple to solve for v3 . I assume the sign was causing your problem in the calculations. The mass of the third piece was found by the missing mass, m(total) = m1 + m2 + m3

Because the problem doesn't specify, i took the assumption that the collision only occurs in a single direction, along the x-axis, and that the velocity of each piece doesn't have to be broken down into individual x- and y-components.