Conservation of Momentum in an Explosion

1. Jul 29, 2004

Beers Law

Hi all,

I have become frusturated at a conservation of momentum problem. A space ship(ss) with a mass of 2.0*10^6 [kg] is crusing at a speed of 5.0*10^6 [m/s], when it blows up. One section(s1), with mass 5.0*10^5 [kg] is blown straight backwards at a speed of 2.0*10^6 [m/s]. The second section(s2), with mass 8.0*10^5 [kg], continues forwards at a speed of 1.0*10^6 [m/s]. A third section moves at a certain speed.

Trying to find the speed of the third section, I initally approached the problem as a kinetic energy problem, considering that I might be able to use KE=(1/2)mv^2 and simply subtract s1+s2 from ss because the explosion is internal and the change in kinetic energy will be 0:

First, the third section has a mass of 7.0*10^5[kg]

KE(ss)=2.5*10^19 [J]
KE(s1)= 1.0*10^18 [J]
KE(s2)= 4.0*10^17 [J]

KE(s1+s2)= 1.4*10^18[J]

So I subtracted this from KE(ss):

2.5*10^19 [J]-1.4*10^18[J]= 2.36*10^19 [J], which is the KE for the third section

Then I set up 2.36*10^19 [J] = (1/2)*(7.0*10^5)(v_s3)^2, and I obtained a velocity for the third section of v_s3= 7.93*10^6 [m/s]! The solution in the back of the book reads 1.46*10^7 [m/s].

So, ok, this is obviously not the correct way to set up this problem, so I did vector sum for momentum using p(vector)=m*v(vector).

The space ship has a vector 9.45*10^12 [kg*m/s]
Section1 has vector 1.144*10^12 [kg*m/s]
Section2 has vector 1.105*10^12 [kg*m/s]

Section 3 has mass of 7.5*10^5 [kg]. I subtracted section1+section2 vectors from the space ship vector, and the result was 7.201*10^12 [kg*m/s]. Then I inserted this back into the p(vector)=m*v(vector):

7.201*10^12 [kg*m/s]= 7.5*10^5[kg] * v

I solved for v and recieved 9.6*10^6[m/s].

Im obviously not doing something right. I think I may not be factoring in that these are vectors? Thanks for any help.

2. Jul 29, 2004

Staff: Mentor

Forget about KE. It's not conserved in an explosion.

Right.

...

Right, but realize that vectors have direction. (Or at least a sign.)

Why 9.45? 2.0 x 5.0 = 10. And in what direction? Call it the positive x-direction.
Again, check your arithmetic. Section 1 moves backwards: that's the negative x-direction; Section 2 moves forward: the positive x-direction.

Yet another arithmetic error!
Careful! Signs matter.
Two main mistakes: (1) arithmetic errors (2) not properly treating the sign (which represents the direction, in this case) of the vectors.

3. Jul 29, 2004

Beers Law

Thanks for the help. Here is what I have done.

The Ship has a vector of +1.0*10^13.
Section 1 has a vector of -1.0*10^12.
Section 2 has a vector of +8.0*10^11
So, section 3 has a vector of 1.02*10^13, which I set up this way:

1.02*10^13= 700000*v, which returns 1.457*10^7, which is most excellent.

My mistakes are fairly embarassing, but not keeping things organized and thinking too far ahead are my achilles heel with physics.

Thanks again.