# Help with classical mechanics question

1. Oct 30, 2009

### henryc09

1. The problem statement, all variables and given/known data
A nucleus of mass 20mu (where mu is the atomic mass unit 1.66*10^-27) is moving with a velocity of 3*10^6 ms^-1 when it breaks in to two fragments. In the course of this process internal energy is released from within the nucleus and the kinetic energy consequently increases by an amount $$\Delta$$E=10^-12J. The heavier fragment has mass 16mu and is emitted at 90o to the original line of flight. What is the speed of the lighter fragment? (since the energy released is small compared with the rest-mass energy of the nucleus you may assume that mass is conserved and that both fragments remain non-relativistic)

2. Relevant equations

3. The attempt at a solution
The part confusing me is whether or not you can assume momentum is conserved, ie. does the release of internal energy count as an "external" force or not? If you assume that momentum is conserved my attempt at the solution is:

Initial KE= 0.5*10mu*(3*10^6)^2 = 1.494*10^-13J
so the final KE of the system is this + 10^-12J which equals 1.1494 * 10^-12J

which means that 16mu*v1^2 + 4mu*v2^2 = 1.1494*10^-12J

Initial momentum of system is 3*10^6*20mu = 9.96*10^-20 kgms^-1

if we say that the nucleus is initially moving to the right, and the larger fragment moves upwards after the "break", the velocity of the smaller fragment would have to be south east.

The vertical component of v2 would equal 16mu*v1 / 4mu so that the overall momentum in the vertical plane remains 0, and the horizontal component would equal 9.96*10^-20 / 4mu (which is 1.5*10^7ms^-1) so that the momentum to the right of the system is the same as before. This means that v2 = $$\sqrt{}(1.5*10^7)^2+(4muv1)^2$$ however when you substitute that back in to the equation for KE and try to solve for v1 you get a math error as v1^2 = a minus number.

If I am wrong with assuming that momentum is conserved then I am unsure as to how to go about solving this problem as surely the change in momentum could be in any direction and so it would not be possible to work out the speed of the smaller fragment.

Any ideas?

2. Oct 30, 2009

Momentum is always conserved in a closed system-that is, if there are no external forces that could effect the situation. Hope that helps!

3. Oct 30, 2009

### kuruman

Momentum is indeed conserved. The force that one fragment exerts on the other is equal and opposite to the force that is exerted on it by the other. The two forces are internal to the two fragment system.

Why do you say that the initial kinetic energy is

KE= 0.5*10mu*(3*10^6)^2 ?

The mass of the parent nucleus is 20mu not 10mu. Would that make a difference?

4. Oct 30, 2009

### henryc09

ok well now that I know that for sure I think I'll get there eventually. That was actually just a typo from me but I expect I've just made a mistake somewhere in the calculation and will go over it. Thanks!