Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The information to the question is, "A $59.0~kg$ boy and his $38.0~kg$ sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with a velocity $3.00~m/s$ toward the west. Ignore friction."

The specific question I am working on is, "How much potential energy in the girl's body is converted into mechanical energy of the boy–girl system?"

Well, I said that $E_{mech,~i}=0$, because no one has kinetic energy (no one is moving), and neither of them have the potential to move, that is, until the girl commences with the push; but, $E_{mech,~f}=PE=KE$, when she uses her potential energy to apply a force over a distance, on her brother, and by doing so she changes the energy of the system, by changing the speed of both of them, meaning the potential energy will convert to kinetic energy.

I've tried to plug every piece of data given to me, but I still couldn't find how much $PE $ was converted to mechanical energy. How do I find it? Is my analysis correct; is there anyway to improve it?


I have another question: why can't internal forces of a system cause momentum to not be conserved?

share|cite|improve this question
up vote 0 down vote accepted

You have this equation:

$m_bv_b+m_gv_g=0$ (conservation of momentum)

and $m_b,v_b,m_g$ are given. You can easily solve for $v_g$. Once you have this, $|\Delta PE|=\Delta KE_f=\frac12m_bc_b^2+\frac12m_gv_g^2$

I have another question: why can't internal forces of a system cause momentum to not be conserved?

Who said that they can't? Momentum is always conserved when there are no external forces on the system.

share|cite|improve this answer
@Manshearth No, I was asking why internal forces don't affect the total momentum of a system. – Mack Nov 17 '12 at 17:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.