The problem I'm currently trying to solve is the following.

Suppose you have a rocket weighing $20kg$ flying completely horizontally at $50m/s$. At some point during its flight an explosion happens which splits the rocket into two separate parts, each weighing $10kg$. After the explosion, part $1$ has a velocity of $0$. What is the velocity of part $2$.

So the simpler solution would be to use the law of conservation of momentum to gain the following answer:

$P_{before} = 20*50 = 1000,\quad P_1 = 10v_1,\quad P_2=10v_2$.

By the law of conservation of momentum, we then know that $P_1 + P_2 = > P_{before}$. Since $v_1 = 0$, this means $v_2 = 1000/10 = 100.$

However, we also tried doing the same thing using the law of conservation of energy, where we get the following result:

Since we're looking at an instantaneous explosion, we need not take gravitational potential energy into account. We'll attempt to calculate kinetic energy for both situations:

  1. (Before explosion) $E_{k,\;before} = \frac12 (m_1 + m_2) v^2 = 10v^2 = 10\cdot 50^2$.
  2. (After explosion) $E_{k,\;after} = \frac12 (m_1 v_1^2 + m_2v_2^2) = \frac12(0+10v_2^2) = 5v_2^2$

Knowing that these must be equal by the law of conservation of energy, we get that $v_2 = \sqrt2\cdot50$.

Now, I'm more confident in the momentum solution; it fits best with my intuition on the subject. But I have to wonder why we get such differing results. And in fact, if we apply the momentum solution, we find that a lot of energy has beeen added to the system. It could be argued that this added energy comes from the explosion, but it still doesn't sit quite right with me.

Could someone please elaborate a little on what would be happening in a system like this?


1 Answer 1


You are absolutely right in stating that the “added energy comes from the explosion”. In order to use conservation of energy, the initial energy calculation need to captures all forms of energy present in the system; in particular, we need to consider the chemical potential energy stored within the rocket, as the chemical energy is converted into the kinetic energy of the separated parts after the explosion.

So your calculation using momentum conservation is correct. After obtaining the final velocities, you can then use conservation of energy to reverse determine the initial chemical potential energy.

  • $\begingroup$ Note that even without any explosion or deformation in the traditional sense, the "collision" will still be inelastic. For example, if a powerful spring is used to separate the two parts of the rocket then the stored elastic energy in the spring is released as work when the rocket splits. In the rest frame of the rocket, a force has to be applied between the two halves of the rocket (over some distance) to result in the two halves ending up with different velocities. So work has to have been done - thus potential energy has been transferred to kinetic in any such situation. $\endgroup$
    – Penguino
    Nov 3, 2019 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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