Perhaps a dumb question to ask, but I was given the following problem to solve:

A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?

When I saw this problem my first attempt to solve this was to use conservation of energy: $m_b$=mass of bullet

$m_d$=mass of door

$L$=length of the door

$I=I_{bullet}+I_{door}= \frac 13m_dL^2+ m_bL^2$

$\frac 12m_bv^2=\frac 12Iw^2$

$w=\sqrt{\frac {m_bv^2}I}$

which is far off from the actual answer. Now, I know that the correct method of solving this type of problem is to use conservation of angular momentum, which will give:


$w=\frac {m_bvL}{I}$

Why is the first method wrong? Thanks in advance.

  • $\begingroup$ Because it's angular momentum rather than angular energy? $\endgroup$ Commented Jun 5, 2015 at 6:44
  • $\begingroup$ Yes I know angular momentum is conserved in this scenario and it is probably obvious to use conservation of angular momentum to solve the problem. But I was wondering why conservation of energy does not work in this case. $\endgroup$
    – whoami
    Commented Jun 5, 2015 at 6:47
  • $\begingroup$ where did you get the moments of inertia of the bullet and the door? $\endgroup$ Commented Jun 5, 2015 at 6:47
  • $\begingroup$ I used moment of inertia of a plane $I=\frac 13Ma^2$ for the door and moment of inertia of a point mass $I=mr^2$ for the bullet $\endgroup$
    – whoami
    Commented Jun 5, 2015 at 6:51
  • $\begingroup$ Angular momentum is only conserved when net torque is zero, but in this case, why is it zero? I think that the collision with the door changes the momentum of the system during the short period of time the collision takes place, which means that there is a force perpendicular to the axis of rotation during that period of time. I this reasoning correct? Its just I don´t understand why angular momentum is conserved. $\endgroup$
    – nonameable
    Commented Nov 18, 2015 at 16:23

1 Answer 1


In any situation, momentum (linear or rotational) is conserved. So if there are no external forces, you can use that.

But energy is only conserved in certain situations (such as elastic collisions). The deformation of the door by the bullet is inelastic. You can consider it similar to energy losses from friction. In this case, the (mechanical) energy in the final state is less than the initial state because of losses.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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