I'm new here. this is in reference to the video posted by veritasium on the bullet block experiment. i realise there is already a thread on this but i want to ask a variation of the question. so basically we have two identical blocks and we shoot identical bullets at them one at the centre and another off centre. which one will rise higher? now the answer assuming the collision is inelastic is already discussed. but what if the collision were perfectly elastic? i know this is practically impossible but still we have momentum telling us that both blocks must rise to the same height but since no energy is lost as heat, energy tells us that the rotating one must rise to a lower height. What am i missing??

  • $\begingroup$ Note that an elastic collision certainly implies a different momentum transfer. Not just different from the fully inelastic case, but also different between the two cases that you propose. Finally, while I haven't seen the video that you mention without even bothering to link to it is likely that the situation you're talking about is properly called a "ballistic pendulum", and has been a common teaching scenario since long before the internet much less veritasium. $\endgroup$ – dmckee --- ex-moderator kitten May 1 '16 at 4:29
  • $\begingroup$ thanks for repling by the way. also could you please clarify on the ballistic part? sorry but i have only started learning physics $\endgroup$ – Juras Kumar May 1 '16 at 6:12
  • $\begingroup$ @dmckee The video appears to show a paradox in that when a bullet hits a block on a line through the block's centre of mass the block and bullet gain a certain amount translational kinetic energy. Yet when an identical bullet with the the same initial kinetic energy as before hits an identical block along a line which is not through the centre of mass of the block, the block and bullet gain exactly the same amount of translational kinetic energy and some rotational kinetic energy as well. Explaining why the ballistic pendulum will work whether or not the projectile is aimed at the CoM. $\endgroup$ – Farcher May 1 '16 at 9:56
  • $\begingroup$ youtube.com/watch?v=vWVZ6APXM4w $\endgroup$ – Juras Kumar May 1 '16 at 10:14

but still we have momentum telling us that both blocks must rise to the same height

That's not true here. In the first case where the bullet is embedded, the final velocity of the bullet and the block must be identical. Since initial momentum of the two shots were the same, then the final momentum will be the same as well. Because they are connected, the distribution of that momentum cannot change.

But in the elastic case, the final velocity of the bullet and the block are not linked. Conditions may distribute the initial momentum between both objects in different ways. This means that in the two cases, the block may rise to different heights.

  • $\begingroup$ Thankks a lot for taking the time to answer, that makes sense. however is there any way say mathematically to quantify this momentum distribution so that it could be more clear? $\endgroup$ – Juras Kumar May 1 '16 at 9:28
  • $\begingroup$ @JurasKumar For the non elastic example in the video $mu = (m+M)V$ $m,u=$ mass & initial velocity of bullet, $M=$ mass of block, $V=$ final velocity of block and bullet and the value of $V$ is fixed. For the elastic case $mu = MV + mv$ where now the final velocity of the block $V$ depends on the the final velocity of the bullet $v$ which in turn depends on there the bullet hit the block. $\endgroup$ – Farcher May 1 '16 at 10:26
  • $\begingroup$ Yeah i finally understood it! thank you so much for clearing my doubts! $\endgroup$ – Juras Kumar May 1 '16 at 13:01

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.