1
$\begingroup$

When a cat or any body falls over to the ground, how is momentum conserved?

I was working on a problem of a cat falling on top of a skateboard, and the system travels together with a new velocity. That seemed intuitive enough for me. This is how I was thinking through:

The cat had momentum that became zero after the impact. Should not the skateboard have recoiled in some way, due to the conservation of momentum? After all, the change in momentum for the board should have been something measurable.

I guess there is something wrong with the way I am approaching the problem. Could you please help me identify this?

EDIT: I apologise: but the situation is like this - a skateboard moves on the ground with constant speed, until the cat is dropped from a tree. The cat lands on the skateboard and then proceeds with a new speed.

$\endgroup$
  • $\begingroup$ Could you clarify the situation? Perhaps with a diagram? Does the cat fall on the skateboard and then they move together as a pair? $\endgroup$ – Jake Rose Nov 25 '18 at 1:07
  • $\begingroup$ You could also ask why the cat's momentum increased as it was falling from the tree. It started at zero and became non-zero during the fall. What objects are exerting forces on it during and after the fall? $\endgroup$ – BowlOfRed Nov 25 '18 at 1:26
  • $\begingroup$ The skateboard does not have to move, especially if the force is perpendicular. The ground (or earth) can absorb the momentum. In other words the earth gains a slight amount of velocity! $\endgroup$ – PhysicsDave Nov 25 '18 at 1:28
  • $\begingroup$ Just for completeness, since the comments and answers (so far) have (understandably) concentrated on the vertical component of momentum. The horizontal component of momentum is conserved, and not affected by gravity at all. Momentum before $=$ momentum after, where "before" is initial momentum of skateboard and "after" is momentum of cat+skateboard together. Final velocity is easily calculated if you know the masses and initial velocity of skateboard. I'm guessing you know all that already, and that your question is about the vertical part, but as I say, just for completeness .... $\endgroup$ – user197851 Nov 25 '18 at 16:02
  • $\begingroup$ Yeah, I do know about the horizontal part. It was the vertical momentums that I did not understand. $\endgroup$ – KV18 Nov 25 '18 at 17:56
1
$\begingroup$

The skateboard does not have to move, especially if the force is perpendicular. The ground (or earth) can absorb the momentum. In other words the earth gains a slight amount of velocity!

$\endgroup$
1
$\begingroup$

If the cat & skateboard are the system then the external force is the gravitational attraction of the Earth on the system and the momentum of the system is not conserved.

If the cat & skateboard & Earth are the system and there are no external force then the momentum of the system is conserved.
As the cat & skateboard fall downwards then the Earth “rises upwards” at a speed which is much less than that of the cat & skateboard.

When the cat & skateboard hit the Earth then the momentum of the system is exactly the same as it was when the cat & skateboard started to fall.
If the skateboard rebounds then so will the Earth and the cat but in the opposite direction and at a much smaller speed than the skateboard.

$\endgroup$
1
$\begingroup$

Both @Farcher and @PhysicsDave answers are correct.

The key point is the momentum of a system is constant only if the system is isolated (for example, not subjected to forces external to the system). So far as we know there are not truly isolated systems- you cannot shield a system against gravity. This is why @Farcher needed to combine the cat, skateboard and Earth as the system in order for the momentum of the system to be conserved in the case of the falling cat.

I don’t mean to imply that gravity must be included in all conservation of momentum problems. Only when the force of gravity influences the momentum exchanges.

Hope this, in addition to the other answers, helps.

$\endgroup$

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.