-1
$\begingroup$

Suppose you are pushing a block on a horizontal smooth surface and giving it a constant force F( Imagine you can walk on it). The force F is removed just before it collides with another identical block kept on the same smooth surface.

The momentum of the blocks can only be conserved if the force gets removed just before the collision. Well, my doubt is, if you consider the whole earth including the blocks as a system, the internal forces should cancel out. So, can the momentum of the blocks be conserved even when the force F is not removed since it should get canceled out?

$\endgroup$
1
  • $\begingroup$ Why am I getting a downvote for this question? What is wrong with this question? $\endgroup$
    – Piano Land
    Jan 18, 2018 at 16:02

1 Answer 1

2
$\begingroup$

Yes. If you imagine walking on that block, and consider earth and yourself (source of force on block) as a part of the system, it is still treated as an internal force and momentum is conserved. Momentum is always conserved when all the forces are internal, i.e., if you take the source of force also as a part of your system of observation.

$\endgroup$
4
  • $\begingroup$ By walking I meant walking on the frictionless floor. Not on the block. And you are pushing the block. $\endgroup$
    – Piano Land
    Dec 13, 2017 at 16:27
  • $\begingroup$ Still, no matter what, IF you take the source of the force F as a part of system , momentum gets conserved. But actually, you can't walk on a frictionless ground. If somehow you still manage to walk, it still effects the momentum of the ground( earth) . Earth is a part of our system. So momentum is conserved. $\endgroup$
    – 0xVikas
    Dec 13, 2017 at 16:37
  • 1
    $\begingroup$ So, if I keep pushing the block before, during and even after the collision occurs, if I consider the earth and the block as the system, the momentum would be conserved? $\endgroup$
    – Piano Land
    Dec 14, 2017 at 11:45
  • $\begingroup$ Not just the earth, but consider yourself too. $\endgroup$
    – 0xVikas
    Dec 14, 2017 at 12:59

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.