Conservation of Momentum in a Frictionless System with External Forces

[milkyway11]

question states: a man of mass 140 kg standing on a frictionless surface throws a 10kg rock horizontally away from himself. what is the momentum of the system immediately after the throw.


After I read the problem, I thought it was regarding the complete inelastic collision, in which the conservation of momentum would work backward. However, the more I think about the problem and try to solve it, I couldn't convince myself that this question had anything to do with collision since nothing really collided. Moreover, if it is collision, then the momentum would be 0 due to 0 initial velocity of both the man and rock. I am really stuck on this one, please help!

 

[PhanthomJay]

It is of sorts a reverse collision problem, but moreso, it is a conservation of momentum problem. Momentum is always a conserved quantity in the absence of net external forces. In which case, you have the correct answer!

 

[ashishsinghal]

momentum would be zero. why do you think you are stuck?

 

[milkyway11]

Because I don't quite get why is it zero since there will be acceleration required to throw the rock out by the player and that means there would be velocity

momentum would be zero. why do you think you are stuck?

 

[milkyway11]

But isn't there external force on the rock exerted by the player?

It is of sorts a reverse collision problem, but moreso, it is a conservation of momentum problem. Momentum is always a conserved quantity in the absence of net external forces. In which case, you have the correct answer!

 

[ashishsinghal]

When you say momentum is conserved, you talk about net momentum. This means momentum of the system. Here the system is the man and rock. Man applies force on rock, rock applies force on man. These are internal forces

 

[ashishsinghal]

External forces are those which are caused by object outside the system. Hence there is no external force and momentum is conserved.