To stop a moving object -- Momentum or Kinetic Energy

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bugatti79
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How should I look at the problem at stopping a moving object with the following conditions

1) mass m and speed v

2) 0.5 m and 2v

3)0.5m and sqrt 2 v

Simple math tells me the number 2 would require more energy to stop it. I can relate to energy better in terms of how to stop a moving mass but how does one visualise in terms of momentum?

Thanks
B
 
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You should look at it using Newton's 2nd law, F=ma.

Both momentum mv, and kinetic energy ##\frac{mv^2}{2}## will reach zero at the same time , when v=0.
 
bugatti79 said:
H
Simple math tells me the number 2 would require more energy to stop it. I can relate to energy better in terms of how to stop a moving mass but how does one visualise in terms of momentum?
Number 2 will require you to dissipate more energy when you stop it. It does not take more energy -- it releases more.

More energy means that the object covers more distance while stopping against a constant force. It does more work.
More momentum means that the object takes more time while stopping against a constant force. It delivers more impulse/recoil.
 
jbriggs444 said:
Number 2 will require you to dissipate more energy when you stop it. It does not take more energy -- it releases more.

More energy means that the object covers more distance while stopping against a constant force. It does more work.
More momentum means that the object takes more time while stopping against a constant force. It delivers more impulse/recoil.

So in laymans terms if I was to describe a situation to my friend about stopping a 18stone rugby player in his tracks by a sufficient tackle..which would be more appropriate...the energy approach?
Or perhaps both are equivalent because to stop him covering more distance instantly is equivalent to reducing the time to bring him to a halt...? :-/
 
bugatti79 said:
So in laymans terms if I was to describe a situation to my friend about stopping a 18stone rugby player in his tracks by a sufficient tackle..which would be more appropriate...the energy approach?
Or perhaps both are equivalent because to stop him covering more distance instantly is equivalent to reducing the time to bring him to a halt...? :-/
A bullet has less energy, but will go right through you before you can stop it.

Edit: Less flip, the question as posed is not answerable. How difficult it is to tackle someone is not a physics question. It is an engineering question. If you could specify a mechanism for the tackle and quantify the difficulty then it might be reduced to a physics question.
 
To help the op, your questions are a bit too simplified. They would work for say colliding balls.

A rugby tackle has a lot of techniques that would require more complex equations.
Your simple equations might apply more to American football where tackles are more pure collisions and less wrestling moves. Hence one wears body armour, the other doesn't.