Kinetic Energy problem

[ntukza]

Please help me with this question:

 

A man sits on a trolley, which runs without friction on horizontal rails. The mass of the trolley (with the man included) is M. The man throws horizontally a stone of mass m, in the direction of the rails. Find the subsequent speed of the trolley if: (i) The momentum imparted by the man to the stone is J; (ii) The total work done by the man is W.

 

The answers given are (i) v = J/M and (ii) v = squareroot[2mW/M(M+m)]

 

For the first part I got J/m and I assumed the trolley's speed did not change after the stone was thrown, which doesn't make sense to me but that's the only way I could proceed with it. For the second part I got v = squareroot[2W/m]

[Sisyphus]

Remember, the trolley’s momentum is its own mass times its velocity. It looks like you were using the rock’s mass.

[swansont]

For part 1 it seems that M does not include the mass of the rock.

 

For part 2: The work will manifest itself as kinetic energy. You have 1/2 mv^2 = W, i.e. only the rock has kinetic energy. What else has kinetic energy in the problem?

[ntukza]

Remember, the trolley’s momentum is its own mass times its velocity. It looks like you were using the rock’s mass.

 

1/2(M+m)v^2 = 1/2Mv^2 + 1/2m(J/m)^2

 

Mv^2 + mv^2 - Mv^2 = m(J/m)^2

 

mv^2 = m(J/m)^2

 

v^2 = (J/m)^2

 

v = J/m

 

This assumes v is the same before and after collision, an assumption that I don't fully agree with. Without this assumption the equation yields another variable, which then becomes hard to solve. Maybe the given answer of J/M is wrong.

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Merged post follows:

Consecutive posts merged

For part 1 it seems that M does not include the mass of the rock.

 

For part 2: The work will manifest itself as kinetic energy. You have 1/2 mv^2 = W, i.e. only the rock has kinetic energy. What else has kinetic energy in the problem?

 

Wouldn't the trolley still have Kinetic Energy since it was said to be moving originally?

[Sisyphus]

For the first question,

 

Concepts you need:

 

Momentum = mass*velocity

Momentum is conserved (the trolley will have the same momentum in the opposite direction as the rock)

 

So, if the rock is thrown with momentum J, the trolley will then have momentum J as well.

 

So,

 

J = M*v

 

Therefore,

 

v = J/M

 

Where you went wrong (aside from making things more complicated than necessary) is using the same term "v" for both the rock's velocity and the trolley's velocity.

Edited March 18, 2009 by Sisyphus

[ntukza]

Momentum is conserved (the trolley will have the same momentum in the opposite direction as the rock)

 

But what about the initial momentum that also needs to be conserved? Surely your statement applies if there was no momentum to begin with, because the momentum before the throw must equal the momentum after the throw - The sum of the trolley's and stone's momentum must equal the momentum before the throw, ratheer than them equalling each other...

 

Having said that I think you've helped me understand a little better. Would it be correct to say:

 

1/2(M+m)u^2 = [1/2(M+m)u^2 - 1/2M(J/M)^2] + 1/2m(J/m)^2 ?

[Sisyphus]

The wording of the question is a bit ambiguous, but I'm sure the intended meaning is that the trolley isn't moving before the throw. If it were moving, you wouldn't have enough information to answer the question. The "it runs without friction" part simply means you don't have to take friction into account in the question, i.e. that it accelerates in direct proportion to the force applied. However, it doesn't really matter either way, since you can just read it as, "how much has the trolley's velocity changed after the throw," in which case it doesn't matter how fast it was moving to begin with. The physics of a motionless object and an object moving at constant velocity with zero friction are exactly the same.

 

And yes, it is the case both that the trolley/stone's momentum before the throw must equal the sum of the two after, and that the trolley and rock, after the throw, will have an equal and opposite change in momentum. In fact, that's saying the same thing. In order for the total system to have the same momentum before and after, any change in one part has to be cancelled out by a change in another. It's just simplest to consider the trolley to initially be motionless (momentum zero), so the rock and trolley will afterwards simply have the same momentum in opposite directions.

 

That equation is incorrect. You’re using kinetic energy instead of momentum, which is just mv. The total kinetic energy is not going to be the same before and after, since you're using potential energy in your arm to throw the rock!

Edited March 24, 2009 by Sisyphus

[ntukza]

//"The wording of the question is a bit ambiguous, but I'm sure the intended meaning is that the trolley isn't moving before the throw."//

 

Yes you're right. I didn't realise that the trolley could actually be motionless initially. It makes sense now that you've mentioned it. Thanks a lot!

 

 

 

//"That equation is incorrect. You’re using kinetic energy instead of momentum, which is just mv. The total kinetic energy is not going to be the same before and after, since you're using potential energy in your arm to throw the rock!"//

 

Yeah I was kind of worried about that. Thanks for clarifying everything.