I'm Lost

[nx2000]

I'm real lost with this problem, probably because I have no clue how to approach it, but if you can help, that would be nice.

Two skaters with masses of 62 kg and 50 kg, respectively, stand 7.5 m apart, each holding one end of a piece of rope.

(a) If they both pull themselves along the rope with equal force until they meet, how far does each skater travel? (Neglect friction.)

[OlderDan]

What do you think is conserved in this problem? Give us some idea how you think you might solve it, or how you tried, then we will help you.

[nx2000]

Well Im sure energy is conserved in the problem because I think when they come together, its an elastic collision. so then I thought is energy is conserved, than KE=.5mv^2. Then I thought that posed a problem because 1) I dont know the energy, 2) I dont know the velocity. And then I was just confusing myself some more. So I went to look for some outside help.

[OlderDan]

Well Im sure energy is conserved in the problem because I think when they come together, its an elastic collision. so then I thought is energy is conserved, than KE=.5mv^2. Then I thought that posed a problem because 1) I dont know the energy, 2) I dont know the velocity. And then I was just confusing myself some more. So I went to look for some outside help.
OK I can tell you that mechanical energy is not conserved. We can talk more about why later. What else could be conserved?

[nx2000]

Momentum??

[OlderDan]

Momentum??
Absolutely right. So can you write an equation involving the motion of the two skaters based on the fact that momentum is conserved?

[nx2000]

Momentum = m1v1+m2v2?

[OlderDan]

Momentum = m1v1+m2v2?
Right. And how much momentum did they have before they started pulling?

[nx2000]

Um they had 0 momentum?

[OlderDan]

Um they had 0 momentum?
Exactly. Set the total momentum in your equation to zero. This relation will be true for their instantaneous velocities at all times, and that means it will be true for their average velocities. If it is true for their average velocities, what does that say about their displacements from their starting points to where they meet?

[nx2000]

i dont really know

[OlderDan]

i dont really know
Momentum = 0 = m1v1+m2v2
m1v1 = -m2v2 <== one of these velocities is negative; they move toward each other
m1(d1/t) = m2(d2/t) <== d1 and d2 are positive distances; that is why the - sign is gone
m1d1 = m2d2 <== because the time is the same for both of them
You know how far apart they were, so what is d1 + d2?
Can you solve the two equations for d1 and d2 using the known masses and the given distance?

[nx2000]

i think im getting more confused, so no not really

[OlderDan]

i think im getting more confused, so no not really
d1 is how far m1 moves and d2 is how far m2 moves from where they started to where they come together

d1 + d2 = 7.5m <== given information
m1d1 = m2d2 <== from the momentum conservation
d1 = d2(m2/m1) <== dividing previuos equation by m1
d2(m2/m1) + d2 = 7.5m <== by substitution

Put in the values given for m1 and m2 (choose them either way you want, but keep track of which is which). Solve for d2. Then go back to d1 + d2 = 7.5m and solve for d1. Let us know what you get.