Simple mass ratio problem

[PhunWithPhysics]

Two ice skaters masses M1 and M2 push off against each other and move in opposite directions with different speeds. As they glide to a halt their magnitudes of their accelerations are equal and skater M1 glides TWICE as far as skater M2. What is the ratio of M1/M2 of their masses. Is it just 2/1 because skater M1 goes twice as far?

 

[Fermat]

It's a bit more complicated than that.

Do a balance of momentum at the start to get a ratio of the two separation speeds.

Then apply the eqns of motion to the movement of each skater and solve for the unknown ratio.

 

[PhunWithPhysics]

how is it more complicated?

There is no momentum at the start so initial momentum is 0. So how do I solve for the masses if there are no mases?

 

[whozum]

The center of mass of the couple can't move, because the total momentum will still be zero. Thats the key point here.. so you'll need to find masses such that if one was twice as far away from the starting point at the other, that the total CoM would still be in the starting point.

 

[alias25]

0 = M1v-M2v (because they are going in opposite directions..so..
M1v = M2v ?

 

[dekoi]

There is no momentum at the start so initial momentum is 0. So how do I solve for the masses if there are no mases?

Of course there is initial momentum if the two skaters are both moving.
p = mv
And we know that both skaters do have velocity and mass.

There are no numerical masses so you are solving for the variables M1 and M2, using equations. It will most likely turn out that the ratio will be expressed without varibles, and as a number.

 

[whozum]

Of course there is initial momentum if the two skaters are both moving.
p = mv
And we know that both skaters do have velocity and mass.
There are no numerical masses so you are solving for the variables M1 and M2.

They arent moving at the beginning.

 

[dekoi]

Sorry, I misread the question.

 

[Fermat]

0 = M1v-M2v (because they are going in opposite directions..so..
M1v = M2v ?

They are moving at different velocities, so it's more a case of,

M1.V1 = M2.V2
M1/M2 = V2/V1

Let M1/M2 = λ, say

then,

V2 = λ.V1
========

Now do the eqns of motion bit, where both masses have the same deceleration, you know the start and end velocities of each mass, and M1 travels twice the distance that M2 does.
Solve for λ.