Two skaters pushed apart by spring: find Al's mass

[shawonna23]

Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.81 m/s, while Jo moves off in the opposite direction at a speed of 1.5 m/s. Assuming that friction is negligible, find Al's mass.

I am stuck on this problem. I think I have to use this equation: m1*v1 + m2*v2=(m1+m2)vf. But, I don't know how I would use it to solve this problem. Can someone please help me figure out how to solve this problem?

 

[stunner5000pt]

this is not a momentum conservation question becasue there is no collision of any sorts going on this question. There is however, energy conservation

the initial spring's energy is converted to the kinetic energies of the two skaters.

also think forces
on one side the spring is being compressed by a force of (168-M)a and it is turn exerts a force of k(x/2) in the oppsite direction.
From the other side a force of Ma and spring force of k(x/2)
now you have one of their weight (pick one doesn't matter) to be 168 - M and other to be M, sub ito the energy conservation equation, and solve using the force relation

 

[xanthym]

Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.81 m/s, while Jo moves off in the opposite direction at a speed of 1.5 m/s. Assuming that friction is negligible, find Al's mass.

I am stuck on this problem. I think I have to use this equation: m1*v1 + m2*v2=(m1+m2)vf. But, I don't know how I would use it to solve this problem. Can someone please help me figure out how to solve this problem?

SOLUTION HINTS:
When released, compressed spring applies the same force F(t), which is a function of time, to both skaters over the same time interval, so that {∫ F(t) dt} is the same for both skaters. Therefore, both skaters gain equal amounts of momentum. Hence, their final momentums are related by:
m₁*v₁ = m₂*v₂
It is also known that:
m₁ + m₂ = (168 kg) ::: ⇒ m₁ = 168 - m₂
Place 2nd equation into the 1st, and solve for required "m" in terms of velocities "v" (evaluate using velocity values given in problem statement).


~~

 

[shawonna23]

Thanks for the help guys. I got the answer which was 109kg. It was correct! Thanks again!