Help: Physics Homework Problem-Conservation of Linear Momentum

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Homework Help Overview

The problem involves two individuals, Al and Jo, who are on skates at an ice rink with a compressed spring between them. They release their hold on each other, causing them to move in opposite directions. The task is to determine Al's mass based on their combined mass and their respective velocities after release, while assuming negligible friction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation laws, questioning whether momentum conservation applies in this scenario. Some suggest that energy conservation may be more relevant due to the spring's role.

Discussion Status

Some participants have provided insights into the relationship between the forces exerted by the spring and the resulting motion of the skaters. There is an exploration of different approaches, including momentum and energy conservation, but no explicit consensus has been reached.

Contextual Notes

There is a mention of the combined mass of Al and Jo being 168 kg, and the velocities at which they move after releasing the spring. The discussion also reflects on the assumptions regarding friction and the nature of the forces involved.

shawonna23
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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?
 
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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
 
Last edited:
shawonna23 said:
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).


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

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