Mass Ratio Problem: Solving for M1/M2 in Equal Acceleration Scenario

  • Thread starter Thread starter PhunWithPhysics
  • Start date Start date
  • Tags Tags
    Mass Ratio
Click For Summary

Homework Help Overview

The problem involves two ice skaters with masses M1 and M2 who push off against each other, resulting in different speeds and distances traveled before coming to a halt. The scenario requires determining the mass ratio M1/M2 given that their accelerations are equal and one skater travels twice the distance of the other.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial momentum and its implications, with some suggesting a balance of momentum to find the mass ratio. Others question the absence of initial momentum and how to approach the problem without numerical values for the masses.

Discussion Status

There is an ongoing exploration of the relationship between the skaters' velocities and distances traveled. Some participants have offered insights into using equations of motion and the concept of center of mass, while others are clarifying misunderstandings about initial conditions and momentum.

Contextual Notes

Participants note that the problem lacks numerical values for the masses and that the initial momentum is zero, which raises questions about how to derive the mass ratio from the given conditions.

PhunWithPhysics
Messages
24
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
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?
 
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.
 
0 = M1v-M2v (because they are going in opposite directions..so..
M1v = M2v ?
 
Last edited:
PhunWithPhysics said:
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.
 
dekoi said:
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.
 
Sorry, I misread the question.
 
alias25 said:
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
12K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
37
Views
6K