Solving for masses in an elastic collision

[mruss]

This isn't an actual homework problem, but it feels like it could be.

Homework Statement

Two masses, m1 and m2 are involved in an elastic collision. The initial velocities v1_0 and v2_0 are 1 and -2, respectively. The final velocities v1_1 and v2_1 are -3 and 0, respectively. Solve for m1 and m2.

Homework Equations

Given the collision is elastic, here are the two relevant equations:
1. m1*v1_0 + m2*v2_0 = m1*v1_1 + m2*v2_1
2. 1/2 (m1*v1_0^2 + m2*v2_0^2) = 1/2 (m1*v1_1^2 + m2*v2_1^2)

The Attempt at a Solution

I'm not sure what I'm doing wrong, but when I try to solve these 2 equations with 2 unknowns, I end up getting m1 = m2 = 0. For example, using equation 1 & 2 from above:
1. m1 - 2 * m2 = -3 * m1
→ 4 * m1 = 2 * m2
→ 2 * m1 = m2
2. m1 + 4 * m2 = 9 * m1
→ 4 * m2 = 8 * m1
→ 2 * m1 = m2

At this point, you only really have one equation so you can't solve for two unknowns, but if you do substitute 2 * m1 for m2 in the other equation you get 2 * m1 = 2 * m1 --> 0 = 0.

Any help would be appreciated, thanks.

 

[gneill]

What you've shown is that the ratio of the masses, m1/m2, is 1/2.

 

[mruss]

Yes, I agree that I've done that. What I'm trying to do is solve for 2 masses in an elastic collision. Do you have advice on how to do that?

 

[Tanya Sharma]

This isn't an actual homework problem, but it feels like it could be.

Homework Statement

Two masses, m1 and m2 are involved in an elastic collision. The initial velocities v1_0 and v2_0 are 1 and -2, respectively. The final velocities v1_1 and v2_1 are -3 and 0, respectively. Solve for m1 and m2.

Homework Equations

Given the collision is elastic, here are the two relevant equations:
1. m1*v1_0 + m2*v2_0 = m1*v1_1 + m2*v2_1
2. 1/2 (m1*v1_0^2 + m2*v2_0^2) = 1/2 (m1*v1_1^2 + m2*v2_1^2)

The Attempt at a Solution

I'm not sure what I'm doing wrong, but when I try to solve these 2 equations with 2 unknowns, I end up getting m1 = m2 = 0. For example, using equation 1 & 2 from above:
1. m1 - 2 * m2 = -3 * m1
→ 4 * m1 = 2 * m2
→ 2 * m1 = m2
2. m1 + 4 * m2 = 9 * m1
→ 4 * m2 = 8 * m1
→ 2 * m1 = m2

At this point, you only really have one equation so you can't solve for two unknowns, but if you do substitute 2 * m1 for m2 in the other equation you get 2 * m1 = 2 * m1 --> 0 = 0.

Any help would be appreciated, thanks.

You have calculated m1/m2 =1/2 and as long as the two masses are in the ratio 1:2 say (4,8) (3,6),they all are valid solutions to the problem.

Nevertheless, 2(m1) =2(m1) doesn't mean m1=0 .

Take both the terms on one side,

2m1-2m1 =0 .
m1(2-2) = 0
m1(0) = 0

This means value of m1 is indeterminate .

 

[mruss]

Thanks Tanya, I appreciate the response. I guess conservation of momentum and conservation of (kinetic) energy are not enough to solve for the masses in an elastic collision.

Do you know what additional information/formulas I would need to actually nail down the two masses, rather than just the relationship between them?

Thanks