Solve 2-Particle Elastic Collision: Mass m & Final Velocity -v/5

[I Like Pi]

Hey, just doing some homework, and can't figure this one out

Homework Statement

Two particles go through an elastic collision. One particle has mass, m, and is initially at rest, while the other particle has initial velocity, v, and final velocity, -v/5. What is the mass of the second particle relating to m.

Answer: 2m/3

Homework Equations

The Attempt at a Solution

What I'm confused about is because you have two unknowns, the final velocity of the resting particle, and the mass of the second particle..

i tried using

m1v1 + m2v2 = m1v1' + m2v2'
and .5m1(v1)² + .5m2(v2)² = .5m1(v1)²' + .5m2(v2)²'

that seems waay too confusing for a question with no values..

 

[ehild]

You can imagine that m and v are numerical values. Just plug in m1=m, v1=0 (initially in rest), v2=v, v2' = -v/5 into your equation and solve for m2/m.

ehild

 

[I Like Pi]

You can imagine that m and v are numerical values. Just plug in m1=m, v1=0 (initially in rest), v2=v, v2' = -v/5 into your equation and solve for m2/m.

ehild

yes i figured it out last night, and when I have time I'll post the answer up here, it's really long haha

thanks ehild!

 

[ehild]

Is it long really?

m₂ v=m v'-m₂v/5
m₂ v²=m v'²+m₂v'²/25

mv'=6/5 m₂ v *
mv'²=24/25 m₂v² **

Divide equation ** by eq. *

v' = 4/5 v

Substitute for v' in eq.*.

ehild

 

[rwisz]

Hello,

Thank you for reaching out for help with your homework problem. I am happy to assist you in finding the solution to this problem.

First, let's define some variables to make the problem easier to understand:

m1 = mass of first particle (initially at rest)
m2 = mass of second particle (initially moving with velocity v)
v1 = velocity of first particle after collision
v2 = velocity of second particle after collision

We can use the conservation of momentum and conservation of kinetic energy equations to solve for the unknowns:

Conservation of momentum: m1v1 + m2v2 = m1v1' + m2v2'
Since m1 is initially at rest, v1 = 0:
m2v2 = m1v1' + m2v2'
m2v2 - m2v2' = m1v1'
m2(v2 - v2') = m1v1'

Conservation of kinetic energy: 0.5m1v1^2 + 0.5m2v2^2 = 0.5m1v1'^2 + 0.5m2v2'^2
Since m1 is initially at rest, v1 = 0:
0 + 0.5m2v2^2 = 0.5m1v1'^2 + 0.5m2v2'^2
0.5m2v2^2 - 0.5m2v2'^2 = 0.5m1v1'^2
0.5m2(v2^2 - v2'^2) = 0.5m1v1'^2

Now, we can substitute the expression we found for m1v1' into the kinetic energy equation:
0.5m2(v2^2 - v2'^2) = 0.5(m2(v2 - v2')/(v2 - v2'))m1v1'^2
0.5m2(v2^2 - v2'^2) = 0.5(m1v1')m1v1'^2
0.5m2(v2^2 - v2'^2) = 0.5m1^2v1'^2

Now, we can solve for m2:
m2 = (0.5