# Homework Help: Momentum problem -- The collision of two particles with different masses

1. Apr 30, 2015

### A.MHF

1. The problem statement, all variables and given/known data
So I have this practice problem with the solution, but I don't understand how:

"A particle of mass m1 and speed v1 in the +x direction collides with another particle of mass m2. Mass m2 is at rest before the collision occurs, thus v2 = 0. After the collision, the particles have velocities v'1 and v'2 . in the xy plane with directions θ1 and θ2 with the x axis as shown below. There are no external forces acting on the system. Express all of your answers in terms of m1, m2, v1, θ1 and θ2.

Q:What is the ratio of the speeds v'2/v'1 ?"

"Conservation of momentum in the y direction: m1v'1 sin θ1 = m2v'2sin θ2
and by solving for ratio we get:

m1sinθ1/m2sinθ2"

Why did we choose the conservation in the y not the x direction?
Can this problem be solved any other way?

2. Relevant equations

-

3. The attempt at a solution

When I first attempted the problem, I thought of solving it like this:
v1'=v1'(sqrt[(cos(θ1)+sin(θ1))^2])
v2'=v2'(sqrt[(cos(θ2)+sin(θ2))^2])
And since momentum is conserved,
m1v1=m1v1'(sqrt[(cos(θ1)+sin(θ1))^2])+m2v2'(sqrt[(cos(θ2)+sin(θ2))^2])

But this feels wrong, and I don't know what's with it.
Please feel free to correct me and give me more info about momentum and collision if needed, I understand it but I don't feel like having a full grasp of the concept.

2. Apr 30, 2015

### paisiello2

I don't understand how you got these equations?

3. Apr 30, 2015

### Staff: Mentor

With no external forces acting the momentum of a system is conserved. Thus the momentum of the system after the collision must be identical to the momentum before the collision.

Momentum is a vector quantity (magnitude and direction). It is conserved separately in both x and y directions and in total. That means if the initial momentum had no y-component (as in this problem), the final momentum must also have no y-component when you sum up the contributions from all the moving parts. Similarly, the final momenta in the x-direction must sum to what it was before the collision.

The given solution for determining the ratio of the speeds V2'/V1' took advantage of the fact that the total momentum in the y-direction happens to be zero.

4. Apr 30, 2015

### A.MHF

Sorry ignore that, I realised I wasn't thinking correctly.