Off-center elastic collision with m2=2m1

[Rubidium]

Homework Statement

The mass m1 has the velocity (v1i)\hat{i} and makes an off-center collision with m2=2m1. The final velocities are v1f=a1\hat{i}+b1\hat{j}, and v2f=a2\hat{i}+b2\hat{j}. Assuming elastic collision and v2i=0m/s, obtain the values of a1, a2, and b2 for the given value of b1. Also obtain the angles \theta1 and \theta2 of v1f and v2f with the x-axis. Retain the solutions for a1>0.
m1=1.24 kg
v1i=6.00 m/s
b1=1.80 m/s

Homework Equations

The Attempt at a Solution

I have tried using the equations derived from the conservation of linear momentum in several forms but every time I end up with too many unknowns. How do I solve this problems without knowing either angle of the final velocities. All he gives us is the y component of the final velocity of mass 1. I'm stuck. Please help.

 

[Doc Al]

The collision is elastic. What does that tell you?

 

[Rubidium]

That the kinetic energy of the system is the same before as it is after the collision.

 

[Rubidium]

I also did find b2. I used the relation: 0=m1v1fsin(theta1)+m2v2fsin(theta2), and got b2=-0.9 m/s. But without either angle of deflection, a1, or a2 I am stuck from there.

 

[Doc Al]

That the kinetic energy of the system is the same before as it is after the collision.

Right, kinetic energy is conserved. That, along with the two conservation of momentum equations (horizontal and vertical), should allow you to solve for the three unknown parameters. (Once you find the constants you can use them to find the angles.)

 

[Rubidium]

I used the vertical conservation of momentum equation but I don't know how to use the horizontal momentum equation without knowing a1, a2, or either angle. Can you tell me more about that?

 

[Doc Al]

What's the horizontal component (\hat{i}) of the initial momentum? Of the final momentum? Set them equal!

You don't have to know the values of a1 or a2 ahead of time--you're going to solve for them. Since you have three unknowns, you'll need three equations. Conservation of momentum will give you two equations.