Particle collision at an angle

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Saxby
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Homework Statement


A particle of mass m traveling at a velocity u makes a perfect elastic collision with a stationary particle. After the collision both particles are observed to be traveling in directions making angles of 30 degrees to the original path of the first particle.

a) Use the laws of the conservation of energy and momentum to write down three equations relating mass and velocities of the particles involved in the collision described above.

b) Solve the equations to find the mass of the seocnd particle and the final velocities of the two particles.


2. relevant equations
Kinetic energy: Ek = (1/2)*m*v2
Conservation of momentum: (m1*u1) + (m2*u2) = (m1*v1) + (m2*v2)
m1 = Mass of particle that was intially moving
m2 = Mass of particle that was intially stationary
u1 = Intial velocity of m1
v1 = Final velocity of m1
v2 = Final velocity of m2

3. The attempt at solution
I believe the first equation in the problem is the conservation of momentum, which for this problem i have written as:

m1*u1 = [m1* ((v1sinθ)2 + (v1cosθ)2)1/2] + [m2 * ( (v2sinθ)2 + (v2sinθ)2)1/2]

I believe the second equation in the problem is kinetic energy:

(1/2)*m1*u12 = (1/2)*m1*(v1sinθ2 + v1cosθ2) + (1/2)*m2*(v2sinθ2 + v2cosθ2)

For the third equation i have no idea, i don't believe it's rotational energy or anything like that. I think it may have somthing to do with the angles but frankly i don't know. Any help would be much apprietiated :)
 
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Saxby said:
I believe the first equation in the problem is the conservation of momentum, which for this problem i have written as:

m1*u1 = [m1* ((v1sinθ)2 + (v1cosθ)2)1/2] + [m2 * ( (v2sinθ)2 + (v2sinθ)2)1/2]
Momentum is a vector so you have to treat is as such. Write separate momentum conservation equations for components parallel and perpendicular to the original direction. (That's how you'll end up with three equations.)
 
Doc Al said:
Momentum is a vector so you have to treat is as such. Write separate momentum conservation equations for components parallel and perpendicular to the original direction. (That's how you'll end up with three equations.)

Thank you, that makes sense. But how do i know what percent of the original momentum goes in the y-direction of both particles and how much goes in the x-direction?
 
Saxby said:
Thank you, that makes sense. But how do i know what percent of the original momentum goes in the y-direction of both particles and how much goes in the x-direction?
You don't need to know anything. Just set up an equation for the x-components and another for the y-components. You have the angles. I would choose the original direction to be along the +x axis.