Elastic Glancing Collision: Final Speeds and Scattering Angle

[kkuhnert]

Homework Statement

Two masses (m and 3m) are moving towards each other along the x-axis with the same initial speed (v). Mass m is traveling to the left while mass 3m is traveling to the right. They undergo elastic glancing collision such that mass m is moving downwards after the collision at right angles from its initial direction. a) Find the final speed of the two masses and b) The angle at which the 3m mass is scattered.

Homework Equations

Ei=Ef
E=(0.5)mv^2
pi=pf
p=mv

The Attempt at a Solution

Pi=pf (x-direction)
3mv-mv = 3mv
2mv = 3mv
v=3mv/2m
v=mv
0=m

pi = pf (y-direction)
0=-mv+3m(1\3)v
mv=mv

 

[Delphi51]

Pi=pf (x-direction)

3mv-mv = (3m)*Vx where Vx is the velocity of 3m in the x direction after collision
2mv = 3mv*Vx
Vx = (2/3)*v

pi = pf (y-direction)
0=-mv+3m(1\3)v
mv=mv

Oh, I made that mistake, too! But you can't use "v" for the downward speed of mass m after the collision as that letter is used for something else. Use another letter, perhaps "u". And Vy for the upward speed of 3m. Then you see that the equation can't be solved since you have two unknowns, u and Vy.

You can get another equation from the fact that the collision is elastic. That should be delightfully complicated.