Please help me solve the following:
An atom of mass m moving in the x direction with speed v collides elastically with an atom of mass 3m at rest. After the collision, the first atom moves in the y direction. Find the direction of motion of the second atom and the speeds of both atoms (in terms of v) after the collision.
I came up with:
1. m1v1i=m2v2f*cos(theta) ---------> x-direction
2. 0 = m1v1f + m2v2f*sin(theta) ----> y-direction
3. (1/2)m1v1i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2
I tried to square the momentum equations and add them (in order to get cos^2 + sin^2 = 1) and I also tried to take the ratio of the momentum equations to solve for theta, but I can't get the correct answers.
Correct answers: v1f = v/sqrt(2); v2f = v/sqrt(6); theta = -35.3
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1. m1v1i=m2v2f*cos(theta) ---------> ...
I have provided detailed explanation in finding the final velocities and angles of two atoms which collide in elastically in two dimensional space.