Three balls of masses m1, m2 and m3 are suspended in a horizontal line by light wires and are almost touching. The mass m1 is given a horizontal velocity v so that it collides head-on with the mass m2.
Find an expression for the final kinetic energy of m3?
What value of m2 results in the maximum energy transfer to the mass m3?
The Attempt at a Solution
I considered the problem as two successive two body elastic collisions, with one body initially stationary.
For a collision between mass m1 with initial velocity v and stationary mass m2, I transformed in the zero momentum frame (velocity m1*v/(m1*m2) ), found the respective velocities before and after the collision (the balls keep their velocities form before the collision but swap directions) and then transformed back into the lab frame. So after the collision I got:
v1 = (m1-m2)*v/(m1+m2), for the first ball
v2 = 2m1*v/(m1+m2) for the second ball
So the second ball receives a fraction of 4*m1*m2/((m1+m2)**2) kinetic energy from the first ball
Similarly, the third ball receives a fraction of 4*m2*m3/((m2+m3)**2) from the second ball, so the kinetic energy of the third ball at the end is:
And to get m2 for the maximum energy transfer I differentiated this expression with respect to m2 and got m2=sqrt(m1*m3).
Now I am told neither of these results is correct, and I'm not sure where I've gone wrong, or if the answer I'm comparing to is wrong. Also, apologies for the messy maths, I'm not sure how to make the equations display nicer.