Two unequal masses m1 and m2 are located in a large, frictionless bowl, and both start at rest. Mass m1 is released at height H and slides to the bottom, where it collides with mass m2. For parts (a) and (b), your answers will be algebraic expressions.
Express your answers only in terms of variables m1, m2, H, g, and mathematical constants, and only in simplest algebraic form.
a. If the two masses stick together in the collision at the bottom of the bowl, find the amount of kinetic energy lost in the collision.
b. If they then slide together up the right-hand side of the bowl, find the maximum height that the two masses reach before sliding back down.
Instead of sticking together, suppose that m1 and m2 collide at the bottom of the bowl perfectly elastically. This applies both to part (c) and (d).
c. Which one of the following statements is now true for the elastic collision?
A. The total K of all masses immediately before collision is less than the total K of all masses immediately after collision.
B.The total K of all masses immediately before collision is the same than the total K of all masses immediately after collision.
C. The total K of all masses immediately before collision is greater than the total K of all masses immediately after collision.
D. Net ∆K of all masses in this perfectly elastic case is the same as in the stuck-together case (part (a)).
d. Immediately after the colision, suppose the velocity of m1 rebounds to the left. Which one of the following must be true about the relationship between the two masses?
A. m1 < 1/2m2
B. m1 < m2
C. m1 < 2m2
D. m1 > 1/2m2
E. m1 > m2
F. m1 >2m2
The Attempt at a Solution
I thought it was Vf = m1v1i + m2v2i/m1 +m2 because it is stated that the kinetic energy lost equals the total kinetic energy before the collision (part a).
Didn't attempt parts b, c, and d.