Question

A block of mass m1 = 4.15 kg is released from circled A. It makes a head-on elastic collision at circled B with a block of mass m2 = 10.5 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

Answer 1

an elastic collision both momentum and energy are conserved. So, first thing's first - find the potential energy of the block at its starting point:
PE = mgh
PE = (5.65)(9.81)(5) = 277.13
From there, we can find the velocity of block of mass m1:
KE = 277.13 = (1/2)mv^2
277.13 = (1/2)(5)v^2
v^2 = 110.85
v = 10.53

Now, we set up a system of equations that will allow us to solve for the velocities of the blocks. We know that momentum is conserved, so we write that the momentums of the blocks will have to add up to 10.53. (However, one of the blocks' individual momentums may be greater than the starting momentum because one may have negative momentum, due to directions of the blocks.)
(m1)(v1) + (m2)(v2) = (10.53)(5.65)
(5.65)(v1) + (8.5)(v2) = 59.5

Also, energy of the blocks is conserved, so we find that:
(1/2)(m1)(v1)^2 + (1/2)(m2)(v2)^2 = 277.13 (<---- 277.13 comes from PE earlier in the problem).
(2.825)(v1)^2 + (4.25)(v2) = 277.13

Now, if we solve the momentum equation for v2 in terms of v1, we can plug our solution into the energy equation to find a numerical value for v1:
(8.5)(v2) = 59.5 - (5.65)(v1)
v2 = (59.5 - [5.65][v1])/8.5

(2.825)(v1)^2 + (4.25)[(59.5-[5.65][v1])/8.5]^2 = 277.13

Working out the algebra gives you a value of -1.48084 (or 9.8907, but that value is extraneous) for v1. So, to find how high the block goes, we find the kinetic energy after the collision (using our new v1) and convert it to potential energy.
KE = PE
(1/2)(m1)(v1)^2 = (m1)(g)(h)
(1/2)(5.65)(-1.48084)^2 = (5.65)(9.81)(h)
h = 0.111768m ~ 0.112m