1. The problem statement, all variables and given/known data Three masses are positioned on a frictionless surface, as shown. Initially, mass m1 (1.0 kg) moves with a velocity of 2.0 m/s to the right, mass m2 (2.0 kg) is at rest, and mass m3 (3.0 kg) moves to the left with a velocity of 0.50 m/s. First, mass m1 collides elastically with mass m2 and recoils to the left. Afterwards, mass m3 collides with mass m2 and sticks. Calculate, the speeds of masses m1 and m2 after the first collision. the speeds of masses m2 and m3 after the second collision. 2. Relevant equations M1V1 + M2V2 = M1V1' + M2V2' 1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2 3. The attempt at a solution Part a). First I put in the values and narrowed the conservation of momentum equation to: 2 = v1' + (2) * v2' which rearranges to v1' = 2 - (2)v2' (equation 1.1) For the Conservation of Energy equation I did the same and got, 2 = (0.5) v1'^2 + v2'^2 (equation 1.2) I then substituted the value of v1' from equation 1.1 into 1.2 and simplified to, v2' (3v2' -4) = 0 v2' = 0 or 1.33 since it is not 0 it must be 1.33 m/s I then substitute this answer back into equation 1.1 and get v1' = -0.667 Therefore, the velocity of m1 after the collision is 0.67 m/s to the left and the velocity of m2 = 1.3 m/s to the right. Part b). for this part it is an inelastic collision and therefore only momentum is conserved. m2v2 + m3v3 = (m2 +m3) v2and3 m2 = 2.0 kg v2 = 1.3 m/s m3 = 3.0 kg v3 = -0.50 m/s but these into the above equation and i get ((2.66 kg * m/s) + (-1.5 kg * m/s) / 5.0 kg) = v2and3 v2and3 = 0.23 m/s (positive indicates to the right) Does this look correct? I made an earlier attempt but with a different answer but it didn't sit right so I retried this way.