Answer:
Step-by-step explanation:
elastic collisions conserve momentum and kinetic energy
To conserve momentum means that the center of mass of the system (CoM) does not change velocity
The CoM has velocity
0.0118(0.191) + 0.0174(-0.282) / (0.0118 + 0.0174) = -0.090856... call it -0.091 m/s
To conserve kinetic energy we can think of the CoM as an Ideal spring returning to each mass that hits it an identical speed in the opposite direction.
The CoM sees the 11.8 g object approach at 0.191 - (-0.091) = 0.282 m/s
and will see it depart at -0.282 m/s.
A ground based observer would see it depart at the sum of the CoM velocity and the relative velocity
v₀₁₁₈ = - 0.091 - 0.282 = - 0.373 m/s or 37.3 cm/s left.
The CoM sees the 17.4 g object approach at -0.282 - (-0.091) = - 0.191 m/s
and will see it depart at +0.191 m/s.
A ground based observer would see it depart at the sum of the CoM velocity and the relative velocity
v₀₁₇₄ = - 0.091 + 0.191 = 0.100 m/s or 10.0 cm/s right.
to check we look that the relative velocity of approach equals the relative velocity of departure
before collision
19.1 - (- 28.2) = 47.3 cm/s
after collision
10.0 - (-37.3) = 47.3 cm/s
so checks out