With the interaction of two carts, their speeds changed by 20 and 60 cm / s.

With the interaction of two carts, their speeds changed by 20 and 60 cm / s. The weight of the larger cart is 0.6 kg. What is the mass of the smaller cart?

According to the law of conservation of momentum:
m1V1 + m2V2 = m1U1 + m2U2; m1 is the mass of the larger cart (m1 = 0.6 kg), m2 is the mass of the smaller cart; V1, V2 are the speeds of the bogies before the collision, U1, U2 are the speeds of the bogies after the collision.
U1 = V1 + ∆V1, U2 = V2-∆V2; ∆V1 is the change in the speed of the larger carriage (∆V1 = 20 cm / s), ∆V2 is the change in the speed of the smaller carriage (∆V2 = 60 cm / s).
m1V1 + m2V2 = m1 (V1 + ∆V1) + m2 * (V2-∆V2).
m1V1 + m2V2 = m1V1 + m1∆V1 + m2V2-m2∆V2.
0 = m1∆V1-m2∆V2.
m1∆V1 = m2∆V2.
m2 = m1∆V1 / ∆V = 0.6 * 20/60 = 0.2 kg.
Answer: The weight of the smaller cart is 0.2 kg.