During the interaction of two trolleys weighing 4 kg and 6 kg, it was found that the first trolley acquired

During the interaction of two trolleys weighing 4 kg and 6 kg, it was found that the first trolley acquired a speed of 0.5 m / s in a certain time interval, what speed the second trolley would have.

m1 = 4 kg.
m2 = 6 kg.
ΔV1 = 0.5 m / s.
ΔV2 -?
Since the carts only interact with each other, they can be considered closed.
For a closed-loop system, the momentum conservation law is valid: the sum of the impulses of a closed-loop system does not change.
m1 * V1 + m2 * V2 = m1 * V1 “+ m2 * V2”, where m1 * V1, m1 * V1 “are the pulses of the first cart before and after the collision, m2 * V2, m2 * V2” are the pulses of the second cart before and after the collision.
m1 * V1 “- m1 * V1 = m2 * V2 – m2 * V2”.
m1 * (V1 “- V1) = – m2 * (V2” – V2).
m1 * ΔV1 = – m2 * ΔV2.
ΔV2 = – m1 * ΔV1 / m2.
The “-” sign indicates that if the speed of the first cart increases, the speed of the second cart decreases. If the speed of the first carriage decreases, then the speed of the second carriage increases.
ΔV2 = 4 kg * 0.5 m / s / 6 kg = 0.33 m / s.
Answer: the speed of the second carriage has decreased by ΔV2 = 0.33 m / s.