Two carts pressed against each other and compressing the spring were released. When the spring straightened

Two carts pressed against each other and compressing the spring were released. When the spring straightened, the cart weighing 5 kg. acquired a speed of 3m / s. Determine the speed of the second trolley weighing 2.5 kg?

m1 = 5 kg.

m2 = 2.5 kg.

V1 = 3 m / s.

V2 -?

The spring acts on different bogies with the same force: F2 = F1.

According to Newton’s 2 law, the force F, which acts on a body, is determined by the formula F = m * a, where m is the mass of the body, and is the acceleration of the body.

F1 = m1 * a1.

F2 = m2 * a2.

We express the acceleration of the body a according to its definition: a = (V – V0) / t.

Since the initial speeds of the cart are V01 = V02, then a1 = V1 / t and a2 = V2 / t.

m1 * V1 / t = m2 * V2 / t.

m1 * V1 = m2 * V2.

V2 = 5 kg * 3 m / s1 / 2.5 kg = 6 m / s.

Answer: the second carriage will acquire a speed of V2 = 6 m / s.