Question

Two stationary carts are tied together by a thread with a compressed spring held between them. When the thread is cut, the two carts move apart. After the spring is released, one cart m = 3.00 kg travels east with a velocity of 0.82 m/s. What is the magnitude of the velocity of the second cart (m = 1.70 kg) after the spring is released?​

Answer 1

Answer:

1.45 m/s

Explanation:

The spring exerts equal and opposite forces on the carts.

Therefore, the impulses on the carts are equal and opposite.

J₁ = -J₂

m₁Δv₁ = -m₂Δv₂

(3.00 kg) (0.82 m/s − 0 m/s) = -(1.70 kg) (v − 0 m/s)

v = -1.45 m/s

The magnitude of the second cart's velocity is 1.45 m/s.