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?

Question

Physics

Missy

22 April, 08:25

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?

+1

Answers (1)

Know the Answer?

Jackie · Answer 1

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.