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.

Answer 2

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

The spring exerts equal and opposite forces on the carts.

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

Using conservation of momentum,

                 $m_{1}v_{1}=-m_{2}v_{2}$

Given that, $m_{1}=3kg,v_{1}=0.82,m_{2}=1.7kg$

Substitute values in above equations.

                  $3*0.82=-1.7*v_{2}\\\\v_{2}=-\frac{2.46}{1.7}=-1.45m/s$

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

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