Question

Math Practice: Cart A loaded with blocks (600g) moving left at 0.7m/s hits stationary cart B (200g).

Answer 1

Answer:

The velocity of cart B after the collision is 1.29 m/s.

Explanation:

We can find the velocity of cart B by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{A}v_{A_{i}} + m_{B}v_{B_{i}} = m_{A}v_{A_{f}} + m_{B}v_{B_{f}}$

Where:

$m_{A}$ is the mass of cart A = 600 g = 0.6 kg            

$m_{B}$ is the mass of cart B = 200 g = 0.2 kg

$v_{A_{i}}$ is the inital velocity of cart A = 0.7 m/s

$v_{A_{f}}$ is the final velocity of cart A = 0.27 m/s

$v_{B_{i}}$ is the initial velocity of cart B = 0

$v_{B_{f}}$ is the final velocity of cart B =?

Taking the left direction as the positive horizontal direction:

$0.6 kg*0.7 m/s + 0 = 0.6 kg*0.27 m/s + 0.2 kg*v_{B_{f}}$

$v_{B_{f}} = 1.29 m/s$

                       

Therefore, the velocity of cart B after the collision is 1.29 m/s.

I hope it helps you!