Question

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

After the collision, both carts move left, Cart A at speed 0.27m/s.

What is the velocity of cart B after the collision? Show your work.

Answer 1

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

Step-by-step 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!