Question

A 10.0 kg cart and a 15 kg cart are locked together with a compressed spring between them. They are then released so that the spring pushes the two carts apart. The 10.0 kg cart is moving at 3.0 m/s afterward. How fast is the 15 kg cart moving

Answer 1

Answer: The velocity of the cart-2 is 2 m/s moving in direction opposite of the cart-1's direction.

Step-by-step explanation:

Mass of cart-1 ,
`m_1=10.0 kg`

Velocity of cart-1 before the release of the spring,
`u_1=0m/s`

Mass of cart-2 ,
`m_1=15kg`

Velocity of cart-2 before the release of the spring,
`u_2=0m/s`

After releasing of the spring:

Velocity of cart-1 after the release of the spring,
`v_1=3 m/s`

Velocity of cart-2 after the release of the spring,
`v_2=?`

According to law of conservation of momentum:

`m_1u_1+m_2u_2=m_2v_2+m_2v_2`

`10.0 kg* 0m/s+15 kg* 0 m/s=10.0 kg* 3m/s+15v_2`

`0+0=30 kg m/s+15 mv_2`

`v_2=-2 m/s`

Negative sign depicts that direction of cart-2 which is moving in the opposite direction of the cart-1's direction. So, the velocity of the cart 2 is 2 m/s.

Answer 2

For the answer to the question above,
conservation of momentum. Before release, the system momentum is 0; after release the system momentum is m1*v1 - m2*v2, which = 0 so v2 = m2*v2/m1 = 10*4.5/15 = 3.0 m/s

I hope my answer helped. Have a nice day!