Question

A small 0.1 kg cart is moving at 1.20 m/s on a frictionless track when it collided with a larger 1.00 kg cart at rest. After the collision, the same cart recoils at 0.850 m/s (backwards). What is the speed of the larger cart after the collision

Answer 1

Answer: 0.205 m/s

Step-by-step explanation:

According the conservation of momentum law, the total momentum before the collision (
`p_(i)`) is equal to the total momentum after the collision (
`p_(f)`):

`p_(i)=p_(f)` (1)

Being:

`p_(i)=m_(1)V_(1)+m_(2)V_(2)` (2)

`p_(f)=m_(1)U_(1)+m_(2)U_(2)` (3)

Where:

`m_(1)=0.1 kg` is the mass of the first cart

`V_(1)=1.20 m/s` is the initial velocity of the first cart

`m_(2)=1 kg` is the mass of the second cart

`V_(2)=0 m/s` is the initial velocity of the second cart

`U_(1)=-0.85 m/s` is the final velocity of the first cart

`U_(2)` is the final velocity of the second cart

Substituting (2) and (3) in (1):

`m_(1)V_(1)+m_(2)V_(2)=m_(1)U_(1)+m_(2)U_(2)` (4)

Isolating
`U_(2)`:

`U_(2)=(m_(1)(V_(1)-U_(1)))/(m_(2))` (5)

`U_(2)=(0.1 kg(1.20 m/s-(-0.85 m/s)))/(1 kg)` (6)

Finally:

`U_(2)=0.205 m/s` This is the speed of the second cart after the collision