Question

A 7 kg cart is moving down a track at a speed of 9 m/s and hits a 11 kg cart sitting still. The two have an inelastic collision. What is the resulting speed after the collision?

Answer 1

Answer: 3.5 m/s

Step-by-step explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum
`p_(i)` must be equal to the final momentum
`p_(f)`, and taking into account this is an inelastic collision:

Before the collision:

`p_(i)=mV_(o)+MU_(o)` (1)

After the collision:

`p_(f)=(m+M)V_(f)` (2)

Where:

`m=7 kg` is the mass of the first cart

`V_(o)=9 m/s` is the velocity of the first cart

`M=11 kg` is the mass of the second cart

`U_(o)=0 m/s` is the velocity of the second cart

`V_(f)` is the final velocity of both carts

`p_(i)=p_(f)` (3)

`mV_(o)+MU_(o)=(m+M)V_(f)` (4)

Since
`U_(o)=0 m/s`:

`mV_(o)=(m+M)V_(f)` (5)

`V_(f)=(mV_(o))/(m+M)` (6)

`V_(f)=((7 kg)(9 m/s))/(7 kg + 11 kg)` (7)

Finally:

`V_(f)=3.5 m/s`