Question

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

cars collide and entangle. What is the resulting velocity of the wreckage?

Answer 1

Answer:-2.61 m/s

Step-by-step explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum
`p_(o)` must be equal to the final momentum
`p_(f)`:

`p_(o)=p_(f)` (1)

Where:

`p_(o)=mV_(o)+MU_(o)` (2)

`p_(f)=(m+M)V_(f)` (3)

`m=1200 kg` is the mass of the first car

`V_(o)=20 m/s` is the velocity of the first car, to the North

`M=1400 kg` is the mass of the second car

`U_(o)=-22 m/s` is the mass of the second car, to the South

`V_(f)` is the final velocity of both cars after the collision

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

Isolating
`V_(f)`:

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

`V_(f)=((1200 kg)(20 m/s)+(1400 kg)(-22 m/s))/(1200 kg+1400 kg)` (6)

Finally:

`V_(f)=-2.61 m/s` (7) This is the resulting velocity of the wreckage, to the south