Question

a 282 kg bumper car moving +3.70 m/s collides with a 155 kg bumper car moving -1.38 m/s. afterwards, the 282 kg car moves at +1.10 m/s. find the velocity of 155 kg car afterwards.(unit=m/s) Please help me !! I have been stuck on this question for 1 hour

Answer 1

3.35

Step-by-step explanation:

Answer 2

Answer: 3.35 m/s

Step-by-step explanation:

This problem is related to the Conservation of Momentum principle, which states the following:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision".

Of course, the momentum of each object may change after the collision, however, the total momentum of the system does not change. In addition, according to this law it is established that the initial momentum
`p_(o)` must be equal to the final momentum
`p_(f)`:

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

Where:

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

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

Being:

`m_(1)=282 kg` the mass of the first car

`V_(1)=3.70 m/s` the initial velocity of the first car

`m_(2)=155 kg` the mass of the second car

`V_(2)=-1.38 m/s` the initial velocity of the second car

`U_(1)=1.10 m/s` the final velocity of the first car

`U_(2)` the final velocity of the second car

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)V_(2))/(m_(2))` (5)

Solving:

`U_(2)=(282 kg(3.70 m/s-1.10 m/s)+(155 kg)(-1.38 m/s))/(155 kg)` (6)

Finally:

`U_(2)=3.35 m/s`