Question

A 1,700kg car is being used to give a 1,400kg car a push start by exerting a force of 140N the impulse on the smaller car during the 30.0s of contact is +670kg*m/s. What is the impulse of the smaller car on the larger car?

-814 kg*m/s
0kg *m/s
-670kg*m/s
-550kg*m/s

Answer 1

-670kg*m/s

Step-by-step explanation:

According to Newton's third law of motion, the force that the smaller car exerts on the bigger car is equal and opposite to the force that the bigger car exerts on the smaller car:

`F_(BS)=-F_(SB)`,

where is
`F_(BS)` is the force that the bigger care exerts on the smaller car, and
`F_(SB)` is the force that the smaller care exerts on the bigger car.

And if the force exerted has the same magnitude, then so should the Impulse
`I`, because

`I=F*\Delta t`

The impulse on the smaller car due to the force exerted by the bigger car is:

`I_(BS)=F_(BS)*\Delta t=670kg*m/s`

and the impulse on the bigger car due to the smaller car is

`I_(SB)=F_(SB)*\Delta t`.

Since

`F_(BS)=-F_(SB)`

then

`I_(SB)=-F_(BS)*\Delta t=-670kg*m/s`

`\boxed{I_(SB)=-670kg*m/s}`

The impulse of the smaller car on the larger car is -670kg*m/s.