Question

A 1000kg car has a speed of 32m/s. If it takes 7s to stop the car, what is the impulse and the average force acting in the car

Answer 1

1) Impulse: -32,000 kg m/s

The impulse acting on the car is equal to the change in momentum of the car:

`I= \Delta p = m (v-u)`

where in this problem we have

m = 1000 kg is the mass of the car

v = 0 m/s is the final velocity of the car

u = 32 m/s is the initial velocity of the car

Substituting values into the equation, we find

`I=(1000 kg)(0-32 m/s)=-32,000 kg m/s`

2) -4751 N

The impulse exerted on the car is also equal to the product between the average force, F, and the duration of the collision, t:

`I=Ft`

where in this situation we know

`I=-32,000 kg m/s` is the impulse

t = 7 s is the duration of the collision

Solving the formula for F, we find the average force:

`F=(I)/(t)=(-32,000 kg m/s)/(7 s)=-4,571 N`

and the negative sign means that the force is in the opposite direction to the motion of the car.