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:

$F=\frac{I}{t}=\frac{-32,000 kg m/s}{7 s}=-4,571 N$

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