Answer to Question #154175 in Classical Mechanics for Max

Question

In the event of a collision between vehicles, the severity of injuries sustained by drivers can be reduced with an air bag. In a car initially traveling at +100 km / h, an air bag immobilizes a driver with a mass of 62 kg in 0.090 s. The magnitude of the average force exerted by the air bag on the driver is?

Answer 1

Let's first convert km/hr to m/s:

$$v_i=(100\ \dfrac{km}{h})\cdot(\dfrac{1000\ m}{1\ km})\cdot(3600\ \dfrac{1\ h}{3600\ s})=27.77\ \dfrac{m}{s}.$$

We can find the average force exerted by the air bag on the driver from the impulse-momentum change equation:

$$F_{avg}\Delta t=m\Delta v,$$ $$F_{avg}\Delta t=m(v_f-v_i),$$ $$F_{avg}=\dfrac{m(v_f-v_i)}{\Delta t},$$ $$F_{avg}=\dfrac{62\ kg\cdot(0\ \dfrac{m}{s}-27.77\ \dfrac{m}{s})}{0.090\ s}=-1.9\cdot10^4\ N.$$

The magnitude of the average force exerted by the air bag on the driver equals $1.9\cdot10^4\ N.$ The sign minus means that the average force exerted by the air bag on the driver directed in the opposite direction to the motion of the car.

Answer:

$F_{avg}=1.9\cdot10^4\ N.$