Question

A car moving at 20.0 m/s (72.0 km/h) crashes into a tree. Find the
magnitude of the average force acting on a passenger of mass 70 kg in each
of the following cases. (a) The passenger is not wearing a seat belt. He is
brought to rest by a collision with the windshield and dashboard that lasts
2.0 ms. (b) The car is equipped with a passenger-side air bag. The force due
to the air bag acts for 45 ms, bringing the passenger to rest.​

Answer 1

Given:

A car moving at 20.0 m/s (72.0 km/h) crashes into a tree.

To find:

Average force when

• Passenger collides with windscreen

• Passenger collides with air-bag

Calculation:

Force is defined as the instantaneous rate of change of momentum with respect to time.

In the 1st case:

Let force be F1;

$\therefore \: \sf{F1 = \dfrac{momentum \: change}{time} }$

$= > \: \sf{F1 = \dfrac{mv - mu}{t} }$

$= > \: \sf{F1 = \dfrac{70(0 - 20)}{2 \times {10}^{ - 3} } }$

$= > \: \sf{F1 = \dfrac{70 \times ( - 20)}{2 \times {10}^{ - 3} } }$

$= > \: \sf{F1 = \dfrac{70 \times ( - 10)}{ {10}^{ - 3} } }$

$\boxed{ = > \: \sf{F1 = - 7 \times {10}^{5} \: N }}$

In the 2nd case:

Let force be F2;

$\therefore \: \sf{F2 = \dfrac{momentum \: change}{time} }$

$= > \: \sf{F2 = \dfrac{mv - mu}{t} }$

$= > \: \sf{F2 = \dfrac{70(0 - 20)}{45 \times {10}^{ - 3} } }$

$= > \: \sf{F2 = \dfrac{70 \times ( - 20)}{45 \times {10}^{ - 3} } }$

$\boxed{ = > \: \sf{F2 = - 31.1 \times {10}^{3} \: N }}$

So, force is applied much less in case of air-bag , hence they are used as safety devices in vehicles in case of accidents.

Hope It Helps.