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 airbag. The force due to the airbag acts for 45 ms, bringing the passenger to rest.
((a) 7.0 x 10' N (b) 3.1 x 10' N)


Given:


The initial Velocity of the passenger is the same as the car = vi = 20 m/s
Final Velocity of the car = vf = 0 m/s
Mass of the passenger = m = 70 kg

Case (a) time = t =2 ms = 2 ｘ10⁻³ s
Case (b) time = t =45 ms = 45 ｘ10⁻³ s

To Find:

(a) Force acting on the passenger = F = ? (when t =2 ms)
(b) Force acting on the passenger = F = ? (when t =45 ms)

Solution:

Case (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.


Method-1:

First to find the deceleration (retardation)

vf = vi + a t 

0 m/s =  20 m/s + a ｘ 2 ｘ10⁻³ s

after simplifying we get

 a = -10,000 m/s² = -1 ｘ10⁴ m/s²


Now according to the Newton's second law, we can calculate the force by the passenger on the windshield and dashboard after the crash.

Force by passenger = F = m a

F = 70 kg ｘ-1 ｘ10⁴ m/s²

F = - 7ｘ10⁵ N

By using Newton's 3rd Law, the Force by the passenger on the windshield and dashboard will be equal in magnitude but opposite in direction to Force on (received by) the passenger. So, In case (a) the force on the passenger will be  7ｘ10⁵ N

Method-2:

We know that a force in terms of momentum is defined as the rate of change in momentum. Mathematically it is written as

 F = `\frac {ΔP}{t}` -----------(1)

ΔP = m vf - m vi 

ΔP = 70 kg (0 m/s) - 70 kg (20 m/s) 

ΔP =  0 - 1400 kg m/s

ΔP = - 1400 kg m/s

now using equation (1)

F = `\frac {- 1400 kg m s^{-1}}{2ｘ10⁻³ s}`

After simplifying we get

F = - 7ｘ10⁵ N

By using Newton's 3rd Law, the Force by the passenger on the windshield and dashboard will be equal in magnitude but opposite in direction to Force on (received by) the passenger. So, In case (a) the force on the passenger will be  7ｘ10⁵ N


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Case (b) The car is equipped with a passenger-side airbag. The force due to the airbag acts for 45 ms, bringing the passenger to rest.


Method-1:

First to find the deceleration (retardation)

vf = vi + a t 

0 m/s =  20 m/s + a ｘ 45 ｘ10⁻³ s

after simplifying we get

 a = - 444.4 m/s² = - 4.44 ｘ10² m/s²


Now according to the Newton's second law

F = m a

F = 70 kg ｘ-4.44 ｘ10² m/s²

F = - 31,111.1 N

F = -3.1 ｘ10⁴ N

By using Newton's 3rd Law, the Force by the passenger on the windshield and dashboard will be equal in magnitude but opposite in direction to Force on (received by) the passenger. So, In case (a) the force on the passenger will be  3.1ｘ10⁴ N


Method-2:

We know that a force in terms of momentum is defined as the rate of change in momentum. Mathematically it is written as

 F = `\frac {ΔP}{t}` -----------(1)

ΔP = m vf - m vi 

ΔP = 70 kg (0 m/s) - 70 kg (20 m/s) 

ΔP =  0 - 1400 kg m/s

ΔP = - 1400 kg m/s

now using equation (1)

F = `\frac {- 1400 kg m s^{-1}}{45 ｘ10⁻³ s}`

After simplifying we get

F = -3.1 ｘ10⁴ N

By using Newton's 3rd Law, the Force by the passenger on the windshield and dashboard will be equal in magnitude but opposite in direction to Force on (received by) the passenger. So, In case (a) the force on the passenger will be  3.1ｘ10⁴ N



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