Problem 3: 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.
Think of the man as a mass of 70 kg moving with the speed of the car, i-e, 20.0 m/s. So he has a momentum, mv. In case (a), it directly collides with the dashboard and comes to stop in 2.0 milliseconds. In case (b), the passenger stops in a longer time of 45 milliseconds. Clearly, in case (b) the rate of change of momentum is slower than the case (a). (Do you know what is meant by “rate of change”)?
vi = 20 m/s vf = 0 m/s
mass = 70 kg Δt1 = 2 ms
Δt2 = 45 ms
Average force on the mass for Δt1 and Δt2.
Put values in (2) from the given data.
J = 70 × 0 – 70 × 20 = – 1400 kg ms-1 ….. (3)
Case (a) Put Δt = 2 ms = 2 × 10-3 s and value of J from equation 3 in equation 1.
Case (b) Put Δt = 45 ms = 45 × 10-3 s and value of J from equation 3 in equation 1.
Note: The negative signs show the force is retarding. Its magnitude is the same.
A very good and brief explanation with easy solutions
Thanks for the comments.
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