Consider the physics of a rear-end automobile collision in which both cars are e

Question

Consider the physics of a rear-end automobile collision in which both cars are engineered to absorb as much energy as possible on impact in order to lessen injury to its occupants. One of the vehicles is stopped at a light, the other runs into it from the rear at 50 km/h. Assume both vehicles have a mass of 1000 kg and that there are unrestrained occupants in both with masses of 60 kg.   

1. What is the resulting motion of the pair of cars?

2. Within the frame of reference of each car, how much momentum does an occupant have after the collision?

3. If the occupant of the incident car is not protected by an air bag, with what velocity in m/s does he strike the windscreen following the collision?

4. Assuming the collision duration is 0.2 seconds, what acceleration backward does the occupant of the stationary car feel on impact? (Compare to the acceleration of gravity.)

Explanation / Answer

1. Applying momentum conervation,

(1060 x 0) + (1060 x 50) = (1060 + 1060) v

v = 25 km/h

2. speed of each occupant = 25 km/h

3. v = 25 km/ h Or 25 x 1000 / 3600 m/s

or 6.94 m/s

4. v = vi + a t

6.94 = 0 + a(0.2)

a = 34.7 m/s^2

compare to g , a = (34.7 / 9.8) g = 3.5 g

so acc is 3.5 times of acc. due to gravity.

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