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In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 seconds. When the car rebounds from the barrier [in the opposite direction], the magnitude of its velocity is 1.5 mi/h. Use the principle of impulse and momentum to determine the magnitude of the average horizontal force (lb) exerted on the car during the impact.

Respuesta :

Answer:

F = 2074.13 lb

Explanation:

Given that,

Mass of car, m = 2800 lb = 1270.059 kg

Initial speed, u = 5 mi/h = 2.2352 m/s

Final speed, v = - 1.5 mi/h = -0.67056 m/s  (in opposite direction)

Time, t = 0.4 s

We need to find the magnitude of the average horizontal force (lb) exerted on the car during the impact. It can be calculated as :

[tex]F=m\times \dfrac{v-u}{t}\\\\F=1270.059\times \dfrac{-0.67056 -2.2352 }{0.4}\\\\F=9226.21\ N[/tex]

or

F = -2074.13 lb

So, the required force is 2074.13 lb.

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