Question

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.

Answer 1

F = 2074.13 lb

Step-by-step 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 :

`F=m* (v-u)/(t)\\\\F=1270.059* (-0.67056 -2.2352 )/(0.4)\\\\F=9226.21\ N`

or

F = -2074.13 lb

So, the required force is 2074.13 lb.