Question

A car of m = 1200. Kg collides with a tree while traveling 60.oo mph. The collision occurs over a time perido of 0.0500 seconds. Determined the margnitude of the force exerted on the car as it comes to rest during the collision (1 m/s= 2.24 mph

Answer 1

The answer is 6.4331 × 10^5N.

Answer 2

Force exerted on the car,
`F=-6.43* 10^5\ N`

Step-by-step explanation:

It is given that,

Mass of the car, m = 1200 kg

Initial velocity of car, u = 60 mph = 26.8224 m/s

Time of collision, t = 0.05 s

We need to find the magnitude of the force exerted on the car as it comes to rest during the collision (final velocity = 0)

We know that,

impulse = change in momentum

`F.t=m(v-u)`

`F=(m(v-u))/(t)`

`F=(1200\ kg(0-26.8224\ m/s))/(0.05\ s)`

F = -643737.6

or

`F=-6.43* 10^5\ N`

So, the magnitude of force exerted on the car is
`F=-6.43* 10^5\ N`. Hence, this is the required solution.