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

Answer:

Force exerted on the car, $F=-6.43\times 10^5\ N$

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

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

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{1200\ kg(0-26.8224\ m/s)}{0.05\ s}$

F = -643737.6

or

$F=-6.43\times 10^5\ N$

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

Answer 2

The answer is 6.4331 × 10^5N.