Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 30.0 m. During the collision at the bottom of the elevator shaft, a 85.0 kg passenger is stopped in 3.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision

a) impulse ft = 2062.2Ns

b) average force F = 6.9×10^5N

Explanation: This question is solved by using

1. Conservation of energy

2. Impulse = change in momentum.

3 Newton 2nd law of motion.

Please find the attached file for the solution

A) Impulse = 2062.1 Ns

B) Average Force = 687,366.67 N or 687.367 KN

Explanation:

We are given that ;

- elevator cab free-falls from a height of 30.0 m

- mass of passenger = 85kg

- collision time = 3 ms = 0.003 s

A) Impulse is given by;

J = Δp = m(Vf - Vi)

Where,

m = mass

Vf = final velocity before elevator hits ground

Vi = initial velocity which is zero in this case

Now, from conservation of energy, we know that;

(1/2)m(Vf)² = mgh

g is gravitational acceleration and h is the height of fall.

Thus;

Vf² = 2gh

Vf = √(2gh)

So, Vf = √(2 x 9.81 x 30) = √588.6 = 24.26 m/s

Since we know Vf and Vi, thus;

Impulse = m(Vf - Vi) = 85(24.26 - 0) = 2062.1 Ns

B) Average force is given by;

F(av) = J/Δt

Where J is Impulse and Δt is time of collision.

Thus, F(av) = 2062.1/0.003 = 687,366.67 N

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