Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of36 m. During the collision at the bottom of the elevator shaft, a 90 kgpassenger is stopped in5.0 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) What are the magnitudes of the average force on the passenger during the collision? If the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab hits the bottom of the shaft, What are the magnitudes of the (c) impulse and (d) Average force (assuming the same stopping time)?

1. The mass of the passenger, m = 90kg .
2. The height from which the elevator falls, h = 36 m .
3. The time of collision,$∆t=0.5\mathrm{ms}$.
4. The speed of passenger relative to cab floor,$v_{pc}=7.0\mathrm{m}/\mathrm{s}$ .

The elevator and the passenger are in free fall motion till they collide at the bottom. Apply the impulse-linear momentum theorem in order to determine the magnitude of the impulse acting on the passenger. It can also determine the average force acting on the passenger using the relation between the impulse and average force which is given as

$$\begin{gathered} \overrightarrow{\mathbf{J}}\mathbf{=}\mathbf{∆}\overrightarrow{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{∆}\overrightarrow{\mathbf{v}} \\ \mathbf{J}\mathbf{=}{\mathbf{F}}_{\mathbf{avg}}\mathbf{∆}\mathbf{t} \end{gathered}$$

You need to determine the speed of the passenger and the cab as they fall through the heightunder gravity. Since the passenger starts from rest, we use the kinematical equation to determine the final velocity as,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{as} \\ {\mathrm{v}}^2=0+2(-9.8)(-36) \\ =705.6 \\ =26.6\mathrm{m}/\mathrm{s} \end{gathered}$$

To determine the magnitude of the impulse on the passenger, we apply the impulse-linear momentum theorem.

$$\begin{gathered} \overrightarrow{\mathrm{J}}=∆\overrightarrow{\mathrm{p}} \\ =\mathrm{m}∆\overrightarrow{\mathrm{v}} \\ \mathrm{J}=∆\mathrm{p} \\ =\mathrm{m}(\mathrm{v}-\mathrm{u}) \\ =90\times \left(26.6-0\right) \\ =2.39\times {10}^3\mathrm{N}.\mathrm{s} \end{gathered}$$

The average force acting on the passenger can be determined as

$$\begin{gathered} \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}∆\mathrm{t} \\ {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{∆\mathrm{t}} \\ =\frac{2.39\times {10}^3}{5.0\times {10}^{-3}} \\ =4.78\times {10}^5\mathrm{N} \end{gathered}$$

As the passenger jumps up in the elevator cab, its effective velocity relative to ground is

$$\begin{gathered} \overrightarrow{V_{pg}}=\overrightarrow{V_{cg}}+\overrightarrow{V_{pc}} \\ =-26.6+7.0 \\ =-19.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, the impulse on the passenger will be

$$\begin{gathered} \overrightarrow{J}=∆\overrightarrow{p} \\ =m∆\overrightarrow{v} \\ =∆p \\ =m(v-u) \end{gathered}$$

The average force acting on the passenger can be determined as

$$\begin{gathered} \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}∆\mathrm{t} \\ {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{∆\mathrm{t}} \\ =\frac{1.76\times {10}^3}{5.0\times {10}^{-3}} \\ =3.52\times {10}^5\mathrm{N} \end{gathered}$$