91Ó°ÊÓ

Given Data:

• The mass of the capsule, $m=210\mathrm{kg}$.
• The speed of the capsule on the ground, $u=311\mathrm{kg}/\mathrm{h}$.
• The penetration in the soil by capsule, $d=81\mathrm{cm}$.

Acceleration of Spacecraft and Force on the capsule:

The acceleration o the spacecraft is found by using the third equation of motion, and the force on the capsule is calculated by using Newton’s second law of motion.

(a)

The acceleration of the spacecraft during the crash is given as follows;

$v^2=u^2-2ad$

Here v is the final speed of the spacecraft, and its value is zero, d penetration in soil by capsule.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} {\left(0\right)}^2={\left[\left(311\mathrm{km}/\mathrm{h}\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\left(\frac{1\mathrm{h}}{3600\mathrm{s}}\right)\right]}^2-2\mathrm{a}\left(81\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ a=4607\mathrm{m}/{\mathrm{s}}^2 \\ a=\left(4607\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{\mathrm{g}}{9.8\mathrm{m}/{\mathrm{s}}^2}\right) \\ a=470g \end{gathered}$$

Therefore, the acceleration of spacecraft during the crash is $4607\mathrm{m}/{\mathrm{s}}^2$or$\left(470\mathrm{g}\right)$ .

(b)

The force exerted on the capsule by the ground is calculated as:

$F=ma$

Here, a is the acceleration of the capsule during the crash and m is the mass of the capsule.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} F=\left(210kg\right)\left(4607m/s^2\right) \\ F=967470N \\ F=m\left(470g\right) \\ F=470mg \\ F=470w \end{gathered}$$

Therefore, the force exerted on the capsule by the ground is 967470 N or (470w) .

(c)

The first equation of motion to calculate the duration of exerted force is given as:

$v=u-at$

Here, $t$is the duration for which the ground exerts a force on the capsule.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} 0=\left(311\mathrm{km}/\mathrm{h}\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\left(\frac{1\mathrm{h}}{3600\mathrm{s}}\right)-\left(4607\mathrm{m}/{\mathrm{s}}^2\right)t \\ t=0.0187s \end{gathered}$$

Therefore, the duration of force is 0.0187 s .