Q13E - University Physics with Modern Physics

Question

On September 8, 2004, the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The 210-kg capsule hit the ground at 311 km/h and penetrated the soil to a depth of 81.0 cm. 

(a) What was its acceleration (in m/s ² and in g’s), assumed to be constant, during the crash? 

(b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule’s weight. 

(c) How long did this force last?

Step-by-Step Solution

Answer

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

(b) The force exerted on the capsule by the ground is $967470 \mathrm{N} \mathrm{or} \left(470\mathrm{w}\right)$ .

(c) The duration of force is 0.0187 s .

Step 1: Acceleration of Spacecraft and Force on the capsule

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.

Step 2: Determine the acceleration of spacecraft during the crash

(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 $\left(470\mathrm{g}\right)$ .

Step 3: Determine the force exerted on the capsule by ground

(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,

  

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

 

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

Step 4: Determine the duration of the force

(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,

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

 

Therefore, the duration of force is 0.0187 s .