Question

On September 8, 2004, the genesis spacecraft crashed in Utah desert because it's 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.


1) Assuming it to be constant, what is its acceleration (in m/s^2) during the crash?

2) Assuming it to be constant, what is its acceleration (in g's) during the crash?

3) What force did the ground exert on the capsule during the crash? Express the force in newtons.

4) What force did the ground exert on the capsule during the crash? Express the force as a multiple of the capsule weight.

5) For how long did this force last?

Answer 1

Answer:

Explanation:

mass of capsule, m = 210 kg

initial velocity, u = 311 km/h = 86.39 m/s

final velocity, v = 0 m/s

distance of soil penetrated, d = 81 cm = 0.81 m

1.

Let a is the acceleration.

Use third equation in motion.

v² = u² + 2as

0 = 86.39² + 2 x a x 0.81

a = - 4606.9

a = - 4607 m/s²

Thus, the acceleration during the crash is 4607 m/s².

2. g = 9.8 m/s²

a / g = 4607 / 9.8

a = 470 g

Thus, the acceleration is 470 g .

3. Force = m a

F = 210 x 4607

F = 967470 N

Thus, the force is 967470 N.

4.  Weight of capsule = m g = 210 x 9.8 = 2058 N

F / mg = 967470 / 2058

F = 470 times the weight of capsule.

5. Let t be the time.

Use first equation of motion.

v = u + at

0 = 86.39 - 4607 x t

t = 0.0188 second

Answer 2

1. The acceleration during the crash is 4607 m/s².

2. The acceleration during the crash is 470 times the gravitational acceleration (470g) .

3. The force exerted by ground on the capsule is 967470 N.

4. The force exerted on the ground is 470 times the weight of capsule.

5. The time interval for the force is 0.0188 s.

Given data:

The mass of capsule is, m = 210 kg.

The speed at which capsule heat the ground is, u = 311 km/h = 86.39 m/s.

The final speed of capsule is, v = 0 m/s.

The depth of soil penetrated is, d = 81.0 cm = 0.81 m.

1.

Let a is the acceleration.  Then by third equation of motion as,

v² = u² + 2as

0 = 86.39² + 2 x a x 0.81

a = - 4606.9

a = - 4607 m/s²

Thus, we can conclude that the acceleration during the crash is 4607 m/s².

2.

Let g be the gravitational acceleration then, g = 9.8 m/s². So, the acceleration (g') during the crash is,

a / g = 4607 / 9.8

a = 470 g

Thus we can conclude that the acceleration during the crash is 470 g .

3.

Let us use the Newton's Second law, which says that the force equals the product of mass and acceleration,

F = m × a

Solving as,

F = 210 x 4607

F = 967470 N

Thus, we can conclude that the force exerted by ground on the capsule is 967470 N.

4.  

The force exerted by the ground is also due to the weight of capsule. So, the weight of capsule is,

W = m g

W = 210 x 9.8

W = 2058 N

Then,

F / mg = 967470 / 2058

F = 470 times the weight of capsule.

Thus, we can conclude that the force exerted on the ground is 470 times the weight of capsule.

5.

Let t be the time.  Use first equation of motion as,

v = u + at

0 = 86.39 - 4607 x t

t = 0.0188 second

Thus, we can conclude that the time interval for the force is 0.0188 s.

Learn more about the Newton's Second law here:

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