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Ugo [173]
3 years ago
14

NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant

vertical speed of 20 m/s after deploying a parachute with an area of 200 m^2. The drag coefficient of the parachute is 1.855. The acceleration due to gravity on Mars is 3.71 m/s2, and the density of the atmosphere near the surface is 0.01 kg/m^3.
a. Calculate the maximum mass of the Mars lander to ensure it can land safely?
b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m².
c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?
Physics
1 answer:
Viktor [21]3 years ago
3 0

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

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1. The final velocity of the 30.0 g marble is 0.100 m/s to the left.

2. The final velocity of the 10.0 g marble is 0.500 m/s to the right.

3. The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

4. The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

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6. The change in kinetic energy for the 10.0 g marble is 4.5 × 10⁻⁴ J

Explanation:

Hi there!

Since the collision is elastic both the momentum and kinetic energy of the system comprised by the two marbles is conserved, i.e., it remains constant after the collision.

momentum before the collision = momentum after the collision

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Where:

mA and vA = mass and velocity of the 10.0 g marble.

mB and vB = mass and velocity of the 30.0 g marble.

vA´ and vB´ = final velocities of marble A and B respectively.

The kinetic energy of the system is also conserved:

kinetic energy before the collision = kinetic energy after the collision

1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

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mA · vA + mB · vB = mA · vA´ + mB · vB´

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2 × 10⁻³ kg · m/s =  0.010 kg · vA´ + 0.030 kg · vB´

Solving for vA´

0.2 kg · m/s - 3 kg · vB´ = vA´

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1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

0.010 kg · (-0.400 m/s)² + 0.030 kg · (0.200 m/s)² = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

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Replacing vA´:

2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

2.8 × 10⁻³ kg · m/s = 0.010 kg · (0.2 kg · m/s - 3 kg · vB´)² + 0.030 kg · (vB´)²

(I will omit units from this point for more clarity in the calculations)

2.8 × 10⁻³  = 0.010  (0.2 - 3 · vB´)² + 0.03 · (vB´)²

2.8 × 10⁻³ = 0.010(0.04 - 1.2 vB´ + 9(vB´)²) + 0.03(vB´)²

divide by 0.01 both sides of the equation:

0.28 = 0.04 - 1.2 vB´ + 9(vB´)² + 3(vB´)²

0 = -0.28 + 0.04 - 1.2 vB´ + 12(vB)²

0 = -0.24 - 1.2 vB´ + 12(vB)²

Solving the quadratic equation:

vB´= 0.200  m/s

vB´ = -0.100  m/s

The first value is discarded because it is the initial velocity. Then, the final velocity of the 30.0 g marble is 0.100 m/s to the left.

The velocity of the 10.0 g marble will be:

0.2 kg · m/s - 3 kg · vB´ = vA´

0.2 kg · m/s - 3 kg · (-0.100 m/s) = vA´

vA´ = 0.500 m/s

The final velocity of the 10.0 g marble is 0.500 m/s to the right.

The change in momentum of the 30.0 g marble is calculated as follows:

Δp = final momentum - initial momentum

Δp = 0.030 kg · (-0.100 m/s) -(0.030 kg · 0.200 m/s) = -9.00 × 10⁻³ kg · m/s

The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

The change in momentum of the 10.0 g marble is calculated in the same way:

Δp = final momentum - initial momentum

Δp = 0.010 kg · 0.500 m/s -(-0.010 kg · 0.400 m/s) = 9.00 × 10⁻³ kg · m/s

The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

The change in kinetic energy for the 30.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.030 kg · (-0.100 m/s)² - 1/2 · 0.030 kg · (0.200 m/s)²

ΔKE = -4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J

The change in kinetic energy for the 10.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.010 kg · (0.500 m/s)² - 1/2 · 0.010 kg · (-0.400 m/s)²

ΔKE = 4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is 4.5 × 10⁻⁴ J

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