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Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) 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?

Short Answer

Expert verified
The answers can be found by doing the arithmetic calculations.

Step by step solution

01

Convert the given units

First, convert the given units into compatible ones. Convert kilometers/hours into meters/seconds for velocity (1km = 1000m, 1 hour = 3600 seconds), and centimeters into meters for depth (1cm = 0.01m). \[311 \ km/h = 311000\ m/3600\ s = 86.39 \ m/s\] \[81\ cm = 81*0.01\ m = 0.81\ m \]
02

Find acceleration

We will use the kinematic equation (assuming acceleration (a) is constant)\[v_f^2 = v_i^2 + 2as\] Here, the final velocity (\(v_f\)) after the spacecraft has crashed is 0 (it's at rest), the initial velocity (\(v_i\)) is 86.39 m/s (which we found in Step 1), the distance (s) is 0.81 m (depth of penetration which we also found in Step 1). By substituting these values, we can find acceleration.\[0 = (86.39)^2 + 2*a*0.81\] Now solve for 'a' - this will give the magnitude of deceleration (it's deceleration as the spacecraft is slowing down).
03

Find force

To find the force that the ground exerted on the spacecraft during the crash, we shall use Newton's second law of motion: 'force = mass * acceleration'. The mass of the capsule is given to be 210 kg and we've found acceleration in the previous step.
04

How long did the force last?

To find out how long the decelerating force acted on the spacecraft, we use one of the kinematic equations: \(t = (v_f - v_i) / a\). We already know the values of \(v_f\), \(v_i\) and \(a\) from previous steps, so by substiting these values, we can find 't' (time).
05

Calculation

Now, substitute the values into the equations derived in the earlier steps to get numerical answers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
In physics, constant acceleration refers to a situation where an object's velocity changes at a steady rate over time. This is a key concept when analyzing motion because it greatly simplifies calculations and predictions. For example, in the case of the Genesis spacecraft, the problem assumes that the craft experienced constant acceleration, or in other terms, unchanging deceleration, during its unfortunate descent.

This is essential for utilizing the kinematic equations which are designed to work under the assumption of constant acceleration. Under these conditions, acceleration can be calculated using the known values of initial velocity, final velocity (which is zero if the spacecraft has come to a stop), and the distance over which the change in velocity occurred.

Constant acceleration is often denoted by the symbol 'a' in equations and can be positive when the object is speeding up or negative in cases of deceleration when a force is applied in the opposite direction of motion.
Newton's Second Law of Motion
Newton's second law of motion is a fundamental principle in physics that states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law explains the relationship between an object's motion and the forces acting upon it.

When applied to the Genesis spacecraft problem, this law allows us to understand the force the ground exerted on the capsule during impact. Since we know the capsule's mass and can compute its constant acceleration during the crash, we can calculate the force. It's vital to remember that this force is a vector quantity, which means it has both magnitude and direction, providing deeper insight into how the spacecraft came to a halt.
Kinematic Equations
Kinematic equations are a set of four equations that describe the motion of objects under the influence of constant acceleration. These equations relate the displacement, initial and final velocities, acceleration, and time of an object's motion.

For the Genesis spacecraft, the kinematic equation \(v_f^2 = v_i^2 + 2as\) was used to compute its acceleration during the crash. This particular equation is useful for solving problems where time isn't directly involved.

However, to determine the crash's duration, another kinematic equation \(t = (v_f - v_i) / a\) helps us calculate the time over which the deceleration took place. These equations are powerful tools for problem-solving in kinematics, helping to break down complex motion into understandable parts.
Deceleration
Deceleration is the term used to describe negative acceleration, meaning it is the reduction in the velocity of an object over time. It is acceleration in the direction opposite to the direction of the velocity. Deceleration occurs when forces act contrary to the direction of motion, causing an object to slow down.

In the context of the Genesis spacecraft, deceleration is what occurred as the spacecraft hit the ground. It went from a certain speed to a complete stop over a particular distance. This deceleration can be calculated using kinematic equations as long as it remains constant. Not only is this concept crucial for understanding crashes like that of the spacecraft, but it is also applicable in everyday situations, such as a car coming to a stop when the brakes are applied.

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Most popular questions from this chapter

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.

A racetrack curve has radius \(120.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.300 .\) A race car with mass \(900 \mathrm{~kg}\) rounds the curve with the minimum speed needed to not slide down the banking. (a) As the car rounds the curve, what is the normal force exerted on it by the road? (b) What is the car's speed?

Traffic Court. You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled \(192 \mathrm{ft}\) before stopping, and the coefficient of kinetic friction between the road and his tires was \(0.750 .\) He was charged with speeding in a \(45 \mathrm{mi} / \mathrm{h}\) zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

A large crate with mass \(m\) rests on a horizontal floor. The coefficients of friction between the crate and the floor are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}} .\) A woman pushes downward with a force \(\overrightarrow{\boldsymbol{F}}\) on the crate at an angle \(\theta\) below the horizontal. (a) What magnitude of force \(\vec{F}\) is required to keep the crate moving at constant velocity? (b) If \(\mu_{\mathrm{s}}\) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \(\mu_{\mathrm{s}}\)

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

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