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Chapter 7: Problem 65

A \(6090 \mathrm{~kg}\) space probe, moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun, fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

Short Answer

Expert verified
The final velocity of the probe is 109.8 m/s

Step by step solution

01

- Define Initial Momentum

Calculate the initial momentum of the space probe before firing the rocket. Use the formula: \[ p_{\text{initial}} = m_{\text{probe}} \times v_{\text{probe}} \] Given: \( m_{\text{probe}} = 6090 \, \text{kg} \), \( v_{\text{probe}} = 105 \, \text{m/s} \) \[ p_{\text{initial}} = 6090 \, \text{kg} \times 105 \, \text{m/s} = 639450 \, \text{kg m/s} \]
02

- Define Final Momentum

After the rocket engine fires, we have two parts: the remaining space probe and the ejected exhaust. Use the conservation of momentum principle: \[ p_{\text{initial}} = p_{\text{final (probe)}} + p_{\text{exhaust}} \]
03

- Calculate Exhaust Momentum

Find the momentum of the exhaust using its mass and velocity relative to the space probe: \[ p_{\text{exhaust}} = m_{\text{exhaust}} \times v_{\text{exhaust}} \] Given: \( m_{\text{exhaust}} = 80 \, \text{kg} \), \( v_{\text{exhaust}} = 253 \, \text{m/s} \) \[ p_{\text{exhaust}} = 80 \, \text{kg} \times 253 \, \text{m/s} = 20240 \, \text{kg m/s} \]
04

- Adjusted Exhaust Momentum

Since the exhaust velocity is relative to the space probe, adjust its momentum relative to the Sun. Because the exhaust is ejected in the opposite direction to the probe's motion, consider the negative sign: \[ p_{\text{exhaust}} = -20240 \, \text{kg m/s} \]
05

- Set Up Equation for Final Momentum

Using conservation of momentum: \[ p_{\text{initial}} = p_{\text{final (probe)}} + p_{\text{exhaust}} \] Substitute the known values: \[ 639450 \, \text{kg m/s} = (6090 \, \text{kg} - 80 \, \text{kg}) \times v_{\text{final (probe)}} + (-20240 \, \text{kg m/s}) \]
06

- Solve for Final Probe Velocity

Isolate \( v_{\text{final (probe)}} \) and solve: \[ 639450 \, \text{kg m/s} = 6010 \, \text{kg} \times v_{\text{final (probe)}} - 20240 \, \text{kg m/s} \] \[ 639450 \, \text{kg m/s} + 20240 \, \text{kg m/s} = 6010 \, \text{kg} \times v_{\text{final (probe)}} \] \[ 659690 \, \text{kg m/s} = 6010 \, \text{kg} \times v_{\text{final (probe)}} \] \[ v_{\text{final (probe)}} = \frac{659690 \, \text{kg m/s}}{6010 \, \text{kg}} \] \[ v_{\text{final (probe)}} = 109.8 \, \text{m/s} \]

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Key Concepts

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

initial momentum
In physics, momentum measures how difficult it is to stop a moving object. It depends on both mass and velocity, using the formula:

\[ p_{initial} = m_{probe} \times v_{probe} \]

This equation tells us the product of the mass and velocity gives the initial momentum.

The probe here has a mass of 6090 kg and travels at 105 m/s toward Jupiter. So, its initial momentum is:

\[ 6090 \times 105 = 639450 \, kg \cdot m/s \]

This large value shows the probe's significant momentum due to its high mass and velocity.
final momentum
The final momentum combines the momentum of the remaining part of the probe and the ejected exhaust.
According to the conservation of momentum, the total momentum before firing the rocket (initial momentum) equals the total momentum after firing (final momentum).

Let's break it down:
The formula is

\[ p_{initial} = p_{final \, (probe)} + p_{exhaust} \]

Here, the initial momentum is equal to the sum of the probe's final momentum and the exhaust's momentum.
This ensures no momentum is lost or created, only transferred.
velocity calculation
To find the probe's final velocity, isolate the unknown in the momentum equation.
Start by calculating the exhaust's momentum. Use the mass (80 kg) and velocity (253 m/s) given.

\[ p_{exhaust} = m_{exhaust} \times v_{exhaust} \]

This results in:

\[ 80 \times 253 = 20240 \, kg \cdot m/s \]

Since the exhaust is ejected in the opposite direction, we adjust this value:

\[ p_{exhaust} = -20240 \, kg \cdot m/s \]

Then, apply it to the conservation formula:

\[ 639450 = (6090 - 80) \times v_{final} - 20240 \]

Finally, solve for the final velocity:

\[ v_{final} = \frac{659690}{6010} \approx 109.8 \, m/s \]

rocket propulsion
Rocket propulsion relies on the principle of action and reaction (Newton's third law).
When a rocket fires its engine, it expels exhaust gases at high speed in one direction. This ejection provides thrust that propels the rocket in the opposite direction.

The process involves converting chemical energy from the rocket fuel into kinetic energy of the exhaust gases, generating momentum.
For our example, energy released by burning fuel ejects the exhaust at 253 m/s, creating momentum and thus moving the space probe.
The change in the probe's velocity depends on the mass expelled and the speed of the ejection, helping it to maneuver in space by conserving momentum.

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Most popular questions from this chapter

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2}\) ?

In Edmund Rostand's famous play, Cyrano de Bergerac, Cyrano, in an attempt to distract a suitor from visiting Roxanne, claims to have descended to Earth from the Moon and proclaims to have invented six novel and fantastical methods for traveling to the Moon. One is as follows. Sitting on an iron platform- thence To throw a magnet in the air. This is A method well conceived - the magnet flown, Infallibly the iron will pursue: Then quick! relaunch your magnet, and you thus Can mount and mount unmeasured distances! \(^{*}\) In an old cartoon, there is another version of this method. A character in the old West is on a hand-pumped, two-person rail car. After getting tired of pumping the handle up and down to make the car move along the rails, he takes out a magnet, hangs it from a fishing pole, and holds it in front of the cart. The magnet pulls the cart toward it, which pushes the magnet forward, and so on, so the cart moves forward continually. What do you think of these methods? Can some version of them work? Discuss in terms of the physics you have learned.

It is well known that bullets and other missiles fired at Superman simply bounce off his chest (Fig.7-22). Suppose that a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets/min, and the speed of each bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?

Two balls \(A\) and \(B\), having different but unknown masses, collide. Initially, \(A\) is at rest and \(B\) has speed \(v_{B}\). After the collision, \(B\) has speed \(v_{B} / 2\) and moves perpendicularly to its original motion. (a) Find the direction in which ball \(A\) moves after the collision. (b) Show that you cannot determine the speed of \(A\) from the information given.

A space vehicle is traveling at \(4300 \mathrm{~km} / \mathrm{h}\) relative to Earth when the exhausted rocket motor is disengaged and sent backward with a speed of \(82 \mathrm{~km} / \mathrm{h}\) relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?
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