/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 In a science fiction novel two e... [FREE SOLUTION] | 91Ó°ÊÓ

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In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary positions they push against each other. Bonzo flies off with a velocity of \(+1.5 \mathrm{m} / \mathrm{s},\) while Ender recoils with a velocity of \(-2.5 \mathrm{m} / \mathrm{s} .\) (a) Without doing any calculations, decide which person has the greater mass. Give your reasoning. (b) Determine the ratio \(m_{\text {Bonno }}\) \(m_{\text {Ender }}\) of the masses of these two enemies.

Short Answer

Expert verified
Bonzo has the greater mass. The mass ratio \(m_{\text{Bonzo}}/m_{\text{Ender}}\) is 1.67.

Step by step solution

01

Assessing Mass Based on Velocity

In this problem, we can use the concept of conservation of momentum. Since both Bonzo and Ender exert equal and opposite forces on each other, their momenta must be equal in magnitude and opposite in direction. From the velocities given, it is clear that Ender, who recoils with a higher speed of \(-2.5 \ \mathrm{m/s}\), must have a lesser mass compared to Bonzo, who travels at \(+1.5 \ \mathrm{m/s}\). Since they faced the same force, a higher speed implies a lower mass and vice versa.
02

Setting Up the Momentum Equation

According to the law of conservation of momentum, the total momentum before and after the push should be the same. Therefore, \(m_{\text{Bonzo}} \times 1.5 \ = \ m_{\text{Ender}} \times 2.5\).Thus, we have:\[ m_{\text{Bonzo}} \cdot 1.5 = m_{\text{Ender}} \cdot -2.5 \] Sign is not relevant for magnitude, so:\[ m_{\text{Bonzo}} \cdot 1.5 = m_{\text{Ender}} \cdot 2.5 \]
03

Solving for the Mass Ratio

To find the ratio of their masses, rearrange the momentum equation:\[\frac{m_{\text{Bonzo}}}{m_{\text{Ender}}} = \frac{2.5}{1.5} \]Now, calculate this fraction:\(\frac{2.5}{1.5} = \frac{5}{3} = 1.67\). So, Bonzo's mass is 1.67 times Ender's mass.

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Key Concepts

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

Mass Ratio
In this scenario, understanding mass ratio is crucial to find out who is more massive between Bonzo and Ender. According to the conservation of momentum, which states that the total momentum of a closed system remains constant when no external forces are applied. This implies that the momentum exchanged between Bonzo and Ender remains constant.
The momentum can be expressed as the product of mass and velocity and can be denoted as \(p = m \times v\). When Bonzo and Ender push against each other, their momenta must cancel each other out to conserve the total momentum.
  • Bonzo has a velocity of \(+1.5 \ \mathrm{m/s}\) and Ender has a velocity of \(-2.5 \ \mathrm{m/s}\).
  • The change in velocity is indicative of the difference in their masses.
By setting up the momentum equation based on their velocities, we can determine their mass ratio. The equation was established as \(m_{\text{Bonzo}} \times 1.5 = m_{\text{Ender}} \times 2.5\).
So, the mass ratio \(\frac{m_{\text{Bonzo}}}{m_{\text{Ender}}}\) resolves to \(\frac{2.5}{1.5} = 1.67\), highlighting that Bonzo's mass is 1.67 times Ender's mass.
Velocity
Velocity plays a pivotal role in understanding the exchange of momentum between Bonzo and Ender. It offers a direct insight into the dynamics of the movement post-interaction. Unlike speed, which is only a scalar, velocity is a vector quantity because it has both magnitude and direction.
In this particular case:
  • Bonzo moves with a velocity of \(+1.5\ \mathrm{m/s}\).
  • Ender moves with a velocity of \(-2.5\ \mathrm{m/s}\).
The velocities demonstrated here are crucial for calculating the momentum and, consequently, the mass. Higher speed at the same level of force indicates a smaller mass because it reflects how much motion is produced given the same amount of applied force. Hence, Ender's higher recoil velocity is a clear indicator of his lighter mass. Understanding velocity, therefore, allows us to infer mass differences when combined with the law of conservation of momentum.
Outer Space
Outer space provides an interesting setting for this scenario due to its lack of air resistance and gravitational forces that could affect momentum calculations. When two objects interact in space, the law of conservation of momentum is observed more accurately because the absence of friction or air resistance means no energy is lost from the system.
  • Momentum is transferred cleanly from one body to the other with minimal external influence.
  • As a result, the calculations made using velocity and mass are highly accurate, focusing solely on the physics of movement themselves.
For Bonzo and Ender, being in outer space means that their push against one another results in straightforward, uninterrupted motion. This highlights one of the interesting attributes of space: interactions are governed purely by the forces the objects exert on each other. Thus, understanding interactions in space can be a great way to illustrate fundamental physical laws and concepts like momentum conservation without additional interfering variables.

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Most popular questions from this chapter

A student \((m=63 \mathrm{kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.040 \mathrm{s} .\) The average force exerted on him by the ground is \(+18000 \mathrm{N},\) where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

A baseball \((m=149 \mathrm{g})\) approaches a bat horizontally at a speed of \(40.2 \mathrm{m} / \mathrm{s}(90 \mathrm{mi} / \mathrm{h})\) and is hit straight back at a speed of \(45.6 \mathrm{m} / \mathrm{s}\) \((102 \mathrm{mi} / \mathrm{h}) .\) If the ball is in contact with the bat for a time of \(1.10 \mathrm{ms},\) what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

John's mass is \(86 \mathrm{kg}\), and Barbara's is \(55 \mathrm{kg}\). He is standing on the \(x\) axis at \(x_{1}=+9.0 \mathrm{m},\) while she is standing on the \(x\) axis at \(x_{\mathrm{B}}=12.0 \mathrm{m}\) They switch positions. How far and in which direction does their center of mass move as a result of the switch?

A model rocket is constructed with a motor that can provide a total impulse of \(29.0 \mathrm{N} \cdot\) s. The mass of the rocket is \(0.175 \mathrm{kg} .\) What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

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