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Calculate the magnitude of the force required to give a \(0.145 \mathrm{~kg}\) baseball an acceleration \(a=1.00 \mathrm{~m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{~m} / \mathrm{s} ;\) (b) \(0.900 c ;\) (c) \(0.990 c\) (d) Repeat parts (a), (b), and (c) if the force and acceleration are perpendicular to the velocity.

Short Answer

Expert verified
For parts (a), (b), and (c), the force required, when the force is in the direction of the baseball's initial velocity, is \(0.145 \ N\). For part (d), when the force and acceleration are perpendicular to the velocity, the force as a function of radius is given by \( F = \frac {(0.145 \ kg)(v)^{2}}{r} \), where \(v\) is either \(10 \ m/s, \ 0.900c, \ or \ 0.990c\)

Step by step solution

01

Case 1 - Calculate Force in the Direction of Motion

For each part (a), (b) and (c), apply the formula \(F = ma\). The mass (m) is \(0.145 \ kg\) and the acceleration (a) is \(1.00 \ m/s^{2}\). The magnitude of velocity (v) varies in each part:- For part (a): Use \(v = 10 \ m/s\) to get \(F = ma = (0.145 \ kg)(1 \ m/s^{2}) = 0.145 \ N\)- For part (b): Use \(v = 0.900c\) where \(c\) is the speed of light (\(3 \times 10^{8} \ m/s\)) to get \(F = ma = (0.145 \ kg)(1 \ m/s^{2}) = 0.145 \ N\)- For part (c): Use \(v = 0.990c\) and apply the same approach to get \(F = 0.145 \ N\). The force is the same because it still acts in the same direction of the motion.
02

Case 2 - Calculate Force Perpendicular to the Motion

For part (d), apply the formula for centripetal force, \(F = m v^{2} / r\). Here we don't know the radius (r) but we can describe the force as a function of radius (r). The expressions for force (F) when velocity \( v = 10 \ m/s, \ 0.900c, \ and \ 0.990c \) are \( F = \frac {(0.145 \ kg)(10 \ m/s)^{2}}{r}, \ F = \frac {(0.145 \ kg)(0.900c)^{2}}{r}, \ and \ F = \frac {(0.145 \ kg)(0.990c)^{2}}{r} \) respectively.

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Key Concepts

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

Force Calculation
In physics, calculating force is a fundamental concept that involves understanding the relationship between mass and acceleration. The force required to accelerate an object is determined by Newton’s second law of motion, expressed as \( F = ma \). This equation implies that the force needed to move an object is the product of its mass (\(m\)) and the acceleration (\(a\)) it experiences.

Here, we look at example calculations involving a baseball, where the mass is given as \(0.145 \ kg\). The force can be different based on how this force is applied:
  • In the same direction of motion, where \(a = 1.00 \ m/s^2\), the force remains constant as \(0.145 \ N\) regardless of the speed because the formula \( F = ma \) does not depend on the velocity.
  • If the force acts perpendicularly, as observed in some cases where centripetal forces come into play, additional factors such as radius and velocity become relevant in calculating force.
Understanding these basics helps in solving many real-world problems involving motion.
Acceleration
Acceleration is the rate at which an object’s velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. When we say that an object is accelerating, we imply that it's speeding up, slowing down, or changing direction.

For the baseball problem, the constant acceleration given is \(1.00 \ m/s^2\). This tells us that every second, the velocity of the baseball increases by \(1 \ m/s\), assuming the acceleration is in the same direction as its present movement.
  • Acceleration requires an external force, which is calculated using \( F = ma \).
  • The direction of action of this acceleration is significant, impacting whether force calculations involve linear motion or centripetal force, if perpendicular.
Knowing these concepts allows us to solve various scenarios where changes in motion need to be evaluated.
Velocity
Velocity is another core concept fundamental to understanding motion. It describes the speed of an object in a specific direction. Unlike speed, which only considers magnitude, velocity incorporates direction, making it a vector quantity.

In this baseball problem, velocity plays a crucial role:
  • Case (a) uses a velocity of \(10 \ m/s\), which is straightforward since it's a typical speed for calculations in our context.
  • In cases (b) and (c), where velocities approach 90% and 99% of the speed of light, special relativity might influence outcomes significantly, though classical mechanics simplifies the force calculation here to \(0.145 \ N\).
Remember, when velocity is altered in direction or magnitude, it affects how forces engage with the object’s movement.
Centripetal Force
Centripetal force comes into play whenever an object moves in a circular path. Unlike linear force, centripetal force is always directed towards the center of the circle along which the object is moving. This is key for understanding motions such as rotation or orbiting.

When the force and acceleration are perpendicular to velocity, as in the case of the baseball, you encounter centripetal force conditions. The force is calculated not simply with \( F = ma \) but rather using: \\[ F = \frac {mv^2}{r} \]
  • Here, \(v\) stands for velocity.
  • \(r\) refers to the radius of the circular path.
This equation signifies how the velocity and the radius govern the magnitude of this force. Without the radius \(r\), the force remains as a function of \(r\), indicating dependency. Understanding these dynamics is crucial for analyzing circular motion scenarios correctly.

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Most popular questions from this chapter

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{~s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c\), what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

\(\mathrm{As}\) you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of \(0.800 c\) relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{~m}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

A spaceship moving at constant speed \(u\) relative to us broad. casts a radio signal at constant frequency \(f_{0}\). As the spaceship approaches us, we receive a higher frequency \(f\); after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0},\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed \(0.758 c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?\) (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Tell It to the Judge. (a) How fast must you be approaching a red traffic light \((\lambda=675 \mathrm{nm})\) for it to appear yellow \((\lambda=575 \mathrm{nm}) ?\) Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is \(\$ 1.00\) for each kilometer per hour that your speed exceeds the posted limit of \(90 \mathrm{~km} / \mathrm{h}\).

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