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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 \(\mathrm{c}\) (c) \(0.990 c .(\mathrm{d})\) Repeat parts \((\mathrm{a}),\) and \((\mathrm{c})\) if the force and acceleration are perpendicular to the velocity.

Short Answer

Expert verified
For (a) 0.145 N, (b) 0.332 N, (c) 1.028 N; (a) perpendicular: 0.145 N, (c) perpendicular: 1.028 N.

Step by step solution

01

Identify the formula for force

To calculate the force, use Newton's second law, which states that the force exerted on an object is equal to the object's mass times its acceleration: \[ F = m \cdot a \]Where- \( F \) is the force,- \( m \) is the mass of the object (0.145 kg for the baseball),- \( a \) is the acceleration (1.00 m/s²).
02

Calculate force magnitude for part (a)

For part (a), since the velocity is non-relativistic (much less than the speed of light \( c \)), we can directly apply Newton's second law to compute the force:\[ F = 0.145 \times 1.00 = 0.145\, \text{N} \]
03

Apply relativistic dynamics for parts (b) and (c)

For parts (b) and (c), when the baseball's speed approaches a significant fraction of the speed of light \( c \), relativistic effects must be considered. The relativistic force is defined by:\[ F = \frac{dp}{dt} = \frac{d}{dt} \left( \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \]To find the force, we must consider the relativistic mass increase.
04

Compute force magnitude for part (b)

For part (b), the velocity is 0.900c. Compute the relativistic factor, \( \gamma \), and then use the relativistic formula:\[ \gamma = \frac{1}{\sqrt{1 - (0.900)^2}} \approx 2.294 \F = \frac{0.145 \times 1.00}{\sqrt{1 - (0.900)^2}} \approx 0.332 \text{ N}\]
05

Compute force magnitude for part (c)

For part (c), with velocity 0.990c, find \( \gamma \):\[ \gamma = \frac{1}{\sqrt{1 - (0.990)^2}} \approx 7.089 \F = \frac{0.145 \times 1.00}{\gamma} \approx 1.028 \text{ N}\]
06

Calculate force for perpendicular acceleration in part (a)

For perpendicular force, the situation doesn't call for a different calculation in non-relativistic conditions because the direction doesn't change the force needed for acceleration:\[ F = 0.145 \text{ N}\]
07

Relativistic perpendicular acceleration for part (c)

When calculating force perpendicular to the velocity in relativistic conditions, we use the perpendicular relativistic law:\[ F = m \times a \times \gamma \] For velocity 0.990c:\[ F = 0.145 \times 1.00 \times \gamma \approx 1.028 \text{ N}\]

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Key Concepts

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

Relativistic Dynamics
When velocities are a significant fraction of the speed of light, classical physics cannot accurately describe the motion of objects. This is where relativistic dynamics comes in. Unlike classical dynamics, relativistic dynamics takes into account the relativistic mass increase. This means that as an object moves faster, its mass appears to increase from the perspective of an outside observer. In this context, Newton's second law of motion requires modification. The relativistic force is given by the formula: \[ F = \frac{dp}{dt} = \frac{d}{dt} \left( \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \] Here, \( p \) is the momentum, \( m \) is the rest mass, \( v \) is velocity, and \( c \) is the speed of light.
  • As velocity \( v \) approaches \( c \), the factor \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) increases, making the calculations more complex.
  • This factor compensates for the relativistic effects at high speeds.
  • The force is not just mass times acceleration but includes this relativistic correction to consider the increase in momentum.
Understanding how relativistic dynamics modifies our classical perspectives is crucial for grasping how forces behave at high speeds.
Non-Relativistic Velocity
Non-relativistic velocities are those which are much smaller compared to the speed of light. In these cases, we can safely use classical mechanics without any additional factors or corrections. For a baseball moving at 10.0 m/s, the speed is significantly under relativistic conditions.
  • The formula \( F = m \cdot a \) is straightforward and applicable. Here, the acceleration \( a \) directly multiplies with the mass \( m \) of the object to give us the force.
  • Because speeds are relatively low, the effects of relativity can be ignored. This simplification assumes that the object's mass is constant regardless of its velocity.
The calculations are simpler and focus only on factors observable in day-to-day life without worrying about relativistic mass increase or time dilation effects.
Perpendicular Acceleration
When force and acceleration are perpendicular to the direction of velocity, the dynamics change slightly depending on whether the velocity is relativistic or non-relativistic. In non-relativistic contexts, the component of force affects only the acceleration perpendicular to the velocity without any additional effects. Therefore, using Newton's second law without alteration is often sufficient. In relativistic scenarios, however, the situation is more complex. When dealing with perpendicular forces or accelerations:
  • The component of acceleration caused by the force is adjusted by the Lorentz factor, \( \gamma \).
  • Forces acting perpendicular can appear stronger due to the effects of relativistic physics. The formula becomes \( F = m \times a \times \gamma \), where \( \gamma \) accounts for the increased 'resistance' to the change in velocities.
Understanding how forces at angles interact with velocity under these different scenarios can reveal much about motion and energy at both ordinary and extraordinary speeds.

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Most popular questions from this chapter

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.190 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 \(\mathrm{ms}\) . (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth expressed as a fraction of the speed of light \(c\) ?

(a) By what percentage does your rest mass increase when you climb 30 \(\mathrm{m}\) to the top of a ten-story building? Are you aware of this increase? Explain. (b) By how many grams does the mass of a \(120-\mathrm{g}\) spring with force constant 200 \(\mathrm{N} / \mathrm{cm}\) change when you compress it by 6.0 \(\mathrm{cm} \%\) Does the mass increase or decrease? Would you notice the change in mass if you were holding the spring? Explain.

(a) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(m v ?\) Express your answer in terms of the speed of light. (b) A force is apphed to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600 \(\mathrm{c}\) . The pursuit ship is traveling at a speed of 0.800 \(\mathrm{c}\) relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the speed of the cruiser relative to the pursuit ship be positive or negative? (b) What is the speed of the cruiser relative to the pursuit ship?

Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (Hint: Since \(u \ll c,\) you can simplify \(\sqrt{1-u^{2} / c^{2} \text { by a binomial expansion. }}\))

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