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Chapter 2: Problem 26

A muon (an elementary particle) enters a region with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) and then is slowed at the rate of \(1.25 \times\) \(10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) (a) How far does the muon take to stop? (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the muon.

Short Answer

Expert verified
The muon stops after traveling 0.10 m over 4.00 脳 10^{-8} s.

Step by step solution

01

Identify the Known Values

We know that the initial speed \( v_i \) of the muon is \( 5.00 \times 10^{6} \, \mathrm{m/s} \). The acceleration \( a \) is \(-1.25 \times 10^{14} \, \mathrm{m/s^2} \) (negative because it's slowing down). The final speed \( v_f \) will be \( 0 \, \mathrm{m/s} \) when the muon stops.
02

Use the Kinematic Equation to Find Distance

We can use the kinematic equation: \( v_f^2 = v_i^2 + 2a d \). Substituting the known values, we have:\( 0 = (5.00 \times 10^{6})^2 + 2(-1.25 \times 10^{14})d \). Now, solve for \( d \):\[ 0 = 25 \times 10^{12} - 2.5 \times 10^{14}d \]\[ 2.5 \times 10^{14}d = 25 \times 10^{12} \]\[ d = \frac{25 \times 10^{12}}{2.5 \times 10^{14}} = 0.10 \, \mathrm{m} \].
03

Calculate the Time to Stop

To find the time \( t \) it takes for the muon to stop, we use the equation:\( v_f = v_i + at \).Substitute the known values:\( 0 = 5.00 \times 10^{6} + (-1.25 \times 10^{14})t \).Solving for \( t \):\[ t = \frac{5.00 \times 10^{6}}{1.25 \times 10^{14}} = 4.00 \times 10^{-8} \, \mathrm{s} \].
04

Plot the Graph for x versus t and v versus t

For the graph of \( x \) versus \( t \), use the equation \( x = v_i t + \frac{1}{2} a t^2 \) and plot the displacement over time from \( t = 0 \) to \( t = 4.00 \times 10^{-8} \, \mathrm{s} \).For the graph of \( v \) versus \( t \), use the equation \( v = v_i + a t \), starting from \( v = 5.00 \times 10^6 \, \mathrm{m/s} \) to \( v = 0 \, \mathrm{m/s} \), over the same time interval. Both should show a steady decrease to zero.

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Key Concepts

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

Muon
A muon is an elementary particle similar to an electron, but it is much heavier. It is part of a family of particles called leptons. Muons are unstable and usually decay into electrons and neutrinos. They are produced when cosmic rays collide with atoms in the Earth's atmosphere.
This makes them particularly interesting in physics experiments. In this exercise, understanding how muons move and slow down helps us apply kinematic equations effectively. Key aspects of a muon's motion:
  • High initial speeds, often close to the speed of light.
  • Short lifetimes, as they decay into other particles.
When analyzing such particles, it is crucial to apply correct physics principles, including dynamics and kinematics, to predict their behavior accurately.
Kinematic Equations
Kinematic equations are mathematical formulas used to describe motion. They relate the five key variables of motion: initial velocity (\( v_i \)), final velocity (\( v_f \)), acceleration (\( a \)), time (\( t \)), and displacement (\( d \)). These equations assume constant acceleration. In this exercise, the following kinematic equation was used:
  • \[ v_f^2 = v_i^2 + 2ad \]
This equation was used to find how far the muon travels before coming to a stop.
It is important to recognize that the acceleration is negative, indicating deceleration. With the right values substituting into the equation, it helps to determine the distance and time the muon takes to halt, making it a vital aspect of solving the problem.
Acceleration
Acceleration is a critical component in analyzing motion. It is defined as the rate of change of velocity with respect to time. The key aspect here is that it is a vector quantity, meaning it has both magnitude and direction. In this exercise, the muon's acceleration was given as \(-1.25 \times 10^{14}\, \mathrm{m/s^2}\). The negative sign indicates that the muon is slowing down or decelerating.
This is crucial for solving such problems because:
  • Negative acceleration means slowing down.
  • We must correctly account for the direction of movement.
Understanding acceleration aids in predicting motion and solving related physics problems, providing valuable insights into how the muon behaves as it travels.
Graphing Motion
Graphing motion gives a visual representation of how objects move over time. In kinematics, two common graphs are displacement versus time (\( x \) vs. \( t \)) and velocity versus time (\( v \) vs. \( t \)). For the muon:
  • Displacement vs. Time: This graph shows how far the muon travels as it comes to a stop. With the equation \( x = v_i t + \frac{1}{2} a t^2 \), we can calculate the displacement at any given time.
  • Velocity vs. Time: This graph illustrates how the muon's velocity decreases to zero. Starting from its initial velocity, the equation \( v = v_i + a t \) shows the velocity at each moment until it stops.
In aiding learning, these graphs concretely visualize the motion,
making it easier to grasp changes in velocity and displacement over time. This understanding is essential for mastering kinematic problems.

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Most popular questions from this chapter

The position of a particle moving along the \(x\) axis is given in centimeters by \(x=9.75+1.50 t^{3}\), where \(t\) is in seconds. Calculate (a) the average velocity during the time interval \(t=2.00 \mathrm{~s}\) to \(t=\) \(3.00 \mathrm{~s} ;(\mathrm{b})\) the instantaneous velocity at \(t=2.00 \mathrm{~s} ;\) (c) the instantaneous velocity at \(t=3.00 \mathrm{~s} ;(\mathrm{d})\) the instantaneous velocity at \(t=\) \(2.50 \mathrm{~s}\); and (e) the instantaneous velocity when the particle is midway between its positions at \(t=2.00 \mathrm{~s}\) and \(t=3.00 \mathrm{~s}\). (f) Graph \(x\) versus \(t\) and indicate your answers graphically.

At a construction site a pipe wrench struck the ground with a speed of \(24 \mathrm{~m} / \mathrm{s}\). (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of \(y, v\), and \(a\) versus \(t\) for the wrench.

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest (at \(x=0)\) and ending at rest (at \(x=900 \mathrm{~m})\). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x\), velocity \(v\), and acceleration \(a\) versus time \(t\) for the trip.

An object falls a distance \(h\) from rest. If it travels \(0.50 h\) in the last \(1.00 \mathrm{~s}\), find (a) the time and (b) the height of its fall. (c) Explain the physically unacceptable solution of the quadratic equation in \(t\) that you obtain.

A pilot flies horizontally at \(1300 \mathrm{~km} / \mathrm{h}\), at height \(h=35 \mathrm{~m}\) above initially level ground. However, at time \(t=0\), the pilot begins to fly over ground sloping upward at angle \(\theta=4.3^{\circ}\) (Fig. 2-38). If the pilot does not change the airplane's heading, at what time \(t\) does the plane strike the ground?
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