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Chapter 7: Problem 1

A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm}\), what then is (a) its speed and (b) the increase in its kinetic energy?

Short Answer

Expert verified
(a) The final speed is \(2.88 \times 10^7 \, \mathrm{m/s}\). (b) The increase in kinetic energy is \(2.15 \times 10^{-13} \, \mathrm{J}\).

Step by step solution

01

Identify Given Information

We are given the initial speed of the proton \(v_i = 2.4 \times 10^7 \, \mathrm{m/s}\), its acceleration \(a = 3.6 \times 10^{15} \, \mathrm{m/s^2}\), and the distance traveled \(s = 3.5 \, \mathrm{cm} = 0.035 \, \mathrm{m}\). We need to find the final speed \(v_f\) and the increase in kinetic energy.
02

Use Kinematic Equation for Final Speed

Use the equation \(v_f^2 = v_i^2 + 2as\) to find the final speed. \[ v_f^2 = (2.4 \times 10^7)^2 + 2 \times 3.6 \times 10^{15} \times 0.035 \]Calculating gives:\[ v_f^2 = 5.76 \times 10^{14} + 2.52 \times 10^{14} = 8.28 \times 10^{14} \]Therefore, the final speed is:\[ v_f = \sqrt{8.28 \times 10^{14}} \approx 2.88 \times 10^7 \, \mathrm{m/s} \]
03

Calculate Initial and Final Kinetic Energy

The initial kinetic energy \(KE_i\) is given by:\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2.4 \times 10^7)^2 \]Calculating gives:\[ KE_i = 4.8 \times 10^{-13} \, \mathrm{J} \]The final kinetic energy \(KE_f\) is given by:\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2.88 \times 10^7)^2 \]Calculating gives:\[ KE_f = 6.95 \times 10^{-13} \, \mathrm{J} \]
04

Calculate Increase in Kinetic Energy

The increase in kinetic energy \(\Delta KE\) is:\[ \Delta KE = KE_f - KE_i = 6.95 \times 10^{-13} - 4.8 \times 10^{-13} = 2.15 \times 10^{-13} \, \mathrm{J} \]

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Key Concepts

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

Proton acceleration
When a proton is accelerated, it undergoes a change in velocity due to the application of a force. In our example, the force causes the proton to accelerate at a significant rate of \(3.6 \times 10^{15} \, \mathrm{m/s^2}\).
Acceleration can be thought of as how quickly an object's speed changes. In straight-line motion, when acceleration is constant, we can predict how a particle's velocity changes with time.
Protons, being positively charged particles found in the nucleus of an atom, have a small mass of \(1.67 \times 10^{-27} \mathrm{~kg}\). Despite this, a constant acceleration over a short distance can lead to a substantial increase in velocity due to their light mass.
  • Acceleration is caused by force and determines how fast the velocity increases.
  • A small mass will allow for greater acceleration when the same force is applied compared to a larger mass.
  • The resulting change in speed over a specific distance can be calculated using kinematic equations.
Kinetic energy
Kinetic energy describes the energy of an object in motion. It depends on both the mass and the square of the velocity (speed) of the object. The formula for kinetic energy is given by:\[KE = \frac{1}{2} m v^2\]where \(m\) is the mass and \(v\) is the velocity.
In the context of our proton, the initial kinetic energy was calculated with its initial speed of \(2.4 \times 10^7 \, \mathrm{m/s}\), whereas the final kinetic energy was determined using the final speed after the proton was accelerated.
As the proton speeds up, its kinetic energy increases because energy is required to change the proton鈥檚 velocity.
  • Initial kinetic energy: starts from the initial velocity.
  • Final kinetic energy: calculated from the final speed after acceleration.
  • The difference between these energies indicates how much work was done to accelerate the proton.
Kinematic equations
Kinematic equations are invaluable tools in physics for analyzing motion considering initial speed, time, acceleration, and distance. These equations help us calculate unknown variables when certain parameters of motion are known.
To determine the final speed of a proton, one of the kinematic equations was utilized:\[v_f^2 = v_i^2 + 2as\]where \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance traveled.
In our problem, by substituting the given values, we calculated the final speed of the proton after traveling the given distance.
  • Kinematic equations allow computation of the motion of objects with constant acceleration.
  • They help find the final speed of a moving object when initial speed, acceleration, and distance are known.
  • Correct application of these equations requires understanding the motion parameters available.

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Most popular questions from this chapter

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .(\) a) How much work does the force do in \(10 \mathrm{~min} ?(\mathrm{~b})\) What is the average power (in horsepower) of the force?

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The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right)\). Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame, A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m}\). If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?
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