Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 113P

(a)

To determine

To Prove:The experimental magnitude of the free-fall acceleration is given by gexp=2Δy( Δt)2 .

(a)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

The distance Δy travelled by the marble in the measured time interval Δt is given by the following equation of motion.

  Δy=v0yΔt+12gexp(Δt)2  .....(1)

Here, v0y is the initial velocity of the marble.

The marble is dropped from a point at a negligible distance from the first light gate. Therefore, the distance travelled by the marble in the measured time interval Δt can be taken to be equal to the distance between the two light gates and the initial velocity of the marble can be taken to be equal to zero.

Therefore, substitute 0 for v0y in equation (1).

  Δy=12gexp(Δt)2

Simplify for gexp .

  gexp=2Δy( Δt)2  .....(2)

Conclusion:

Thus, it is proved that the experimental value of the free fall acceleration is given by the expression, gexp=2Δy( Δt)2 .

(b)

To determine

The expected value of (Δt) for the marble to pass through the light gates in this experimental setup assuming that the value of free-fall acceleration is equal to the standard value gexp=9.81 m/s2 .

(b)

Expert Solution
Check Mark

Answer to Problem 113P

The expected value of the time (Δt) for the marble to pass through the light gates in this experimental setup is found to be 0.319 s .

Explanation of Solution

Given:

The position of the first light gate (at the edge of the table)measured from the floor

  yi=1.00 m

The position of the second light gate when measured from the floor

  yf=0.500 m

The standard value of the free-fall acceleration

  gexp=9.81 m/s2

Formula used:

The time taken for the marble to fall between the two light gates is given from equation (2) and is written as

  Δt=2Δyg exp  .....(3)

Where, Δy=yiyf.......(4)

Calculation:

From equations (3) and (4)

  Δt=2( y i y f )g exp  .....(5)

Substitute the given values of variables in equation (5).

  Δt= 2( y i y f ) g exp = 2[ ( 1.00 m )( 0.500 m )] ( 9.81  m/s 2 )=0.319 s

Conclusion:

Thus, the expected value of the time (Δt) for the marble to pass through the light gates in this experimental setup is found to be 0.319 s .

(c)

To determine

The value of the gexp that will be determined if the first light gate is positioned 0.5 cm below the table and the percentage difference of this value from the standard value of g .

(c)

Expert Solution
Check Mark

Answer to Problem 113P

The value of the free fall acceleration determined in the current experimental setup would be 10.05 m/s2 and the percentage difference between this value and the standard value of g is 2.5% .

Explanation of Solution

Given:

The height of the edge of the table from the floor

  yi=1.00 m

The position of the first light gate (at the edge of the table) measured from the table

  yi'=0.5×102 m

The position of the second light gate when measured from the table

  yf=0.500 m

The standard value of the free-fall acceleration

  g=9.81 m/s2

Formula used:

The marble is dropped from the edge of the table with a velocity v0y and when it reaches the first light gate, it has a velocity vy which can be calculated using the expression,

  vy2=v0y2+2gyi'  .....(6)

Assume the origin to be located on the table and the initial velocity is zero.

The time Δt' recorded by the light gates can be obtained from the following expression:

  Δy'=vyΔt'+12g(Δt')2  ........(7)

Here, Δy'=yfyi'  .....(8)

The value of the free fall acceleration that would be obtained from this experiment (with errors) is given by the following expression

  gexp'=2Δy( Δt')2  .....(9)

The percentage difference between the calculated and the standard values of the free fall acceleration is given by

  % difference=gexp'gg×100%  ......(10)

Calculation:

Calculate the speed of the marble when it passes through the first light gate by substituting the values of the variables in equation (6).

  vy2=v0y2+2gyi'=(0 m/s)2+2(9.81  m/s2)(0.5× 10 2m)vy=3.31×102m/s

Calculate the distance between the light gates using equation (8).

  Δy'=yfyi'=0.500 m(0.5× 10 2m)=0.495 m

Substitute the values of the variables (calculated and given) in equation (7) and write a quadratic equation in (Δt') .

  Δy'=vyΔt'+12g(Δt')2(0.495 m)=(3.31× 10 2m/s)Δt'+12(9.81  m/s2)(Δt')2

  (4.905 m/s2)(Δt')2+(3.31×102m/s)Δt'(0.495 m)=0

Solve for (Δt') .

  Δt'=( 3.31× 10 2 m/s)± ( 3.31× 10 2 m/s ) 2 4( 4.905  m/s 2 )( 0.495 m )2( 4.905  m/s 2 )=( 3.31× 10 2 m/s)±( 3.116 m/s)( 9.81  m/s 2 )

Taking the positive root alone

  Δt'=0.3154 s

The person performing the experiment does not take into account the difference in the positioning of the light gates. Hence, the value of free fall acceleration gexp' she calculates would be given by equation (9).

Substitute the values of variables in equation (9) and calculate the experimental value of the free fall acceleration.

  gexp'=2Δy ( Δt' )2=2( 0.500 m) ( 0.3154 s )2=10.05 m/s2

Calculate the percentage difference between the measured value and the standard value is found using equation (10).

  % difference=g exp'gg×100%=( 10.05  m/s 2 )( 9.81  m/s 2 )( 9.81  m/s 2 )×100%=2.5%

Conclusion:

Thus, the value of the free fall acceleration determined in the current experimental setup would be 10.05 m/s2 and the percentage difference between this value and the standard value of g is 2.5% .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
1. Two pendula of slightly different length oscillate next to each other. The short one oscillates with frequency 0.52 Hz and the longer one with frequency 0.50 Hz. If they start of in phase determine their phase difference after 75 s.
A mass is connect to a vertical revolving axle by two strings of length L, each making an angle of 45 degrees with the axle, as shown. Both the axle and mass are revolving with angular velocity w, Gravity is directed downward. The tension in the upper string is T_upper and the tension in the lower string is T_lower.Draw a clear free body diagram for mass m. Please include real forces only.Find the tensions in the upper and lower strings, T_upper and T_lower
2. A stone is dropped into a pool of water causing ripple to spread out. After 10 s the circumference of the ripple is 20 m. Calculate the velocity of the wave.

Chapter 2 Solutions

Physics for Scientists and Engineers

Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 66PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 92PCh. 2 - Prob. 93PCh. 2 - Prob. 94PCh. 2 - Prob. 95PCh. 2 - Prob. 96PCh. 2 - Prob. 97PCh. 2 - Prob. 98PCh. 2 - Prob. 99PCh. 2 - Prob. 100PCh. 2 - Prob. 101PCh. 2 - Prob. 102PCh. 2 - Prob. 103PCh. 2 - Prob. 104PCh. 2 - Prob. 105PCh. 2 - Prob. 106PCh. 2 - Prob. 107PCh. 2 - Prob. 108PCh. 2 - Prob. 109PCh. 2 - Prob. 110PCh. 2 - Prob. 111PCh. 2 - Prob. 112PCh. 2 - Prob. 113PCh. 2 - Prob. 114PCh. 2 - Prob. 115PCh. 2 - Prob. 116PCh. 2 - Prob. 117PCh. 2 - Prob. 118PCh. 2 - Prob. 119PCh. 2 - Prob. 120PCh. 2 - Prob. 121PCh. 2 - Prob. 122P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY