Basics of Physics

Measurements and dealing with their uncertainties are some first skills you learn in a lab. In this article, I will go over what you need to know when dealing with data.

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Measurement and Uncertainty

The beginning of physics started with measuring quantities. It soon became apparent that not all data meant the same. It was important to have reliable measurements. However, that was often a problem. Measuring things was often a random affair. The devices used were not all made the same either. Many measuring devices were poorly made and inaccurate at first. So measurements will never be perfect. They will not because humans are error prone and the devices we use are as well.

Uncertainty

Every device, like a ruler, has a built in amount of uncertainty. It is vital to state this when publishing measurements. It also gives the uncertainty as a percentage. This can help make it clear. It is the ratio of the uncertainty to the measured value, multiplied by 100.

If we measure a piece of metal to be 6.5 cm long, we can assume it has an uncertainty of .1 cm. To get the percentage of uncertainty we divide:

\[ \frac{0.1}{6.5} = .015 = 1% \]

Significant Figures

This is the number of reliably known digits in a number. Non-zero digits are usually significant. Zeroes vary according to the situation. In basic lab work, you never want to state more precision than you have. This is just inaccuracy. If your measurements have 3 significant digits, then your calculations should have only 3 significant digits. Many people do this, don’t be one of them, please. If you are dealing with numbers that contain different amounts of significant digits, then always use the least amount. Consider the number 12 and 23.4. I use these two numbers in a calculation. The result should have 2 significant figures in it.

Scientific Notation

When you are dealing with numbers that contain many digits it is convenient to use scientific notation. A number like 100,00,000 can be written as \( 1.0 * 10^7 \). This is just a standard way to display numerical values. Another example is the number .00000383. I would write it as \( 3.83 * 10^-6 \).

Derived Quantities

Derived quantities come from the known base quantities. Base quantities are length, time, mass, electric current, temperature, substances, and luminosity. So a derived quantity uses these to form a different unit. Speed uses time and length units, for example.

Conversion of Units

Anything that we can measure comprises, a number, and the associated unit. Without the unit, the number means nothing. Sometimes there are multiple units involved. When this happens, we must convert to something meaningful to us. Lets convert 15 inches to centimeters. 1 in = 2.54 cm.

\[ 15 in * \frac{2.54 cm}{in} = 38.1 = 38 cm \]

You see that the inches cancelled out. This is the trick to knowing if you set it upright. The correct units will cancel out and it will leave you with the one unit that you want.

Example

We consider the age of the universe to be 14 billion years old. Let’s write this in scientific notation.

\[ 14,000,000,000  = 1.4 * 10^10  \]

Example

How many significant figures do the following numbers have? 214, 81.6, and .0086?

 214 = 3 significant digits, 81.6 = 3 significant digits, and .0086 has 2 significant digits 

Example

Write out the following numbers in scientific notation. 21.8, .0068

 21.8 = 2.18 * 10^1 and .0068 = 6.8 8 10^-3 

Write out the following number. \( 8.69 * 10^4 \)

\[ 8.6900 \]

This section was about the basics of measurements and uncertainty. When measuring, you must always take into account the uncertainty of your tool. As you state your calculation, give the correct number of significant digits.

Vectors

Quantities like mass and temperature are described with a number called a scalar. Other quantities like displacement, velocity, or force have a direction associated with them and are called vectors. When looking at a vector on paper, a vector is an arrow with the length of the arrow representing the number and direction of the vector. 

A vector can be described with a number and an angle, like \( A=19 \angle 25 \deg \). When doing basic math operations, you should use the component form to write vectors. So, if a line were placed with the tail of the vector at the origin of the coordinate system, x and y portions can be calculated like this:

\( y = 19 \sin 25^{\circ} = 8.02 \)

\( x = 19 \cos 25^{\circ} =  17.21 \)

This is how you calculate vectors. 

If you have the x and y vectors, you can calculate the length of the corresponding line by:

\( \sqrt{x^{2} + y{2}} \)

\( \sqrt{8.02^{2} + 17.21^{2}} \)

\( \sqrt{64.32 + 296.18} \)

\( \sqrt{360.5} = 19  \)

To find the original angle we use:

\( \theta = \tan^{-1} \frac{y}{x} \)

So:

\( \theta = \tan^{-1} \frac{8.02}{17.21} \)

\( \theta = 25^{\circ} \)

Example 1

Write out the components of the vector \( c = 47 \angle 193^{\circ} \)

The x component is:

\( x = 47 \cos 193^{\circ} = -45.8 \)

The y component is:

\( y = 47 \sin 193^{\circ} = -10.6 \)

The use of unit vectors simplifies the mathematical operations on vectors. In two dimensions, unit vectors are vectors of unit value and in the +x and +y directions. 

So the first vector would be written as \( 18.4i + 13.8j \) and the second vector is 

\( -45.8i - 10.6j \).

You can add vectors by adding the individual components. If you have:

\( (18.4i + 13.8j) + (-45.8i - 10.6j) \)

You just add the individual components together. The combined vector becomes:

\( -27.4i + 3.2j \)

Subtracting vectors can also be done. It is a bit quirky, however. You will use the opposite sign of the second vector.

\( (18.4i + 13.8j) - (-45.8i - 10.6j) \)

This gives the result of:

\( 64.2i + 24.4j \)

We should also know how to work backward when given the components. When we do this properly it will give us the length of the vector and its angle.

Let us use the last vector:

\( 64.2i + 24.4j \)

\( V = \sqrt{64.2^{2} + 24.4^{2}} = 68.7 \)

\( \theta = \tan^{-1}\frac{24.4}{64.2} = 20.8^{\circ} \)

When put in vector form:

\( V = 68.7 \angle 20.8^{\circ} \)

Example

Add the vectors: 

\(A=13 \angle 50^{\circ}\)

\(B=15 \angle -60^{\circ} \)

\(C=17 \angle 20^{\circ} \)

The first step is to put each vector in its component form.

\( A = \sin 50 * 13 = 10 = y \)

\( A = \cos 50 * 13 = 8.4 = x \)

\( A = 8.4i + 10.0j \)

\( B = \sin(-60) * 15 = -13 = y \)

\( B = \cos(-60) * 15 = 7.5 = x \)

\( B = 7.5i -13.0j \)

\( C = \sin 120 * 17 = -8.5 = y \)

\( C = \cos 120 * 17 = -14.7 = x \)

\( C = -14.7i -8.5 j \)

Sum all the i’s and all the j’s. You get:

\( 1.2i -11.5j \)

That is your x and y dimensions

Now use the Pythagorean theorem with those values

\( \sqrt{11.5^{2} + 1.2^{2}} = 11.6 \)

That gives you the length of the new vector

Now, let’s find the new angle

We use the inverse tangent to do this

\( \tan^{-1} \frac{-11.5}{1.2} = -84^{\circ} \)

So the new vector is:

\( 11.6 \angle -84^{\circ} \)

Use as many diagrams as possible when doing problems like this. Diagrams help keep your thought processes straight and help prevent silly errors. 

Example

Find the resultant of the two forces \(800(N) \angle 47^{\circ}\) and \(600(N) \angle 140^{\circ}\)

I will write out all the calculation steps first, before I actually compute them

\(800 \cos 47^{\circ} = 545.6 \)

\(800 \sin 47^{\circ} = 585.1 \)

\(600 \cos 140^{\circ} = -460.0 \)

\(600 \sin 140^{\circ} = 385.7 \)

Vector 1 is: \(545.6i + 585.1j\)

Vector 2 is: \(-460.0i + 385.7j\)

Add the vectors together

We get:

\(85.6i + 970.8j\)

Use Pythagorean theorem

\(\sqrt{85.6^{2} + 970.8^{2}}\)

The length of our new vector is:

\(= 974.6 \)

The new angle of our vector is:

\(\tan^{-1}\frac{970.8}{85.6} = 85^{\circ} \)

So:

\(974.6(N) \angle 85^{\circ}\)

Scalars

The dot product of two vectors produces a scalar. Scalars, as you know, do not have a direction associated with them. There are two ways to compute a dot product.

\(A*B = AB \cos \theta\)

And 

\(A*B = a_xb_x + a_yb_y\)

Example

Form the dot product of \(A = 23 \angle 37^{\circ}\) and \(B = 14 \angle -35^{\circ}\)

Use the first definition

\(A*B = A*B \cos \theta = 23*14 \cos 72^{\circ} = 100\)

Now use the second definition

This requires the components of both vectors

We learned how to do this earlier, it is the same process

We get:

\(18.4i + 13.8j\) and \(11.5i + (-8.0)j\)

\(A*B=100\)

Cross Product

The cross product of A*B also produces a vector. 

\(A*B \sin \theta\)

Example

Form the cross product of \(A = 23 \angle 37^{\circ}\) and \(B = 47 \angle 193^{\circ}\)

This is A*B* sine of the difference in the angles.

The difference in angles is 156 degrees.

\(A*B = 23*47*\sin 156^{\circ} = 440\)

We can also use determinants to get the same result

Calculate the components of each vector:

Cross multiply. This is why it is called the cross product

\(A*B = 18.4(-10.6) - 13.8(-45.8) = 440\)

If it is a simple problem then use the first definition any time you can. Otherwise, use the second definition.

Kinematics

The motion of a particle is described by giving position, velocity, and acceleration, usually as a function of time. We start with one dimension. Distance means total distance traveled, while displacement is the actual difference between endpoint and the beginning.

\[Velocity = \frac{displacement}{time}\]

and:

\[Acceleration = \frac{v2-v1}{time}\]

This velocity calculation fits our present definition and is equivalent to drawing a straight line between points on the smooth curve of x versus t. Calculating velocity this way presents a problem, because, depending on the interval, we will get different answers for the velocity. To find the velocity at t=1, we found x at t=1 and x at t=2 and performed the velocity calculation. This is not a good approximation to the velocity at t=1 because the average is between t=1 and t=2. For better approximations, shorten the time interval centered about t=1. Getting it as small as you can will give better results. So, the slope of the straight line tangent to the curve at x=1 represents the velocity at x=1.

Instantaneous Velocity and Acceleration

A more versatile definition of velocity is \(\frac{\Delta x}{\Delta t}\), where the interval \(\Delta t\) is very small and centered about the time where the velocity is desired. This approach leads to a general method for obtaining an expression for velocity that can be evaluated at any point rather than going through the numeric calculation whenever we want a velocity. This definition is stated as the instantaneous velocity.

\[v = \lim\limits_{\Delta t \to \0} = \frac{dx}{dt}\]

This definition of the derivative is the slope of the tangent to the curve evaluated at the point in question. If we want to find the velocity of any particle traveling according to a polynomial relationship between x and t, all we need is a general technique for finding the slope of the tangent to the polynomial at any point. 

The instantaneous acceleration is defined as the value of \(\Delta v / \Delta t\) as \(\Delta t\) approaches zero.

\[a = \lim\limits_{\Delta t \to \0} = \frac{dy}{dt}\]

The instantaneous acceleration is the slope of the tangent to the curve of v versus t. The general expression for the slope of any polynomial of the form \(x = ct^{n}\). The expression for the slope at any point is \(cnt^{n-1}\). Stating this in calculus terms, for any function \(x = ct^{n}\), the derivative of the function is \(cnt^{n-1}\). This can be verified in the case of the parabola by taking successively smaller intervals of \(\Delta t\) and \(\Delta x\) at any point t to verify that the slope at any point is 2t.

Kinematic Equations of Motion

The kinematic equations of motion are derived under the assumption of constant acceleration. While this may seem at first to be a restriction, there are a large number of problems where the acceleration is constant. The simplest and most obvious are falling body problems, or problems involving bodies falling on or near the surface of the earth where the acceleration due to gravity is a constant. 

Starting with the assumption of constant acceleration, we can write:

\[a = \frac{v_f - v_o}{t}\]

This can be rearranged to this form:

\[ta = v_f - v_o \longrightarrow ta + v_o = v_f \longrightarrow v_f = v_o + ta \]

We define the average velocity as:

\[v_{avg} = \frac{v+ v_o}{2}\]

We define displacement as:

\[x = x_o + v_{avg}t\]

We can now substitute \(v_{avg}\) with the previous equation.

\[x = x_o + \frac{v + v_ot}{2}\]

One more substitution for \(v = v_o+ at\) and we get:

\[x= x_o + v_ot + (½)at^2\]

We now have the four kinematic equations of motion.

\[v = v_o + at\]

\[x - x_o = (½)(v + v_o)t\]

\[x - x_o = v_ot + (½) at^2\]

\[v^2 = v_o^2 + 2a(x - x_o)\]

The first three equations relate displacement, velocity, and acceleration in terms of time, while the fourth equation does not contain the time. 

Example 1

A train starts from rest and moves with constant acceleration. On first observation, the velocity is 20 m/s and 80 s later the velocity is 60 m/s. At 80 s, calculate the position, average velocity, and constant acceleration over the interval.

Calculate the acceleration:

\[a =\frac{v_f-v_o}{t} = \frac{60\text{ m/s} - 20\text{ m/s}}{80\text{ s}} = 0.50 \text{ m/s^2}\]

Calculate the distance traveled over this 80 s:

\[x = v_ot + (½)at^2 = (20\text{ m/s})(80\text{ s}) + (½)(0.50\text{ m/s^2})6400\text{ s^2}= 3200\text{ m}\]

The average velocity is:

\[v_{avg} = \frac{v_f+v_o}{2} = {20\text{ m/s}+60\text{ m/s}}{2} = 40\text{ m/s}\]

If the acceleration is constant, then the average velocity is the average of 20 m/s and 60 m/s, or 40 m/s, and at an average velocity of 40 m/s and 80 s, the distance traveled is 3200 m. 

Example 2

Using the above problem, calculate the position of the train at 20 s.

\[x = v_ot + (½)at^2  = (20\text{ m/s})(20\text{ s}) + (½)(0.50\text{ m/s^2})(20\text{ s})^2 = 500 \text{ m}\]

Example 3

Using the above problem, find the time required for the train to reach 100m.

\[x-x_o = v_ot+(½)at^2 \longrightarrow 100\text{ m} = (20\text{ m/s})t + (½)(0.50\text{ m/s^2})t^2\]

\[t = \frac{-80{\pm} \sqrt{6400-4(1)(400)}}{2(1)} = \frac{-80{\pm}89}{2} = 4.5, -85\]

The negative answer is not correct because we do not want negative time, so t=4.5s

Example 4

Using the above problem, find the velocity of the train at 120 m.

\[v^2 = v_o^2 + 2ax = (20\text{ m/s})^2 + 2(0.50\text{ m/s^2})120m = 400\text{ m^2/s^2} + 120\text{ m^2/s^2} = 520\text{ m^2/s^2} = 23\text{ m/s}\]

Example 5

Two vehicles are at position x=0 and t=0. Vehicle 1 is moving at constant velocity of \(30 \text{ m/s}\). Vehicle 2, starting from rest, has acceleration of \(10 \text{ m/s^2}\). Where do they pass? 

Translated into algebra, this is asking what is the value of x when they pass? This can be determined by writing equations for the position of each vehicle and equates

\[ x_1 = v_10 t = 30 \text{ m/s} \text{ t} \]

and

\[x_2=(½)a_2t^2=(5\text{ m/s^2})t^2\]

Setting x1=x2 gives the time when they pass as \(30t=5t^2\) or \(5t(t-6)=0\), so the vehicles pass at t=0 and t=6. Putting t=6 in either equation for x gives 180 m as the distance.

Example 6

In the above situation, when do the vehicles have the same velocity?

In algebra, this means to set the equations for velocity equal (v1=v2) and solve for the time. Remember that three of the four equations of motion are functions of time, so most questions are answered by first calculating the time for a certain condition to occur.

\[30\text{ m/s}=(10\text{ m/s^2})t = 3.0\text{ s}\]

Now that we know when, we can calculate where they have the same velocity. Use either equation for position and t=3.0s

\[x_1|{t=3.0s}=(30\text{ m/s})(3.0s)=90\text{ m}\]

Example 7

For the situation in problem 5, what are the position, velocity, and acceleration of each vehicle when vehicle 2 has traveled twice the distance of vehicle 1?

The time when this occurs is when x2=2x1 so:

\[5t^2=60t\]

Or

\[5t(t-12)=0\]

This gives the time of t=0 and t=12. The time t=0 is correct but t=12 is what we are looking for because we want to know something after some time has passed. At t=12,

\[x_1=(30\text{ m/s})(12\text{ s})=360\text{ m}\]

From the original statement of the problem,

\[v_1 0=30\text{ m/s}\]

and \(a_1=0\). Now solve for the remaining variables for the second vehicle by substitution.

\[x_2=(5\text{ m/s^2})(12\text{ s})=720\text{ m}\]

\[v_2=(10\text{ m/s^2})(12\text{ s})=120\text{ m/s}\]

From the original statement of the problem, \(a_2=10\text{ m/s}\).

Example 8

Two trains are traveling along a straight track, one behind the other. The first train is traveling at \(12\text{ m/s}\). The second train, approaching from the rear, is traveling at \(20\text{ m/s}\). When the second train is 200 m behind the first, the operator applies the brakes, producing a constant deceleration of \(0.20\text{ m/s^2}\). Will the trains collide, and if so, where and when?

First, diagram the situation. Write down the equations for each train and the information provided in the problem.

Take t=0 when the brakes are applied and the first train is 200 m ahead of the second. This makes the position of the first train 200 m at t=0.

The question as to whether the trains collide means, is there a real time solution to the equation resulting from setting x1=x2?

\[200m+(12 \text{ m/s})t=(20 \text{ m/s})t-(0.10 \text{ m/s^2})t^2\]

If there are no real solutions to the equation, then the trains do not collide. Drop the units, and write \(0.10t^2-8t+200=0\) which is solved by the quadratic formula. In a calculator, you can see there are no real solutions, so the trains never collide.

Example 9

Change the previous problem by giving the second train an initial velocity of 25 m/s. This will give a real time for the collision. Find the collision time.

Setting the expressions for x1 ands x2 equal and dropping the units produces

\[200+12t=25t-0.10t^2\]

Or

\[0.10t^2-13t+200=0\]

This equals \(17.8, 112\)

The two times correspond to when x1=x2. The earliest time is the first coincidence and the end of the physical problem. The position at this time can be obtained from either expression for x.

\[x_1=200m+(12 \text{ m/s})17.8\text{ s}=414 \text{ m}\]

Verify this distance by using x_2. The velocity of the second train at collision is 

\[v_2=25 \text{ m/s}-(0.20 \text{ m/s^2})17.8\text{ s}=21.4 \text{ m/s}\]

The relative velocity between the two trains is \(v=1.4 \text{ m/s}\). The two times are the result of the quadratic in t. The two solutions occur when the curves cross. The equation for \(x_1=200+12t\) is a straight line of slope 12 starting at 200. The equation for \(x_2=25t-0.10t^2\) is a parabola that opens down. While the mathematics produces two times, the reality of the problem dictates the earlier time as the one for the collision.