Category Archives: Physics

Impulse and Momentum – Physics Example Problem

Desktop Momentum Balls Toy

This model of a common desktop toy shows the forces acting on the raised ball. The principles of impulse and momentum show how the momentum is transferred to each ball and the process repeats.

Impulse and momentum are physical concepts that are easily seen from Newton’s Laws of Motion.

Start with this equation of motion for constant acceleration.

v = v0 + at

where
v = velocity
v0 = initial velocity
a = acceleration
t = time

If you rearrange the equation, you get

v – v0 = at

Newton’s second law deals with with force.

F = ma

where
F = force
m = mass
a = acceleration

solve this for a and get

a = F/m

Stick this into the velocity equation and get

v – v0 = (F/m)t

Multiply both sides by m

mv – mv0 = Ft

The left side of the equation deals with momentum (often denoted by a lower-case p) and the right side is impulse (often denoted by an upper-case letter J).

Mass times velocity is known as momentum and force applied over time is called impulse.

Impulse and Momentum Example Problem

Question: A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) What is the initial momentum of the mass?
b) What is the final momentum of the mass?
c) What was the force acting on the mass?
d) What was the impulse acting on the mass?

Impulse and Momentum Example Problem

Part a) What is the initial momentum?

Momentum is mass times velocity. Since the mass is at rest, the initial velocity is 0 m/s.

momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s

Part b) What is the final momentum?

After the force is finished acting on the mass, the velocity is 3 m/s.

momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s

Part c) What was the force acting on the mass?

mv – mv0 = Ft

From parts a and b, we know mv0 = 0 kg⋅m/s and mv = 150 kg⋅m/s.

150 kg⋅m/s – 0 kg⋅m/s = Ft
150 kg⋅m/s = Ft

Since the force was in effect over 3 seconds, t = 3 s.

150 kg⋅m/s = F ⋅ 3s
F = (150 kg⋅m/s) / 3 s
F = 50 kg⋅m/s2

Unit Fact: kg⋅m/s2 can be denoted by the derived SI unit Newton (symbol N)

F = 50 N

Part d) What was the impulse acting on the mass?

The impulse is the force multiplied by the time passed. It is also equal to the change in momentum over the same time period.

Ft = 50 N ⋅ 3 s
Ft = 150 Ns or 150 kg⋅m/s

The impulse was 150 kg⋅m/s.

These problems are relatively simple as long as you keep your units straight. Impulse and momentum should have the same units: mass⋅velocity or force⋅time. Check your units when you check your answer.

Another possible way to cause errors is to confuse your vector directions. Velocity and Force are both vector quantities. In this example, the mass was pushed in the direction of the final velocity. If another force pushed in the opposite direction to slow down the mass, the force would have a negative value compared to the velocity vector.

If you found this helpful, check out other Physics Example Problems.

Make a Lichtenberg Figure the Easy Way

This Lichtenberg figure was made by shooting a beam of electrons (~2.2 million volts) through an insulator. The pattern is illuminated by blue LEDs. (Bert Hickman, Wikipedia Commons)

This Lichtenberg figure was made by shooting a beam of electrons (~2.2 million volts) through an insulator. The pattern is illuminated by blue LEDs. (Bert Hickman, Wikipedia Commons)lichte

Lichtenberg figures are fern-like branching structures formed from an electrical discharge on or inside of an insulator. Like snowflakes, every Lichtenberg figure is unique – an intricate and beautiful natural work of art. Lichtenberg figures form naturally, sometimes in the skin of lightning strike victims, in lightning strikes into sand (fulgurites), and potentially from pretty much any high-voltage electrical discharge into an insulator. The structures take their name from Georg Christoph Lichtenberg, the physicist who discovered and studied them.

Make Lichtenberg Figures the Original Way

One way to make your own Lichtenberg figure is to use Lichtenberg’s original method. He used hardened tree sap resin for the insulator and dust to reveal the fractal pattern. You can use polyethylene sheets for the insulator and then talcum powder, powdered sulfur, or lycopodium powder to reveal the fractal pattern.

  1. Place a sharp metal point in the center of the sheet of plastic. A nail is a good choice. All that really matters is that it’s a good electrical conductor.
  2. Zap the metal object with static electricity.  Electricity travels through the metal and across the plastic insulator. The amount of discharge affects how far the pattern extends from the metal point and how deep into the plastic it gets. So, if you shock the metal with your fingertip after shuffling through carpeting, you probably won’t get as big a pattern as if you use a Wimhurst machine.
  3. Blow powder over the surface of the plastic sheet. It will stick to the pattern, revealing the figure.

Burning a Lichtenberg Figure Into Wood

Another easy method wood-burning a fractal pattern into pine. This can be achieved by applying 2-10 kV of voltage to a pair of nails which have been driven into a piece of dampened pine wood. Another option is to snap alligator clips onto pieces of wood. The reason the pine is dampened is so it doesn’t burn as you apply the electricity. You can mix some baking soda into the water used to dampen the wood to enhance its surface conductivity. You’ll need to experiment with the distance between the nails and the duration of the charge. If the wood starts to dry out, turn off the power and spritz the wood with more water before continuing. A little fire is to be expected and is fine. The type of wood you use makes a difference, too. Here’s what to expect:

There is an easier method you may wish to try:

Acrylic and Toner Lichtenberg Figures

  • sharp metal object (e.g., awl)
  • insulator (e.g., sheet of acrylic)
  • photocopier toner

An adaptation of Lichtenberg’s original method is to use hard acrylic and toner. Gather your materials and make the Lichtenberg figure:

  1. Position the metal object so that only its tip is touching the surface of the insulator.
     
  2. If you have a Wimshurst machine or Van de Graaff generator handy, discharge it through the metal point into the acrylic. 
     
  3. If you don’t have a machine, you’ll have to generate static electricity another way, like by dragging your feet through a shag carpet and zapping yourself on the metal object.
     
  4. In either case, you will create a Lichtenberg figure across the surface of the acrylic, radiating outward from the metal point. However, you probably won’t be able to see it. If you (carefully) blow toner powder across the surface of the acrylic, the Lichtenberg figure will be revealed.

Another way to reveal the fractal is to illuminate it with a light source. You want a strong beam of light and may need to play with the angle to catch the figure. No two Lichtenberg figures are identical, so you will get different results each time you try the project.

When you’re ready to get serious, here’s what you can do…

Moving Clocks Run Slower – Time Dilation

Special relativity theory introduced an interesting notion about time. Time does not pass at the same rate for moving frames of reference. Moving clocks run slower than clocks in a stationary frame of reference. This effect is known as time dilation. To calculate this time difference, a Lorentz transformation is used.

Time Dilation Lorentz Formula
where
TM is the time duration measured in the moving frame of reference
TS is the time duration measured from the stationary frame of reference
v is the velocity of the moving frame of reference
c is the speed of light

Time Dilation Example Problem

One way this effect was experimentally proven was measuring the lifetime of high energy muons. Muons (symbol μ) are unstable elementary particles that exist for an average of 2.2 μsec before decaying into an electron and two neutrinos. Muons are formed naturally when cosmic ray radiation interacts with the atmosphere. They can be produced as a by-product of particle collider experiments where their time of existence can measured accurately.

A muon is created in a laboratory and observed to exist for 8.8 μsec. How fast was the muon moving?

Solution

Time Dilation - Relativity Example Problem

A muon forms at t=0 moving at a velocity v. After 2.2 microseconds, the muon decays. A stationary observer measured the lifetime to be 8.8 microseconds. What was the velocity of the muon?

From the muon’s frame of reference, it exists for 2.2 μsec. This is the TM value in our equation.
TS is the time measured from the static frame of reference (the laboratory) at 8.8 μsec, or four times as long as it should exist: TS = 4 TM.

We want to solve for velocity, Let’s simplify the equation a little bit. First, divide both sides by TM.

Time Dilation Example Step 2

Flip the equation over

Time Dilation Step 3

Square both sides to get rid of the radical.

Time Dilation Step 4

This form is easier to work with. Use the TS = 4 TM relationship to get

time dilation step 5
or
Time Dilation step 6

Cancel out the TM2 to leave

Time Dilation Step 7

Subtract 1 from both sides

Time Dilation Example Step 8
Time Dilation Example step 9
Time Dilation Example Step 10

Multiply both sides by c2

Time Dilation Example Step 11

Take the square root of both sides to get v

Time Dilation Example step 12
v = 0.968c 

Answer:

The muon was moving at 96.8% the speed of light.

One important note about these types of problems is the velocities must be within a few orders of magnitude of the speed of light to make a measurable and noticeable difference.

Moving Rulers Are Shorter – Length Contraction Example Problem

Relativity tells us moving objects will have different lengths in the direction of motion, depending on the frame of reference of the observer. This is known as length contraction.

This type of problem can be reduced to two different frames of reference. One is the frame of reference where a static observer is observing the moving object as it passes by. The other frame of reference is riding along with the moving object. The length of the moving object can be calculated using the Lorentz transformation.

Length Contraction Formula
where
LM is the length in the moving frame of reference
LS is the length observed in the stationary frame of reference
v is the velocity of the moving object
c is the speed of light.

Length Contraction Example Problem

How fast would a meter stick have to move to appear half its length to a stationary observer?

Moving Rulers are Shorter

In the above illustration, the top meter stick is measured as it zips by at velocity v. Both meter sticks are the same length (1 meter) in their own frame of reference, but the moving one appears to only be 50 cm long to the stationary observer. Use the Lorentz transformation contraction formula to find out the value of v.

LM is the length in the moving frame of reference. In the moving frame of reference, the meter stick is 1 meter long.
LS is the measured length from the stationary frame of reference. In this case, it is ½LM.

Plug these two values into the equation

Length Contraction step 2

Divide both sides by LM.

Length Contraction step 3

Cancel out the LM to get

Length Contraction Example Step 4

Square both sides to get rid of the square root

Length Contraction Example Step 5

Subtract 1 from both sides

Length Contraction Example Step 6
Length Contraction Step 7
Length Contraction Step 8

Multiply both sides by c2

Length Contraction Step 9

Take the square root of both sides

Length Contraction Example step 10
or
Length Contraction Example Step 11

v = 0.866c or 86.6% the speed of light.

Answer

The ruler is moving 0.866c or 86.6% the speed of light.

Note the moving frame of reference must be moving rather quickly to show any measurable effect. If you follow the same steps as above, you can see the ruler needs to be traveling at 0.045c or 4.5% the speed of light to change the length by a millimeter.

Note too that the meter stick only changes its length in the direction of the movement. The vertical and depth dimensions do not change. Both rulers are as tall and thick in both frames of reference.

How Fast Would You Have To Go To Make A Red Light Look Green? – Relativistic Doppler Effect

Everyone knows about the doppler effect with sounds. When a train approaches, the pitch of its sound increases. After it passes, the pitch seems to drop off. This is because the sound waves are compressed (wavelength shortened/frequency increased) ahead of a moving sound source. The sound waves expand (wavelength increased/frequency decreased) as the source moves away. The faster the sound source moves, the greater the change in pitch.

The doppler effect happens with all types of waves, not just sound. Light waves can be affected by the speed of the observer in the same matter. If you drive fast enough, you can change a red light to appear green to the driver. How fast would you have to be driving to make a red light look green?

Red Light Appears Green

How fast would you have to drive to make a red light appear green?

The speeds necessary to achieve a noticeable change in light are on the order of the speed of light. These velocities need to take account of relativistic transformations of the moving systems. The relativistic doppler effect of wavelength for systems approaching each other can be expressed by the formula

Relativistic Doppler Effect Example Step 1
where
λR is the wavelength seen by the receiver
λS is the wavelength of the source
β = v/c = velocity / speed of light

We can solve this for the velocity in a few steps. First, divide both sides by λS

Relativistic Doppler Effect Example Step 2

Square both sides

Relativistic Doppler Effect Example Step 3

Cross multiply each side

λR2( 1 + β) = λS2( 1 – β)

Multiply out both sides

λR2 + λR2β = λS2 – λS2β

Add  λS2β to both sides

λR2 + λR2β + λS2β = λS2

Subtract λR2 from both sides

λS2β + λR2β = λS2 – λR2

Factor out β from the left side of the equation

β (λS2 + λR2) = λS2 – λR2

Finally, divide both sides by (λS2 + λR2)

Relativistic Doppler Effect Example Step 4

Now we can find the velocity using the relationship: β = v/c.

Now we can plug in some numbers for red lights and green lights. Let’s take the wavelength of a red light to be 650 nm and a green light to be 540 nm. The source light is red and the received light is green. λS = 650 nm and λR is 540 nm. Plug these values into the above equation.

Relativistic Doppler Effect Example Step 5

Relativistic Doppler Effect Example Step 6

Relativistic Doppler Effect Example Step 7

β = 0.183

β = v/c

v = βc

v = 0.183c

If we take the speed of light to be 3 x 105 km/s, then you would have to be driving 54,900 km/s to shift a red light to look green. Another way to view it is that you’d need to be travelling 18.3% of the speed of light.

Multiply this value by 3600 s/hr to convert to km/hr, you get 197,640,000 km/hr. While you won’t get a citation for running a red light, you will get one for speeding.

If you do get pulled over, respect the police officer that managed to catch up to you.

Laws of Thermodynamics

The laws of thermodynamic describe the relationships between matter and energy under different conditions. (image credit: Brenda Clarke)

The laws of thermodynamic describe the relationships between matter and energy under different conditions. (image credit: Brenda Clarke)

Thermodynamics is the study of energy and heat. The laws of thermodynamics describe the relationship between matter and energy and how they relate to temperature and entropy. Many texts list the three laws of thermodynamics, but really there are four laws (although the 4th law is called the zeroeth law, probably to confuse you).

Here’s a list of the laws of thermodynamics and a quick summary of what each law means.

Zeroeth Law of Thermodynamics

If two systems are in thermal equilibrium with a third system, they must be in thermal equilibrium with each other. This law establishes the concept of temperature.

Example: If your car is the same temperature as your house and your car is the same temperature and your office, then your home and your office are the same temperature as each other.

First Law of Thermodynamics

The first law of thermodynamics is also known as the Law of Conservation of Energy. It states energy of a system may change forms, but it is neither created nor destroyed. One way to state this law is “you can’t get something for nothing”.

Second Law of Thermodynamics

The second law states the entropy of a system not in thermal equilibrium increases. Entropy is a measure of the randomness or disorder of a thermodynamic system. As entropy increases, less energy is available for useful work. If the first law states you can’t get something for nothing, you could consider the second law to mean “… and you can’t break even.”

Example: A watch driven by a spring will wind down as its potential energy is converted into kinetic energy. After that, the watch won’t run again until new energy is input into the system by winding the watch.

Third Law of Thermodynamics

The third law states the entropy of a system approaches as constant value as the temperature approaches absolute zero. Absolute zero is the lowest theoretically possible temperature (0K or zero Kelvin). The entropy of a system at absolute zero is nearly zero.

Perpetual Motion Machines Are Impossible

One implication of the laws of thermodynamics is that perpetual motion machines are not possible. While energy may change from one form into another, entropy increases, and a bit of usable energy is lost. Machines are powered by energy sources that eventually are depleted. The closest people can get to perpetual motion is to use an initial power supply that seems endless, such as solar power, as the initial energy input.

Table of Common Constants

Lasers

Series of different colored lasers. No matter the color, the speed of light is a constant. The speed of light is just one constant in this table of common constants. Credit: Pang Ka kit, Creative Commons

Here is a useful table of common constants. These physical constants appear time and again in chemistry and physics homework problems.

Why memorize them when you can just look them up.

I personally prefer to keep a sheet with these constants stuffed in the back of book. You can print up your own sheet with a PDF copy of the Table of Common Constants.

TABLE OF COMMON CONSTANTS

 Name SymbolValue
Speed of lightc299792458 m⋅s-1
Charge of an electrone1.602176565(35)×10−19 C
Universal gravitational constantG6.67384(80)×10-11 N⋅m2/kg2
Planck’s constanth6.626070040(81)×10−34 J⋅s
h4.135667662(25)×10−15 eV⋅s
Dirac’s constantħ1.054571800(13)×10−34 J⋅s
ħ6.582119514(40)×10−16 eV⋅s
Boltzmann’s constantk1.3806488(13)×10−23 J⋅K-1
8.6173324(78)×10−5 eV⋅K-1
Avogadro’s constantNA6.022140857(74)×1023 mol−1
Faraday’s constantF96485.3365(21) C⋅mol−1
Ideal gas constantR8.3144621(75) J⋅K−1⋅mol−1
Mass of an electronme9.10938291(40)×10−31 kg
0.510998928(11) MeV/c2
Mass of a protonmp1.672621777(74)×10−27 kg
938.272046(21) MeV/c2
Mass of a neutronmn1.674927351(74)×10−27 kg
939.565378(21) MeV/c2
Atomic mass unitu, amu or Da1.660538921(73)×10−27 kg
931.494061(21) MeV/c2
Permittivity of free spaceε08.854187817×10−12
Coulomb’s constant1/4πε08.9875517973681764×109 N⋅m2⋅C-2
Permeability of free spaceμ04π×10−7 N⋅A-2
Rydberg constantR1.0973731568539(55)×10−7 m-1
Bohr radiusa05.2917721092(17)×10−11 m