Category Archives: Physics Example Problems

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

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

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.

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
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?


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
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 


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
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
Length Contraction Example Step 11

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


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.

Wavelength and Energy Example Problem

Human Eye

You can use this wavelength and energy example to calculate the number of photons needed to “see”. Credit: Petr Novák, Wikipedia

This wavelength and energy example problem will show how to find the energy of a photon from its wavelength.

First, let’s look at the physics of the problem. The energy of a single photon of light is dependent on its frequency. This relationship is expressed in the equation

E = hƒ

E is the energy of the photon
h is Planck’s constant = 6.626 x 10-34 m2kg/s
ƒ is the frequency of the photon

The wavelength of a photon is related to the frequency by the equation

c = ƒλ

c is the speed of light = 3.0 x 108 m/s
ƒ is the frequency
λ is the wavelength

Solve this for frequency, and you get

ƒ = c / λ

Substitute this equation into the energy equation and get

E = hc / λ

With this equation, you can now find the energy of a photon when the wavelength is known. You can also find the wavelength if the energy of the photon is known.

Wavelength and Energy Example Problem

Question: The human eye’s optic nerve needs 2 x 10-17 joules of energy to trigger a series of impulses to signal the brain there is something to see. How many photons of 475 nm blue light is needed to trigger this response?

Solution: We are given the amount of energy needed to trigger the optic nerve and the wavelength of light.

First, let’s figure out how much energy is in a single photon of the blue light. We are given the wavelength as 475 nm. Before we go any further, let’s convert this to meters.

1 nm = 10-9 m

Using this relationship, convert 475 nm to meters

x m = 4.75 x 10-7 m

Now we can use the energy formula from above

E = hc / λ

Plug in the variables

E = (6.626 x 10-34 m2kg/s)(3 x 108 m/s) / 4.75 x 10-7 m

Solve for E

E = 4.18 x 10-19 J

This is the energy of a single photon of 475 nm blue light. We need 2 x 10-17 J of energy to begin the process.

x photons = 2 x 10-17 J

1 photon = 4.18 x 10-19 J

Divide one equation into the other to get

photon ratio

Solve for x

x = 47.8 photons

Since you can’t have partial photons, we need to round this answer up to the nearest whole photon count. 47 photons isn’t enough, so one more is needed to get over the threshold energy.

x = 48 photons

Answer: It takes 48 photons of 475 nm blue light to trigger the optic nerve.

Conservation of Momentum Example Problem

Momentum is a measurement of inertia in motion. When a mass has velocity, it has momentum. Momentum is calculated by the equation

momentum = mass x velocity
momentum = mv

This conservation of momentum example problem illustrates the principle of conservation of momentum after a collision between two objects.


Consider a 42,000 kg train car travelling at 10 m/s toward another train car. After the two cars collide, they couple together and move along at 6 m/s. What is the mass of the second train car?

Momentum Example Problem 1


In a closed system, momentum is conserved. This means the total momentum of the system remains unchanged before and after the collision.

Before the collision, the total momentum was the sum of the momentums of both train cars.

The first car’s (blue freight car) momentum is

momentumBlue = mv
momentumBlue = (42,000 kg)(10 m/s)
momentumBlue = 420,000 kg·m/s

The second car’s (tanker car) momentum is

momentumtanker = mv
momentumtanker = m(0 m/s)
momentumtanker = 0 kg·m/s

Add these two together to get the total momentum of the system prior to collision.

Total momentum = momentumBlue + momentumtanker
Total momentum = 420,000 kg·m/s + 0 kg·m/s
Total momentum = 420,000 kg·m/s

After the collision, the two cars couple together into one mass. This new mass is

massBlue + masstanker
40,000 kg + masstanker

The velocity the new pair of cars is travelling is 6 m/s. Because momentum is conserved, we know the total momentum of the cars after the collision is equal to the momentum prior to the collision.

Total Momentum = 420,000 kg·m/s
Total Momentum = mv
Total momentum = (42,000 kg + masstanker)·(6 m/s)

420,000 kg·m/s = (42,000 kg + masstanker)·(6 m/s)

Divide both sides by 6 m/s

70,000 kg = 42,000 kg + masstanker

Subtract 40,000 kg from both sides

70,000 kg – 40,000 kg = masstanker
30,000 kg = masstanker


The mass of the second car is equal to 30,000 kg.

Remember, the momentum of a system is conserved. The momentum of the individual masses may change, but the net momentum of the system does not change.

Equilibrium Example Problem – Balance

Consider a string with a weight tied to the end. Gravity pulls the string down while the string pulls the weight. These two forces are equal and opposite to each other. Together, they cancel each other out. The weight is “at rest” with no net forces acting on it. This state is known as mechanical equilibrium.

Mathematically, this is expressed as

ΣF = 0

or the sum of all forces acting on a mass is equal to zero. For the weight on a string, the force up is the tension of the string and the force down is the force of gravity pulling the weight. Together, they add up to zero.

Example Problem

A very large cat sits at the center of a swing. Two scales are attached to the ropes.

Equilibrium Cat 1One scale reads 5 kg.
A. What does the second scale read?
B. What is the combined weight of the cat and swing?


The cat in this situation is in mechanical equilibrium. The force of gravity pulls down on the cat and swing. The force up to counteract this pull must equal the force down.

Since the cat is sitting in the center of the swing, the force up is equally distributed between both ropes. Since one scale reads 5 kg, the other rope should carry the same amount of weight. The second scale will read 5 kg.

The weight of the cat and swing will be the sum of these two up forces.

5 kg (right scale) + 5 kg (left scale)
10 kg

The cat and swing weighs 10 kg.

Let’s look at what happens if the cat scoots to the left a bit.

Equilibrium Cat 2Now the left hand scale reads 7 kg. What will the right hand scale read?

The cat and swing did not change their weight. The weight was just moved around a bit. The cat and swing still pulls down with a weight of 10 kg. Since they are in equilibrium, the forces up will equal the forces down.

10 kg = 7 kg (left scale) + ? kg (right scale)

10 kg – 7 kg = ? kg (right scale)

3 kg = right scale reading.

It is important to remember when a system in mechanical equilibrium the total forces acting on the system are equal to zero.

Coulomb Force Example Problem

Coulomb force is the force of either attraction or repulsion between two charged bodies. The force is related to the magnitude and charge on the two bodies and the distance between them by Coulomb’s Law:

q1 and q2 is the amount of charge in Coulombs
r is the distance in meters between the charges
k is the Coulomb’s Law constant = 8.99×109 N•m2/C2

The direction of the force depends on the positive or negative charges on the bodies. If the two charges are identical, the force is a repulsive force. If one is positive and the other negative, the force is an attractive force.

This Coulomb force example problem shows how to use this equation to find the number of electrons transferred between two bodies to generate a set amount of force over a short distance.

Example Problem:
Two neutrally charged bodies are separated by 1 cm. Electrons are removed from one body and placed on the second body until a force of 1×10-6 N is generated between them. How many electrons were transferred between the bodies?


First, draw a diagram of the problem.

Coulomb Force Example Problem 2

Define the variables:
F = coulomb force = 1×10-6 N
q1 = charge on first body
q2 = charge on second body
e = charge of a single electron = 1.60×10-19 C
k = 8.99×109 N•m2/C2
r = distance between two bodies = 1 cm = 0.01 m

Start with the Coulomb’s Law equation.

As an electron is transferred from body 1 to body 2, body 1 becomes positive and body two becomes negative by the charge of one electron. Once the final desired force is reached, n electrons have been transferred.

q1 = +ne
q2 = -ne

The signs of the charges give the direction of the force, we are more interested in the magnitude of the force. The magnitude of the charges are identical, so we can ignore the negative sign on q2. This simplifies the above equation to:

We want the number of electrons, so solve the equation for n.

Substitute in the known values. Remember to convert 1 cm to 0.01 m to keep the units consistent.

n = 6.59×108

6.59×108 electrons were transferred between the two bodies to produce an attractive force of 1×10-6 Newtons.

For another Coulomb force example problem, check out Coulomb’s Law Example Problem.