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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.

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.

How To Find the Limiting Reactant – Limiting Reactant Example

Ammonia Ball and Stick Model

3D ball and stick model of the ammonia molecule. Todd Helmenstine

Many chemical reactions take place until one of the reactants run out. This reactant is known as the limiting reactant. Often it is straightforward to determine which reactant will be the limiting reactant, but sometimes it takes a few extra steps.

For example, burning propane in a grill. The propane and oxygen in the air combust to create heat and carbon dioxide. You are obviously more likely to run out of propane long before you run out of oxygen in the air. This makes the propane the limiting reactant. Other reactions aren’t quite as easy.

This example problem will show how to use the stoichiometric ratios between the reactants given in the balanced chemical equation to determine the limiting reactant.

Find the Limiting Reactant Example

Question: Ammonia (NH3) is produced when nitrogen gas (N2) is combined with hydrogen gas (H2) by the reaction

N2 + 3 H2 → 2 NH3

50 grams of nitrogen gas and 10 grams of hydrogen gas are reacted together to form ammonia. Which of the two gasses will run out first? (Which gas is the limiting reactant?)

Answer:  The reaction shows us for every mole of N2 consumed, 3 moles of H2 is also consumed. We need 3 moles of hydrogen gas for every mole of nitrogen gas. The first thing we need to find out is the number of moles of each gas is on hand.

N2 Gas: How many moles of nitrogen gas is 50 grams? One mole of nitrogen is 14.007 grams, so one mole of N2 will weigh 28.014 grams.

Limiting Reactant Example Step 1

Limiting Reactant Example Step 2

x moles N2 = 1.78

H2 Gas: How many moles of hydrogen gas is 10 grams? One mole of hydrogen is 1.008 grams so one mole of H2 is 2.016 grams.

Limiting Reactant step 3
Limiting Reactant Example Step 4
x moles H2 = 4.96

Now we know the number moles of each reactant, we can use the ratio from the chemical equation to compare the amounts.  The ratio between hydrogen gas and nitrogen gas should be:

Limiting Reactant Example Step 5

If we divide our moles of H2 into moles of N2, our value will tell us which reactant will come up short. Any value greater than the above ratio means the top reactant is in excess to the lower number. A value less than the ratio means the top reactant is the limiting reactant. The key is to keep the same reactant on top as the step above.

Limiting Reactant Example 6
2.79

Since our value is less than the ideal ratio, the top reactant is the limiting reactant. In our case, the top reactant is the hydrogen.

Answer: Hydrogen gas is the limiting reactant.

It doesn’t matter which reactant you put on top when you do this type of problem as long as you keep it the same throughout the calculations. If we had put nitrogen gas on top instead of hydrogen the ratio would have worked out the same way. The ideal ratio would have been 13 and the calculated ratio would have been 0.358 ( 1.78/4.96 ). The value would have been greater than the ideal ratio so the bottom reactant in the ratio would be the limiting reactant. In this case, it is the hydrogen gas.

 

Calculate Theoretical Yield of a Chemical Reaction – Theoretical Yield Example Problem

Gas Collection Woodcut

Source: Elements of Modern Chemistry, Charles Adolphe Wurtz, 1887.

The theoretical yield of a chemical reaction is the amount of product you expect to get after a chemical reaction takes place from the reagents you have available. To calculate the theoretical yield of a reaction, you must first know the reaction. Let’s look at the following reaction where heating potassium chlorate (KClO3) produces oxygen gas (O2).

2 KClO3 (s) → 3 O2 (g) + 2 KCl (s)

where KCl is potassium chloride. This reaction is fairly common in school laboratories since it is a relatively inexpensive method of obtaining oxygen gas.

The balanced reaction shows that 2 moles of KClO3 will produce 3 moles of O2 and 2 moles of KCl. To calculate the theoretical yield, you use these ratios as a conversion factor.  Here is a typical example problem.

Calculate Theoretical Yield Example Problem

Question: How many moles of oxygen gas will be produced from heating 735.3 grams of KClO3?

Answer:

Step 1. We need to know the amount of KClO3 in moles to use the conversion, so the first step is to convert grams KClO3 to moles KClO3. To make this easier, know the molecular mass of KClO3 is 122.55 g/mol.

Theoretical Yield Example Step 1

Theoretical Yield Example Step 2

Theoretical Yield Example Step 3

6 = x moles KClO3

Step 2: Use the chemical equation to relate moles KClO3 to moles O2.

From above, we see 2 moles of KClO3 will produce 3 moles of O2 gas.

Theoretical Yield Example Problem Step 4

Theoretical Yield Example Problem Step 5

Theoretical Yield Example Problem Step 6

x moles O2 = 3 x 3 moles O2
x moles O2 = 9 moles O2

6 moles of KClO3 (735.3 grams of KClO3) will produce 9 moles of O2 gas.

Right Triangle Trigonometry and SOHCAHTOA

Sohcahtoa isn't actually an Egyptian god, but if it helps to remember him that way, you'll have an easier time recalling right angle trig relationships.

Sohcahtoa isn’t actually an Egyptian god, but if it helps to remember him that way, you’ll have an easier time recalling right angle trig relationships.

Right triangles are extremely common in science homework. Even though they are common, they can be confusing to new students. That is why we have the Egyptian god SOHCAHTOA.

SOHCAHTOA is a handy mnemonic trigonometry students learn to remember which sides of a triangle are used for the three main trig functions: sine, cosine, and tangent.

These functions are defined by the ratios of various lengths of the sides of a right triangle. Let’s look at this right triangle.

Right TriangleThis triangle is made up of three sides of lengths a, b and c. Note the angle marked θ. This angle is formed by the intersection of b and c. The hypotenuse is always the longest of the three sides and opposite of the right angle. The side b is ‘adjacent’ to the angle, so this side is known as the adjacent side. It follows the side ‘opposite’ of the angle is known as the opposite side. Now that we have all our sides labelled, we can use SOHCAHTOA.

SOHCAHTOA

S – Sine
O – Opposite
H – Hypotenuse

C – Cosine
A – Adjacent
H – Hypotenuse

T – Tangent
O – Opposite
A – Adjacent

SOH = sin θ = opposite over hypotenuse = ac
CAH = cos θ = adjacent over hypotenuse = bc
TOA = tan θ = opposite over adjacent = ab

Easy to remember. Now let’s see how easy it is to apply.

Example Problem

Consider this triangle.

trig example for SOHCAHTOAThe hypotenuse has a length of 10 and one angle of the triangle is 40º. Find the lengths of the other two sides.

Let’s start with the side with length a. This side is opposite the angle and we know the length of the hypotenuse. The part of SOHCAHTOA with both hypotenuse and opposite is SOH or sine.

sin 40º = opposite / hypotenuse
sin 40º = a / 10

solve for a by multiplying both sides by 10.

10 sin 40º = a

Punch 40 into your calculator and hit the sin key to find the sine of 40º.

sin 40º = 0.643

a = 10 sin 40º
a = 10 (0.643)
a = 6.43

Now let’s do side b. This side is adjacent to the angle, so we should use CAH or cosine.

cos 40º = adjacent / hypotenuse
cos 40º = b / 10

solve for b

b = 10 cos 40º

Enter 40 and hit the cos button on your calculator to find:

cos 40º = 0.766

b = 10 cos 40º
b = 10 (0.766)
b = 7.66

The sides of our triangle are 6.43 and 7.66. We can use the Pythagorean equation to check our answer.

a2 + b2 = c2
(6.43)2 + (7.66)2 = c2
41.35 + 58.68 = c2
100.03 = c2
10.00 = c

10 is the length of the triangle’s hypotenuse and matches our calculation above.

As you can see, our friend SOHCAHTOA can help us calculate the angles and lengths of the sides of right triangles with very little information. Make him your friend too.