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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 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.

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.

Problem:

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

Answer

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.

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:


where
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?

Solution:

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

Answer:
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.

Coulomb’s Law Example Problem

Coulomb’s Law is a force law between charged bodies. It relates the force to the magnitude and charge on the two bodies and the distance between them by the relationship:


where
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’s Law example problem shows how to use this equation to find the charges necessary to produce a known repulsive force over a set distance.

Example Problem:
The force between two identical charges separated by 1 cm is equal to 90 N. What is the magnitude of the two charges?

Solution:
First, draw a force diagram of the problem.

Setup diagram of Coulomb's Law Example Problem.

Two charges separated by one centimeter experiencing a force of repulsion of 90 N.

Define the variables:
F = 90 N
q1 = charge of first body
q2 = charge of second body
r = 1 cm

Use the Coulomb’s Law equation

The problem says the two charges are identical, so

q1 = q2 = q

Substitute this into the equation

Since we want the charges, solve for q

Enter the values for the variables. Remember to convert 1 cm to 0.01 meters to keep the units consistent.

q = ±1.00×10-6 Coulombs

Since the charges are identical, they are either both positive or both negative. This force will be repulsive.

Answer:
Two identical charges of ±1.00×10-6 Coulombs separated by 1 cm produce a repulsive force of 90 N.

For another Coulomb’s law example problem, check out Coulomb Force Example Problem.

Accelerometer – Inertia Example Problem

An accelerometer is a device to measure acceleration. One of the simplest accelerometers is a small mass hanging from a thin rod or string that can pivot freely as a body accelerates. As the body containing the accelerometer accelerates one direction, the freely hanging weight will swing in the opposite direction. How far it swings is an indication of how much the body accelerates. This example problem shows how to use an accelerometer to determine the force of gravity.

Example Problem:
A rocket ship flies over the surface of a planet. There hangs a mass suspended by a wire to act as a simple accelerometer attached to the underside of the rocket. As the rocket accelerates at 10 m/s2, the accelerometer mass is deflected by an angle of 34°. What is the acceleration due to gravity (g) of this planet?

Solution: Here is an illustration of the problem. AccelerometerThe ship is accelerating to the right at a constant acceleration of 10 m/s². The mass is pulled down by the force of gravity, but held up by the tension in the wire. As the ship accelerates, the mass wants to stay in place but the tension in the wire pulls it along with the ship.

Here are the forces acting on the mass.

accelerometer forcesFirst, let’s find the forces acting in the x-direction.

ΣFx = Tsinθ

The acceleration is acting in the positive x-direction, so

ΣFx = ma.

Set these two to equal each other.

Tsinθ = ma

Now for the vertical forces.

ΣFy = Tcosθ – mg

Since the mass is not accelerating in the vertical direction, the sum of the forces vertically is equal to zero.

Tcosθ – mg = 0
Tcosθ = mg

Now we have two equations and two unknowns.

Tsinθ = ma
Tcosθ = mg

Divide the two equations into each other.

accelerometer math step 1
tanθ = a/g

Since we want to know the gravity on the planet, solve for g.

accelerometer math step 2

Plug in a = 10 m/s2 and θ = 34°

accelerometer math step 3
g = 14.8 m/s2

Answer:
The acceleration due to gravity on this planet is 14.8 m/s2.

Bonus:
If you’d like to know how many times Earth’s gravity, divide your answer by Earth’s gravity.

(14.8 m/s2) ÷ (9.8 m/s2) = 1.5

The gravity on this planet is 1.5x the force of gravity on Earth.

Equations of Motion Example Problem

Motion in a straight line under constant acceleration is a common physics homework problem. The equations of motion to describe these conditions that can be used to solve any problem associated with them. These equations are:

(1) x = x0 + v0t + ½at2
(2) v = v0 + at
(3) v2 = v02 + 2a(x – x0)

where
x is the distance travelled
x0 is the initial starting point
v is the velocity
v0 is the initial velocity
a is the acceleration
t is the time

This example problem shows how to use these equations to calculate position, velocity and time of a constantly accelerating body.

Example:
A block slides along a frictionless surface with a constant acceleration of 2 m/s2. At time t = 0 s the block is at x = 5m and travelling with a velocity of 3 m/s.
a) Where is the block at t = 2 seconds?
b) What is the block’s velocity at 2 seconds?
c) Where is the block when it’s velocity is 10 m/s?
d) How long did it take to get to this point?

Solution:
Here is an illustration of the setup.Equations Of Motion

The variables we know are:
x0 = 5 m
v0 = 3 m/s
a = 2 m/s2

Part a) Where is the block at t = 2 seconds?
Equation 1 is the useful equation for this part.

x = x0 + v0t + ½at2

Substitute t = 2 seconds for t and the appropriate values of x0 and v0.

x = 5 m + (3 m/s)(2 s) + ½(2 m/s2)(2 s)2
x = 5 m + 6 m + 4 m
x = 15 m

The block is at the 15 meter mark at t = 2 seconds.

Part b) What is the block’s velocity at t = 2 seconds?
This time, Equation 2 is the useful equation.

v = v0 + at
v = (3 m/s) + (2 m/s2)(2 s)
v = 3 m/s + 4 m/s
v = 7 m/s

The block is travelling 7 m/s at t = 2 seconds.

Part c) Where is the block when it’s velocity is 10 m/s?
Equation 3 is the most useful at this time.

v2 = v02 + 2a(x – x0)
(10 m/s)2 = (3 m/s)2 + 2(2 m/s2)(x – 5 m)
100 m2/s2 = 9 m2/s2 + 4 m/s2(x – 5 m)
91 m2/s2 = 4 m/s2(x – 5 m)
22.75 m = x – 5 m
27.75 m = x

The block is at the 27.75 m mark.

Part d) How long did it take to get to this point?
There are two ways you could do this. You could use Equation 1 and solve for t using the value you calculated in part c of the problem, or you could use equation 2 and solve for t. Equation 2 is easier.

v = v0 + at
10 m/s = 3 m/s + (2 m/s2)t
7 m/s = (2 m/s2)t
72 s = t

It takes 72 s or 3.5 s to get to the 27.75 m mark.

One tricky part of this type of problem is you have to pay attention to what the question is asking for. In this case, you were not asked how far the block travelled, but where it is. The reference point is 5 meters from the origin point. If you needed to know how far the block travelled, you would have to subtract the 5 meters.

For further help, try these Equations of Motion example problems:
Equations of Motion – Interception Example
Equations of Motion – Vertical Motion
Equations of Motion – Breaking Vehicle
Equations of Motion – Projectile Motion