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An Object is 1.0 cm tall and its inverted image is 5.0 cm tall. what is the exact magnification?
An Object is 1.0 cm tall and its inverted image is 5.0 cm tall. what is the exact magnification?
Video Solution to “An Object is 1.0 cm tall and its inverted image is 5.0 cm tall. what is the exact magnification?”
video solution
Text Solution to “An Object is 1.0 cm tall and its inverted image is 5.0 cm tall. what is the exact magnification?”
Data
To find the exact magnification of the object, we will need to extract our data from the question
Height of an Image that is not inverted will be, v = 5.0 cm
But when a height of an image is inverted, the value will become negative.
Therefore, the height of the Image, v = – 5.0 cm
Object’s height, u = 1.0 cm
Unknown
Magnification, M = ?
Formula
To apply the right formula, we need to remember that magnification is the ratio of the height of the image to the height of the object.
Therefore, we can write the above definition in mathematical form as
Magnification, M = Height of the image (v) / Height of the object (u)
Therefore, the formula that can help us solve the problem is
M = v / u
Solution
We will now put our data into the above formula
Magnification, M = v / u = – 5 / 1 = – 5
Therefore, the exact magnification of the image is negative five (-5).
However, when the magnification is negative, it shows that the mirror is inverted and real. Thus, the lens of the mirror can be classified as a concave mirror.
More Problems on Magnification
Here are more examples to give you a clear understanding of how to solve problems on magnification;
Problem 1
An object 5 centimeters high is placed at a distance of 12 centimeters from a convex lens of a focal length of 8 centimeters. Calculate the position, size, and nature of the image.
Solution
Real is positive
New cartesian
f = +8 cm
u = +12 cm
Using (1/v) + (1/u)=(1/f)
(1/v)=(1/f)-(1/u)
(1/v)=(1/8)-(1/12)=1/24
v=24cm (real image)
M=(v/u)=(24/12)=2
Also, M=(h/5)
2=h/5
h=10
f = +8 cm
u = -12 cm (object left of lens)
Using (1/v)-(1/u)=(1/f)
(1/v)=(1/f)+(1/u)
(1/v)=(1/8)+(1/-12)=1/24
v = 24 cm (on the right real image)
Solution Table
From the solution in the table above, the image is inverted, magnified, and real 24 centimeters away from the lens which is facing the object. Additionally, the image’s height is 10 centimeters.
Problem 2
An object placed 5 centimeters in front of a converging lens of a focal length of 10 centimeters produces a magnified image 8 centimeters high. Find the size and position of the image.
Data
Focal length, f = 10 cm
Object’s height, u = 5 cm
magnified image height = 8 cm
Unknown
Height of the image, v = ?
Magnification, M = ?
height of the object, h = ?
Formula
First formula to find the height of the image, v = ?
(1/v) + (1/u) = (1/f)
Second formula to find the magnification
M = v / u
Third Formula to find the object’s height, h
M = magnified image height / h
Solution
We will apply our first formula (1/v) + (1/u) = (1/f) to find v
(1/v) = (1/10) – (1/5) = – 1 / 10
v = – 10 cm
Which shows that the image is virtual,
Hence, by applying the second formula, M = v / u = 10 / 5 = 2
The answer to this question (is velocity 0 at max height) is yes. The velocity of an object at maximum height is zero.
Explanation to “is velocity 0 at max height”
When you throw an object into the sky, for example, a basketball. There are multiple forces acting on the ball. Here is the list of forces that will be acting on the ball:
is velocity 0 at max height
1. Newton’s second law
According to newton’s second law of motion, “the rate of change of momentum is directly proportional to the applied force and takes place in the direction of the force.”. It shows that you will need to apply force to throw the ball in an upward direction.
Thus, as the ball continues to move up, it will reach a point where the applied force will be exhausted and the ball will start coming back to the ground.
Hence, the height covered before it starts coming back to the ground is the maximum height. And at that height, the body possesses zero acceleration which will lead to deceleration.
Therefore, as the acceleration keeps reducing, so is the velocity. And once the acceleration reaches its peak displacement, it will become zero. Hence, the velocity will also become zero.
Here is another explanation, momentum is the product of velocity and mass. And as the ball moves up, its velocity will keep reducing until it reaches zero.
Hence, the height at which the velocity reached zero is the ball’s maximum height. Therefore, at the maximum height, the velocity is zero.
2. Effect of Weight
Similarly, when you throw a ball upward, a force of gravity pulls it down due to its weight. This is because the earth is surrounded by a gravitational field.
Thus, acceleration due to gravity will continue to act on the ball until it cancels the external you applied on the ball. Hence, it will reach a threshold where it will hang and then start coming back to the surface of the earth.
The position where it hangs is the maximum height, and the velocity at that height is zero. Remember that weight is the product of mass and the force due to gravity. Thus, w = mg
3. Due to Decelaration
Additionally, here is the more accurate answer and simple explanation to “is velocity 0 at max height”. When you throw a ball up, it will gain momentum, and its acceleration will keep decreasing.
As the acceleration keeps reducing, the velocity of the ball will also be affected. The velocity will also continue to reduce along with the ball’s acceleration until it reaches zero (0).
Once the velocity of the ball reaches zero, this shows that the ball has reached its maximum height. It will then return down through the process of deceleration.
Additionally, when the ball reaches its maximum height it will be forced to come back to the ground due to acceleration due to gravity. When an object returns to the ground after reaching maximum height, it will undergo retardation which is negative acceleration.
And for an object to move from positive to negative acceleration, it must cross a zero mark. Therefore, the velocity at maximum height is zero because when an object is falling, its velocity will become negative
4. Kinetic Energy Vs Potential Energy
When you throw an object up, it will gain kinetic energy, its kinetic energy will keep decreasing until it reaches zero and then gets converted into potential energy.
The height at which the kinetic energy reaches zero is called maximum height. Don’t forget that kinetic energy consists of mass and velocity.
Therefore, when the kinetic energy is zero, the velocity is also zero. Hence, the velocity at maximum height is zero.
Displacement Scenario
Velocity is the rate of change of displacement with time. Let’s assume you throw a ball to cover a distance of 20 meters before it returns to the floor in 10 seconds. It means the maximum height is a distance covered in 20 meters.
Also, remember that displacement is the distance covered in a specific direction (vector quantity). Thus, our displacement is 20 meters. However, the ball returned to its initial position. Hence, the displacement covered while returning to its initial position is 20 meters. We can now show that:
This means that the total displacement is 20 m – 20 m = 0
But Velocity, v = total displacement / time = 0 / 10 = 0
Hence, velocity is zero (0)
Therefore, the answer to the question “is velocity 0 at max height” is yes.
We can easily define the gravitational field as a region of space in which every object with a mass experiences a pull due to gravitational force.
For example, when you throw a ball vertically upward, it will go up and then gradually start losing momentum due to deceleration. The ball will reach a point where it will no longer continue to move up again, and then start falling back to the ground. Ever wonder why that happens?
This is because as the ball is moving upward due to the force you applied by throwing it up, another force is also pulling the ball down and it’s called gravitational force. The space where the ball was thrown is called the gravitational field.
Another example is, ever wonder why a mango fruit falls to the ground instead of moving to space? The answer to this question is also very simple. The gravitational force will always pull the mango fruit back to the ground.
What happens to you when you jump up? You will still come back to the ground.
When you drop two objects, say a pencil and a big stone at the same time from the same height. What did you observe? Both the pencil and the big stone will fall on the ground at the same time.
The examples I have mentioned here were thanks to the effect of gravitational force in a gravitational field.
Important Notes on Gravitational Field
Did you know that the Earth exerts a gravitational force on the moon, even though they are 385,000 kilometres away from each other? Yes, the Earth pulls the moon toward its centre. The space between the earth and the moon is known as the gravitational field.
Did you know that the sun pulls the earth due to gravity even though the distance between them is 1.5 x 108 kilometres? Yes, it does pull the earth due to the force of gravity.
Law:Newton’s law of universal gravitation states that the force of attraction (F) between two masses (M) and (m) is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. This law is also known as inverse law.
Gravitational field
F ∝ Mm / r2
Which will become
F = (GMm) / r2
Where
F = Force of attraction between the two masses M and m
M and m are the masses of the two objects
G = Gravitational constant = 6.7 x 10-11 Nm2kg-2 [The unit of gravitational constant is Newton meter square per kilogram square (Nm2kg-2) or meter cube per kilogram per second square (m3kg-1s-2) ]
r = distance between two masses
According to newton’s third law of motion, actions and reactions are equal and opposite. Therefore, mass M attracts m by an equal but opposite force.
A gravitational field is everywhere. It surrounds bodies with mass, and it fills up space. The gravitational field makes it possible for the moon to orbit around the earth and the earth to orbit around the sun.
After throwing a ball, it will return to the earth. This shows that there is a force of attraction between the mass of the ball and that of the earth.
The force of attraction between the two masses may be too weak to notice unless one of the masses is very large. Therefore, the mass of an object creates a force field that is attracted toward another object. The gravitational force on earth is equal to the force of attraction between you and the earth.
More Explanation on Gravitational Force
When a body is at rest, on or near the surface of the earth, the gravitational force is equal to the weight of that body. The gravitational acceleration of the moon and Venus is different from that of earth. Thus, if you are to measure your weight on these three different planets, you will notice different weights.
When two bodies are gravitationally locked, their gravitational force is centered in an area that is not at the center of either body, but at the barycenter of the system. For example, when two people sit at both ends of a seesaw, the lighter person will be pushed up while the heavier person will go down.
To balance the heavier and the lighter person on the same seesaw. The heavier person must sit closer to the center of the mass of the seesaw (fulcrum) just so that it can balance. Barycenter is the balance point between the earth and the moon. The earth and the moon move around the barycenter in their orbits.
The knowledge of the gravitational field has given us the chance to send people to other planets. It has also helped us to keep satellites in any orbit of our choice.
Weight of an Object
When an orange slips from your hand, it will drop on the floor, this is due to the weight of the orange pulling it to the floor. The weight of a body is the product of its mass and gravitational acceleration of the body.
Weight (w) = mass (m) x gravitational acceleration (g)
w = mg
In the 17th century, newton published Newton’s law of universal gravitation together with his three laws of motion.
He extensively studied the works of Galilei Galileo and Kepler on the motion of planets and came up with his contributions.
Relationship Between Gravitational Constant (G) and Gravitational Acceleration (g)
Some students find it difficult to understand the difference between the gravitational constant (G) and the gravitational acceleration (g).
The value of the gravitational constant (G) is 6.7 x 10-11 Nm2kg-2. It’s the relationship between two masses and the distance between them.
While gravitational acceleration is due to a pull on a body because of the force of gravity. The value of gravitational acceleration is 9.81 ms-2.
Now, to understand the relationship between G and g when an object is on the surface of the earth. We will equate the force of attraction between the two masses with weight.
w = F
This implies
mg = (GMm) / r2
and m will cancel each other from both sides to give us
g = GM / r2
Where M = mass of the earth
Therefore, we can say that the mass of the earth is
M = gr2 / G
Therefore, we can describe acceleration due to gravity (g) as force per unit mass on the surface of the earth.
Additionally, the equation g = GM / r2 shows that we can determine the acceleration due to gravity on the surface of the earth from the mass of the earth and its radius.
Therefore, the gravitational acceleration on the top of a mountain is less when we compare it with the gravitational acceleration at sea level.
This is because the radius of the earth from its center is a bit greater at the top of a mountain than at sea level.
The velocity of a Mass, m Moving Around the Earth’s Surface
When we have an object of mass m moving around the earth with a constant speed (v). We can obtain v by equating centripetal force [ (mv2) / r ] with the gravitational force [ (GMm) / r2 ] that keeps it moving in a circular direction.
Thus,
(mv2) / r = (GMm) / r2
and by making v subject of the formula, we will have
v = √(GM / r)
By multiplying √(GM / r) with (r / r). We will get
v= √(GM / r) x (r / r)
And the above equation will become
v = √(GMr / r2) = √(GM / r2) x r
But g = GM / r2
Hence
v = √gr
Gravitational Field Intensity
The gravitational field intensity (I) is defined as the force experienced per kilogram of mass at any point.
Gravitational field intensity, I = Froce (f) / mass (m) = g
Gravitational Potential
Definition: The gravitational potential (V) at a point on the surface of the earth, is the work done in moving a unit mass of a body from infinity to that point. The unit of gravitational potential is in joule per kilogram (jkg-1).
The mathematical expression of the above definition is
Gravitational potential of point mass, V = – (Gm) / r
From the above equation:
m is the mass of the body causing the gravitational field.
r is the distance between the point and the mass.
Note:
As the body changes its direction away from the earth. The radius r will increase which will cause the gravitational potential to reduce. It will get to a point where the gravitational potential will become zero when the radius r is infinitely large.
It is also important to know that the gravitational force will perform a negative work if the body is moving away from the earth.
At the same time, if the body returns to the earth, the radius r will reduce.
When you apply force to raise an object of mass m through a height h, the work done in raising the object is
Work done (W) = Force (mg) x distance (h)
which implies that
W = mgh [where g is the acceleration due to gravity]
Potential Energy of a Mass on the Earth’s Surface
If a mass, m is on the surface of the earth of radius, r. The potential energy possessed by the mass, V = mgh
and since g = GM / r2
and h = r [where r = radius of the earth]
Therefore,
gh = GM / r
Now, multiply both sides by m to obtain
mgh = GMm / r
Which implies that
The potential energy on the earth surface, V = – (GMm) / r
Kepler’s Laws
Johannes Kepler is a german astronomer and mathematician. He laid out the foundation of Newton’s universal law of gravitation. Kepler came up with three laws to describe the motion of the planets in orbit:
Each planet travels along an ellipse with the sun at one focus
The line joining the sun and a given planet sweeps equal areas at equal times.
The square of the period of revolution of the planets is proportional to the cubes of their mean distances from the sun ( T2 ∝ r3).
Solar System
The concept of the solar system simply refers to the sun and other objects that circle it due to the force of gravity. These objects that orbit the sun are called planets.
Gravitational Field: Planetary Motion
The currently known planets around the sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto.
We have two categories of planets, terrestrial planets, and gas giants.
Terrestrial planets are made up of rocks and metals. They also look similar to the earth. They are closer to the sun and their size is smaller. Terrestrial planets include Mercury, Venus, Earth, and Mars.
Gas giants are bigger than terrestrial planets and are made up of gases. Jupiter and Saturn are composed mainly of hydrogen and helium.
The solar system also contains objects like Asteroids, Comets, Meteoroids, and others.
Points to note on the planets:
They revolve around the sun in their orbits
When a planet is close to the sun, its revolution is faster.
The planets revolve in a similar direction
Apart from pluto, the orbits of the planets lie in the same plane.
Satellites
An object that moves around the earth or in space is called a satellite. We have two types of satellites, Artificial and Natural satellites.
An artificial satellite is an object created and launched into space by man. Creating such satellites aims to improve communication, supervision, and security, and to observe the distant universe.
Natural Satellite is synonymous with the moon and is a celestial body that orbits a planet or smaller bodies. They are used to identify non-artificial satellites.
A parking orbit is an orbit in which the period of a satellite in its orbit is precisely equal to the period of the earth as it rotates about its axis.
How to Find the Period and Velocity of a Satellite in Its Orbit
If a satellite of mass m circles the earth at an orbit of radius R, from the center of the earth and in the same direction as the rotation of the earth. Then, the centripetal force that sustains it in that orbit will equal the gravitational force. Hence
(mv2) / R = (GMm) / R2
Remember that Gm = gr2 [where r = radius of the earth]
We can now rewrite mv2 / R = (GMm) / R2 into
mv2 / R = (mgr) / R2
and by making v subject of the formula, we will have
v = √ (gr2 / R)
Now, to determine the period T, of a satellite moving around an orbit, we will say
Velocity (v) = (Displacement of the satellite around an orbit) / Time taken
v = 2πR / T
Which implies that
Period, T = 2πR / v
Parking orbit helps in finding the location of a communication satellite so that the transmission of events will continue to flow across the globe.
Escape Velocity
Definition: Escape velocity is the minimum velocity required by an object to permanently escape the influence of the gravitational field.
To derive the formula for escape velocity, we will consider the work done in carrying the mass of a body away from the center of the earth through a distance r. We will also put into consideration that gravitation is weak.
Hence
Work done (W) = Force (F) x Distance (r)
Which implies that
W = F x r
and F = (GMm) / r2
Therefore, W = ((GMm) / r2) x r = (GMm) / r
Additionally, the work done (W = (GMm) / r) is equal to the kinetic energy (K = (1/2)mv2)
Thus, (GMm) / r = (1/2)mv2
And by making v subject of the formula, we will obtain
v = √(2GM / r)
And if the object is launched from the surface of the earth. r = R
Therefore, our equation will become
v = √(2GM / R)
But M = gr2 /G
Hence, v will become
v = √(2GM / R) x (gr2 /G)
to obtain v = √2gR which is the formula for calculating escape velocity
Gravitational Field Practice Problems
Here are some practice problems to help you understand how to calculate problems involving the gravitational field:
Problem 1
Two 5 kilograms of spherical balls are placed so that their centers are 50 centimeters apart. What is the magnitude of the gravitational force between the two balls? Take G = 6.67 x 10-11 Nm2kg-2.
Data
Both masses are the same = M = m = 5 kg
Distance between the masses, r = 50 cm = 50 / 100 = 0.5 m
Gravitational constant, G = 6.67 x 10-11 Nm2kg-2
Formula
Applying the formula F = (GMm) / r2
Solution
We can now insert our data into the formula F = (GMm) / r2
F = (6.67 x 10-11 x 5 x 5) / 0.52
Therefore, F = 6.67 x 10-9 N
Problem 2
Determine the force of attraction between the sun (ms = 1.99 x 1030 kg) and the earth (me = 5.98 x 1024 kg). Assume the sun is 1.5 x 108 kilometers from the earth.
Data
Mass of the sun, ms = 1.99 x 1030 kg
Mass of the earth, me = 5.98 x 1024 kg
Distance between the sun and the earth, r = 1.5 x 108 km
Gravitational constant, G = 6.67 x 10-11 Nm2kg-2
F = ?
Formula
We will use F = (Gmsme) / r2 to solve the problem
Solution
Insert the data into F = (Gmsme) / r2 to obtain
F = (6.67 x 10-11 x 1.99 x 1030 x 5.98 x 1024) / (1.5 x 108)2
Therefore, the force of attraction between the sun and the earth is F= 3.53 x 10-22
Problem 3
Determine the mass of the earth if the radius of the earth is approximately 6.38 x 106 m, G = 6.67 x 10-11 Nm2kg-2 and g = 9.8 m/s2
Data
Mass of the earth, me = M = ?
Radius of the earth, re = 6.38 x 106 m
Gravitational constant, G = 6.67 x 10-11 Nm2kg-2
Acceleration due to gravity, g = 9.8 m/s2
Formula
M = gr2 / G
Solution
Let us substitute our formula with our data
M = gr2 / G = (9.8 x (6.38 x 106)2) / 6.67 x 10-11 = 5.98 x 1024 kg
The mass of the earth is 5.98 x 1024 kilograms
Problem 4
Find the gravitational potential at a point on the earth’s surface. Take a mass of the earth as 5.98 x 1024 kilograms, its radius as 6.38 x 106 meters and G = 6.67 x 10-11 Nm2kg-2
Data
Mass of the earth, M = 5.98 x 1024 km
Radius of the earth, r = 6.38 x 106 m
G = 6.67 x 10-11 Nm2kg-2
Unknown
Gravitational Potential, V = ?
Formula
V = GM / r
Solution
We will now apply our data to the formula
V = GM / r = (6.67 x 10-11 x 5.98 x 1024) / 6.38 x 106 = 6.25 x 107 Jkg-1
Therefore, the gravitational potential is 6.25 x 107 Jkg-1
a. To find the time of the flight, we will apply the following methods:
Unknown
Time of flight, T = ?
Formula
The formula for solving a problem on time of flight, T is (2usinθ) / g
Solution
We will now insert our data into the above formula
Since T = (2usinθ) / g
T = (2 x 65 x sin600) / 10 = (130 x 0.866) / 10 = 112.58 / 10 = 11.258 = 11.3 seconds
Therefore, the time of flight is 11.3 seconds
b. Here is how to calculate the maximum height
Unknown
Maximum Height, Hmax = ?
Formula
The formula for calculating the maximum height is
Maximum height, Hmax = (u2sin2θ) / 2g
Solution
We can now substitute our formula with the data
The maximum height, Hmax = (652sin2600) / 2 x 10
Therefore,
Hmax = (4225 x (sin600)2) / 2 x 10
And the above equation will give us
Hmax = (4225 x 0.75) / 20
Hence
Hmax = 3168.75 / 20 = 158.4375 m
Our maximum height can be approximated into
Hmax = 158 m
Therefore, our maximum height is 158 meters.
c. We will find the range
Unknown
The range, R = ?
Formula
Here is the formula that will help us find the range
The range, R = (u2sin2θ) / g
Solution
The range, R = (652sin2(600) / g
And the above expression will give us
R = (4225sin1200) / 10 = (4225 x 0.866) / 10 = 3658.85 / 10 = 365.885 m
Therefore, the range of the stone shot out of the catapult is 365.885 meters.
Problem 2
A bullet is fired horizontally with a velocity of 40 meters per second from the top of a building 50 meters high. How far from the feet of the building will the bullet be assumed to touch the ground? (Take the gravitational acceleration, g = 10 m/s2).
Problems of Projectile Motion
Data
Horizontal velocity = Initial velocity = ux = 40 m/s
Therefore, the bullet will cover 120 meters from the foot of the building to the ground.
Problem 3
A tennis ball is thrown vertically upwards from the front with a velocity of 50 m/s. Calculate
a. The maximum height reached
b. The time it takes to reach the maximum height
c. The time of the flight
Data
Initial velocity, u = 50 m/s
a. Here is how to find the maximum height of the tennis ball
Formula
The formula for finding the maximum height is
Hmax = u2 / 2g
Solution
After inserting our data into the formula Hmax = u2 / 2g, we will have
Maximum height, Hmax = (50)2 / 2 x 10 = 2,500 / 20 = 125 m
Therefore, the maximum height reached by the tennis ball is 125 meters
b. The time taken to reach the maximum height can be calculated this way:
Unknown
The time it takes to reach the maximum height, t = ?
Formula
The formula to apply is t = u / g
Solution
We will apply our data to the formula t = u / g
Time, t = 50 / 10 = 5 seconds
c. Time of flight will be
T = 2u / g = (2 x 50) / 10 = 100 / 10 = 10 seconds.
Problem 4
A bullet is fired from a point making an angle of 36 degrees to the horizontal. The initial velocity of the bullet is 43 m/s. Find
a. The greatest height reached
b. The time taken to reach the maximum height
[Take gravitational acceleration, g = 9.8 m/s2]
Data
The angle made, θ = 360
Initial velocity, u = 43 m/s
Gravitational acceleration, g = 9.8 m/s2
a. We can calculate the maximum height, Hmax this way
Unknown
The maximum height, Hmax = ?
Formula
Maximum height, Hmax = (u2sin2θ) / 2g
Solution
Apply the data into Hmax = (u2sin2θ) / 2g to obtain
Hmax = (432sin2360) / 2 x 9.8
Thus
The maximum height, Hmax = (1849 x 0.34549) / 19.6 = 638.8 / 19.6 = 32.59 = 33 m
Therefore, the maximum height reached is 33 meters.
b. The time of flight will be
Time of flight, T = usinθ / g = 43sin360 / 9.8 = (43 x 0.59) / 9.8 = 25.37 / 9.8 = 2.589 s
Therefore, the time of flight is 2.6 seconds
Problem 5
A projectile is fired with an initial velocity of 100 m/s at an angle of 30 degrees with the horizontal. Calculate:
a. The time of flight
b. The maximum height attained
c. The range
Problems of Projectile Motion
Data
The initial velocity of the projectile, u = 100 m/s
Angle of inclination, θ = 300
Gravitational acceleration, g = 10 m/s2
Solution a
To find the time of flight, we will follow the method below:
Formula
we will apply
Time of flight, T = (2usinθ) / g
We will now put our data into the formula
T = (2 x 100 x sin300) / g = (200 x 0.5) / 10 = 100 /10 = 10 s
Therefore, the time of flight is10 seconds
Solution b
To find the maximum height attained, we will use
Formula
Maximum height, Hmax = (u2sin2θ) / 2g
We can now substitute our formula with our data
Hmax = (1002 x sin2300) / (2 x 10)
Hmax = (10,000 x (0.5)2) / 20 = (10,000 x 0.25) / 20 = 2,500 / 20 = 125 m
Therefore, the maximum height attained by the projectile is 125 meters
Solution c
Now, we are going to calculate the range
Formula
The range, R = (u2sin2θ) / g
After inserting our data into the above formula, we will get
R = (1002sin2 x 300) / 10 = (10,000 x sin600) / 10 = (10,000 x 0.866) / 10
And we will have
R = 8660 / 10 = 866 m
Therefore, our range is eight hundred and sixty-six (866) meters.
Problem 6
A bomber on a military mission is flying horizontally at a height of 3000 meters above the ground at 60 kilometers per minute. It drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released.
Problems of Projectile Motion
Data:
The horizontal velocity of the bomber aircraft = 60 km/min = (60 x 1000) / 60 = 60,000 / 60 = 1000 m/s
Initial velocity, u = 0
Gravitational acceleration, g = 10 m/s2
Height above the ground, h = 3000 m
Formula
The first formula is
h = ut + (1/2)gt2 to find t
The second formula
Distance covered in time t, s = horizontal velocity x time to find s
The third formula is
tanθ = s / h to find the angle of inclination (θ)
Solution
We will need to insert our data into the first formula to find t
h = ut + (1/2)gt2
Which can be rewritten as
3000 = (1/2) x 10 x t2 [Remember that u = 0]
By making t subject of the formula, we will have
(3000 x 2) / 10 = t2
t2 = 600
We can now take the square root of both sides to obtain
t = √600
And our final answer for the time will be
Time, t = 24.5 seconds
We will apply our second formula
Distance, s = horizontal velocity x time = 1000 x 24.5 = 24,500 meters
We will now proceed with our third formula
tanθ = s / h = 24,500 / 3000 = 8.2
Thus,
tanθ = 8.2
Hence, we divide both sides by tan to obtain a tan inverse of 8.2
θ = tan-18.2 = 83.0470
We can now approximate our answer into
θ = 830
Therefore, the angle of inclination is eighty-three (83) degrees.
Problem 7
A ball is projected horizontally from the top of a hill with a velocity of 30 meters per second. If it reaches the ground 5 seconds later. The height of the ball is
Data:
The initial velocity of the ball, u = 0
Time it takes to reach the ground, t = 5 s
Gravitational acceleration, g = 10 m/s2
Formula
Maximum height of the ball, h = ut + (1/2)gt2
Solution
To solve the problem, insert your data into the above formula
h = ut + (1/2)gt2 = 0 x 5 + (1/2) x 10 x 52 = (1/2) x 10 x 25 = (1/2) x 250 = 250 / 2 = 125 meter
Therefore, the height of the hill is 125 meters
Problem 8
A stone is projected at an angle of 60 degrees and an initial velocity of 20 meters per second. Determine the time of flight.
Data:
The angle of inclination of the stone, θ = 600
Initial velocity of the stone, u = 20 m/s
Gravitational acceleration, g = 10 ms-2
Unknown
Time of flight, T = ?
Formula
The formula for calculating the time of flight, T = (2usinθ) / g
Solution
We will put our data into the formula T = (2usinθ) / g
Now,
T = (2usinθ) / g = T = (2 x 20 x sin600) / 10 = (40 x 0.866) / 10 = 34.64 / 10 = 3.464 s
Therefore, the time of flight is 3.46 seconds.
Problem 9
A body is projected upward at an angle of 30 degrees with the horizontal at an initial speed of 200 meters per second. In how many seconds will it reach the ground? How far from the point of projection will it strike?
Data
Angle of projection, θ = 200 m/s
Horizontal initial speed, u = 200 m/s
Unknown
The question wants us to solve
Number of seconds for the body to reach ground = Time of flight, T = ?
We were also asked to find out how far from the point of projection will the body strike the ground. Hence, we are going to find the range.
Formula
First formula to find the time of flight, T = (2usinθ) / g
Second formula to find the range, R = (u2sin2θ) / g
Solution
To find the time of flight, we will insert our data into the first formula T = (2usinθ) / g
Time of flight, T = (2usinθ) / g = (2 x 200 x sin300) / 10
Hence
T = (400 x 0.5) / 10 = 200 / 10 = 20 s
Therefore, the time of flight is 20 seconds.
By applying the second formula R = (u2sin2θ) / g and inserting our data, we will have
R = (u2sin2θ) / g = (2002 x sin (2×300)) / 10
Which implies that
R = (40,000 x sin600) / 10 = (40,000 x 0.866) / 10
Thus,
The range, R = 34,640 / 10
Hence
R = 3464 m
Therefore, the range covered is 3,464 meters
Problem 10
A tennis ball projected at an angle θ attains a range of R = 78 meters. If the velocity imparted to the ball by the racket is 30 meters per second, calculate the angle θ.
Data
Range, R = 78 m
The velocity imparted to the ball by racket = initial velocity, u = 30 m/s
Gravitational acceleration, g = 10 m/s2
Unknown
The angle θ = ?
Formula
We will use the formula R = (u2sin2θ) / g
Solution
Substitute your formula with the data
78 = (302sin2θ) / 10
Hence
780 = (900sin2θ)
Sin2θ = 780 / 900
Thus,
Sin2θ = 0.867
After dividing both sides by sin, will now have
2θ = sin-10.867
Which is equal to
2θ = 60.110
After approximating 60.110 to 600. Therefore,
θ = 600 / 2 = 300
Therefore, the angle θ is 30 degrees
Problem 11
An aeroplane, flying in a straight line at a constant height of 500 meters with a speed of 200 meters per second drops a food package. The package takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 meters per second square, and ignoring air resistance, what are the values of t and d?
Data
The height, h = 500 m
Speed of the aeroplane = 200 m/s
Gravitational acceleration, g = 10 ms-2
Unknown
The time it takes the package to reach the ground, t = ?
Distance covered by the package to reach the ground, d = ?
Formula
First formula: We will apply t = √2h/g to find time (t)
Second Formula: We will use s = speed of the aeroplane x time it takes the package to drop on the ground [where s = distance covered by the package to reach the ground]
Solution
By applying our first formula t = √2h/g and inserting our data, we will obtain
While driving in your pickup truck down Highway 280 between San Francisco and Palo Alto, a meteorite lands in your truck bed! Despite its 220 kg mass, the meteor does not destroy your 1200 kg truck.
In fact, it landed so softly it added mass but did not change the total momentum of your truck.
Before the meteor landed you were going 25 m/s. After it landed, approximately how fast were you going? Answer in m/s.
While driving in your pickup truck down Highway 280
Answer
The final answer to this question for the final velocity is 21 m/s
Explanation
We need to start by extracting our data from the question
Data
Our Initial mass which is the mass of the truck is m1 = 1,200 kg
The second mass is the mass of the meteorite, m2 = 220 kg
Your initial velocity before the meteor landed on the truck is u1 = 25 m/s
The velocity after the meteor landed on the truck is u2 = 0
The final velocity is v1 = v2 = v
Formula
We will make v subject of the formula from the formula of momentum, which is
m1u1 + m2u2 = m1v1 + m2v2
But v1 = v2 = v
Thus,
m1u1 + m2u2 = (m1 + m2)v
At the same time, the velocity (u2) after the meteor landed on the truck is zero (0)
Hence,
m1u1 + m2 x 0 = (m1 + m2)v
Which is equal to
m1u1 = (m1 + m2)v
We can now divide both sides by (m1 + m2) to obtain
Therefore, the formula to apply so that we can find the velocity traveled so far of v = m1u1 / (m1 + m2)
Solution
We will now insert our data into the formula v = m1u1 / (m1 + m2) to obtain
v = m1u1 / (m1 + m2) = (1200 x 25) / (1200 + 220)
Thus,
v = 30,000 / 1420 = 21.126 m/s
Therefore, after the meteor landed, the speed covered by the truck is approximately 21 meters per second.
Therefore, the linear momentum of a truck of mass 5000 kg and speed 30 m/s will be 150,000 kilograms meter per second
Explanation
Here is some important points you will need to understand
What is Linear Momentum?
Linear momentum is defined as the product of mass and the velocity of a body. The formula for calculating linear momentum is
Linear momentum (P) = Mass (m) x Velocity (v)
Which can be written as P = mv
Linear momentum is a vector quantity. This is because it has magnitude and direction.
After reading a question on linear momentum, you will need to extract your data from the problem. You will need to identify the mass and the velocity from the question.
After identifying your data, you can go ahead to multiply the mass by the velocity to obtain the linear momentum.
Remember that the S.I unit of linear momentum is kilogram meter per second (kgm/s).
Over the last 300 years, Newton’s laws of motion became one of the most famous laws of physics. The laws were stated by a famous English physicist and mathematician, and his name is Sir Isaac Newton. Below is a video explaining of the three laws of motion.
Newton’s three laws of motion are as follows:
A body will continue to be at rest or in a motion until it’s acted upon by an external force. This is also known as the law of inertia (“Inertia of a body is its reluctance to start moving when at rest or stationary and its reluctance to stop moving once it is moving”).
The rate of change of momentum is proportional to the external force and takes place in the direction of the applied force.
When an external force is applied to a body, there is always an equal and opposite reaction. The law can also be defined as to every action, there is an equal and opposite reaction.
Newton was born in England on 4th of January, 1643 in Woolsthorpe Manor House. During his time, children were trained to rear sheeps to earn a living. However, Newton was uninterested in making a living as a farmer. Instead, he was more attracted to books.
At young age, he found love in the books that were arranged in the shelves of his parent’s library. The famous scientist dedicated his time to observing the possible occurrences that happen between sunrise to sunset, spring to summer, autumn to winter. Newton tried to understand what triggers these changes, and it led him to understanding the gravity.
He observed the effect of sun, wind and water on materials like water clocks, windmills, and kites. Isaac finally realizes that the earth and the sky work on the same principle which is contrary to Aristotle’s understanding.
The influential English scientist devoted himself to reading the works of ancient Greeks, Islamic and medieval views on gravity. Newton asked himself as to why an apple fruit falls on the ground instead of going up to the sky. He provided mathematical evidence to Galilei Galileo and Johannes Kepler’s theory of gravity.
Isaac Newton started his journey of research at the age of 23. He came up with a book on classical mechanics, and the name of the book is “Principia Mathematica Philosophiae Naturalis.”. The meaning of the book in English is “Mathematical principles of natural philosophy”.
This book perfectly explained the 3 laws of motion. The three (3) laws of motion shaped our understanding of gravity and we call them Newton’s laws of motion.
Newton’s Laws of Motion in our Everyday Life
We apply Newton’s laws of motion in our everyday life. For example, while playing soccer, we must apply force for the ball to move. Once we apply force on the ball, it will head towards our target due to momentum. The wide range of applications of Newton’s laws is as follows:
Newton’s First (1st) Law of Motion: Law of Inertia
The first law of motion states that A body will continue to be at rest or in motion, until it’s acted upon by an external force. According to this law, when you hide an object in your room. The object will continue to remain in that place for millions of years, until someone or something applies a force to move the object from its original hiding position.
Therefore, when you keep something in a safe place. We can easily say that the object will remain in that place until you decide to change its position or someone moves it from its original position.
Additionally, the same law state that when you set your car in motion. The car will continue to move in a straight line and will never stop. It will wait for you to apply brakes before it can stop. Additionally, the car may also stop if the engine is bad or the fuel is empty. Hence, there must be an application of external force for the car to stop.
Also, if you don’t control the steering, it will continue to move in a straight line until you decide to go left or right. This scenario also explains Newton’s first law of motion or law of inertia.
Understanding Inertia
Newton’s first law of motion is also known as the law of inertia. As defined in his book, the law of inertia is the power by which an object, if it’s at rest, stays at rest, or if in motion, will travel in a straight line.
To understand this law, try rolling a marble on smooth tiles, the marble will continue to move until an object blocks its movement. Alternatively, the friction of the tiles will make the speed of the marble keep decreasing until it comes to rest.
This scenario of a rolling marble against the tiles perfectly describes Newton’s first law of motion. This law explains why your coffee remains peacefully in its cup until you decide to stir it or why your car continues down the highway unless you hit the brakes.
Everyday Examples
Consider a few everyday examples:
A hockey puck sliding on an ice rink with no friction would continue forever if not for external forces like air resistance or the rink’s boards.
In the vacuum of space, where there is no air resistance, celestial bodies such as planets and moons continue in their orbits with the same speed and direction due to their inertia.
The second law of motion states that when you apply an external force on an object the rate of change of momentum is proportional to the applied force and it takes place in the direction of the force.
Here is the best way to explain this law, when you push a shopping cart in the supermarket. The shopping cart will move because of the external force you apply on it, and the mass of the cart.
Therefore, the mass of the , mass of the body and the external force acting upon it plays a crucial role in setting a body into motion.
Here is another example, before buying anything, push the cart for a while. You will notice that it moves faster. Now, add some of the items you bought from the store into the cart. What did you notice? It’s heavier than it was, right? You may need to add extra force to push it this time around.
Ask yourself again, what brought this sudden change? Here is the answer to your question: “The more mass you add to the cart, the heavier it becomes, and the more force you will need to apply to move the cart”.
Hence, there exists a relationship between the mass of the body and the applied external force.
The mathematical expression (formula) for newton’s second law of motion is
The S.I unit of force is called Newton (Honors Isaac Newton) or kilogram meter per second square. The symbol of Newton is N, while the symbol for kilogram meters per second square is kgm/s2
The unit of mass is in kilograms (kg)
The unit of acceleration is in meter per second square (m/s2)
Practical Examples
In simpler terms, the more force you apply to an object, the faster it will accelerate, while a heavier object requires more force to achieve the same acceleration.
When you push a shopping cart, the harder you push, the faster it accelerates.
In sports, athletes understand this law intuitively. A soccer player applies more force to the ball to make it accelerate quickly.
How to Derive Newton’s Second Law
According to Newton’s second law, the rate of change of momentum is proportional to the applied force. Therefore, this law can be translated into
Force (F) = change in momentum / time (t)
Change in momentum is the product of mass and velocity.
Thus,
Change in momentum = mass x velocity
Which can be further broken down into
Change in momentum = mv – mu [where v = final velocity and u = initial velocity]
Therefore, we can insert (Change in momentum = mv – mu) into [Force = change in momentum / time] to obtain
F = (mv – mu) / t
Hence,
F = m (v – u) / t
And acceleration is the rate of change of displacement
Acceleration, a = (v – u) / t
And we have F = ma which is Newton’s second law of motion.
Momentum
This is the product mass and velocity of an object. The S.I unit of momentum is kilogram meter per second (kgm/s). The formula for calculating momentum is
Momentum (M) = mass (m) x velocity (v)
The rate of change of momentum = (change in momentum) / time
Impulse of a Force
The impulse of a force is the instantaneous response of an object to the force that acts on it. We can also define impulse (I) as the product of the force and the time it takes momentum to change. The S.I unit of impulse is Newton-second (Ns).
Impulse (I) = Force (F) x Time (t)
Calculations Involving Second Law of Motion
Here are some questions to help you understand the applications of Newton’s second law of motion to solve a problem.
Problem 1
An object of mass 7.5 kilograms has a force applied to it that pulls eastward with a magnitude of 50 Newtons and another applied to it that pulls westward with a magnitude of 150 Newtons. What is the westward acceleration of the object? Give your answer to one decimal place.
Data
Mass of the object is m = 7.5 N
Initial Force, F1 = 50 Newtons
Final force, F2 = 150 Newtons
Unknown
Acceleration of the object, a = ?
Formula
The formula we need to apply to solve this problem is
FNet = ma
But FNet = F2 – F1
Therefore, w can further break it down into
F2 – F1 = ma
Solution
We can now insert our data into F2 – F1 = ma
150 – 50 = 7.5 x a
Make a subject of the formula
a = (150 – 50) / 7.5
Hence,
a = 100/7.5 = 13 m/s2
Problem 2
An object of mass 3 kilograms has a force of 9 Newtons applied to it. At what rate does the force accelerate the object?
Data
Mass of the object, m = 3 kg
Applied force, F = 9 N
Unknown
Acceleration, a = ?
Formula
The formula that can help us to solve this problem is
a = F/m
Solution
We can substitute our formula with our data
Acceleration, a = 9/3 = 3 m/s2
Therefore, the acceleration (a) is 3 meters per second square.
Problem 3
A boy pushes his toy car with a cat in inside. The mass of the toy car and the cat put together is 70 kilograms. The toy car accelerates at 0.9 meters per second square. Find the force that the boy applies to pull the toy car and the cat together.
Data
Mass of the toy car with a cat is m = 70 kg
Acceleration of the toy car, a = 0.9 m/s2
Unknown
Force, F = ?
Formula
We will use the formula F = ma to solve the problem
Solution
Insert the data into the above formula
F = 70 x 0.9 = 63 N
Therefore, the applied force is 63 Newtons
Third Newton’s Law of Motion
The third Newton’s law of motion states that action and reaction are equal and opposite. This law explains that when a force is applied to a body, there is a chain of equal and opposite reactions that will follow.
For example, when you blow a balloon, and later release the air entrance. The balloon will run away from you.
What if we decide to insert the balloon inside a toy canoe? We will blow the balloon and position its air entrance to face the opposite direction of the canoe’s motion. After releasing the air entrance of the balloon, it will set the toy canoe in motion on water.
Therefore, we can see that when an object X applies a force on another object Y. Object Y will equally apply the same type of force on X. Both applied forces are equal in magnitude but opposite in direction.
We can demonstrate this mathematically as
Force X = – Force Y [ Force X is the action force, and Force Y is the reaction Force]
Another example of Newton’s third law of motion is how the rocket operates. When a rocket is launched into space, it will accelerate away from the earth surface. This is thanks to the burning fuel of the rocket that kept igniting fire at the back of the rocket to push it in the upward direction.
We have many examples to demonstrate the effect of this law in our everyday life. Among these examples are:
The lift of an airplane
Student’s school bag on the desk
A hammer hitting a nail
A bird flying
A boy pushing against the wall
Applying brakes on your car
A duck swimming in the water
Everyday Examples
In other words, when one object exerts a force on another object, the second object exerts an equal and opposite force on the first.
Walking: When you walk, your foot pushes backward on the ground, and the ground pushes your foot forward, propelling you forward.
Rocket Propulsion: Rockets work by expelling high-speed exhaust gases backward. The force of the exhaust gas going backward propels the rocket forward.
Principle of Conservation of Momentum
We can state the principle of conservation of momentum in three ways:
The principle states that when two bodies that are traveling in a straight line collide, the total force before collision is equal to the total force after collision provided no external force acts on the body.
In any system of colliding bodies, the total momentum is always conserved provided that there is no net external force acting on the system.
The total momentum of an isolated or closed system of colliding bodies remains constant.
Therefore, the mathematical definition of this principle is
m1v1 = m2v2 [Where m1v1 is the initial momentum, and m2v2 is the final momentum
Now, if we have two balls M and N exerting equal forces on each other. We can say that
FM = – FN
And since F = ma
Then, it’s easy to say that
mMaM = – mNaN
Additionally, a = (v – u) / t
Which implies
mM [ (vm – um) / t ] = – mN [ (vN – uN) / t ]
And we will end up with
mMvm – mMum = – ( mNvN – mNuN)
The above equation will give us
mMum + mNuN = mMvm + mNuN
And the above expression shows that momentum before collision is equal to the momentum after collision.
Applications of Newton’s Third Law to Solve a Problem
Here are a few problems to help you understand how to solve problems that are related to newtons third law of motion, momentum, and impulse.
Problem 1
A body (P) has mass of 5 kilograms moving with a velocity of 30 m/s collides with another body (Q), moving in opposite direction with a velocity of 20 m/s. If both bodies now move in the direction of P at a velocity of 10 m/s. Calculate the mass of Q.
Newton’s Laws of Motion
Solution
To solve this problem, we need to remember that
m1v1 = m2v2
Which implies that
(m1 + m2)v = m1u1 – m2u2
and m1u1 – m2u2 = Before collision
(m1 + m2)v = After collision
Therefore, since m1 = 5kg, u1 = 30 m/s, m2 = ? , u2 = 20 m/s, v = 10 m/s
We can now insert the above information into (m1 + m2)v = m1u1 – m2u2 to obtain
(5 + m2) x 10 = (5 x 30) – (m2 x 20)
We will now have
50 + 10m2 = 150 – 20m2
We can now collect like terms to get
150 – 50 = 20m2 + 10m2
And it will become
100 = 30m2
By making m2 subject of the formula, we will have
m2 = 100 / 30 = 3 kg
Therefore, the mass of Q is 3 kilograms.
Problem 2
A bullet of mass 0.05 kg is fired horizontally into a 10 kg block which is fired free to move. If both bullet block move with velocity 0.5 m/s after the impact, find the velocity with which the bullet hit the body.
Data
Mass of the bullet = 0.05 kg, mass of the block = 10 kg, and the mass of both bullet and the block = 10 + 0.05 = 10.05
Initial mass of the bullet = u
The momentum before collision = Momentum after collision
0.05 x u = 10.05 x 0.5
Hence,
u = 5.025 / 0.05 = 100.5 = 101 m/s
Problem 3
A rifle of mass 15 kilograms fires a bullet of mass 60 grams with a velocity of 200 m/s. Calculate the recoil velocity of the rifle.
Solution
m1 = 15 kg, and m2 = 60 g = (60/1000) kg = 0.06 kg,
v1 = ? and v2 = 200 m/s
We need to remember that Momentum before impact = Momentum after impact
Momentum before impact = 0
And the Momentum after impact = m1 v1 + m2 v2
Hence,
m1 v1 + m2 v2 = 0
By inserting our data, we will have
15 x v1 + 0.06 x 200 = 0
Which will become
15 v1 + 12 = 0
After making v1 subject of the formula
v1 = – (12/8) = – 0.8 m/s
Therefore, the recoil velocity is 0.8 meters per second
Isaac Newton Defined the Following Quantities in His Book
Matter: Anything that takes up space
Mass: The measure of a quantity or amount of matter
Momentum: The quantity of motion, which is the product of velocity (speed) and mass
Inertia: The power by which an object, if it is at rest, stays at rest, or if in motion, will travel in a straight line
Force: An action applied upon a body
Centripetal Force: An attraction toward the center of something (as in gravity)
We apply Newton’s laws of motion in our everyday activities.
Importance and Applications
Now that we’ve explored the laws themselves, let’s consider their practical applications in our everyday lives and various fields of science and technology.
Transportation and Engineering
Newton’s Laws are fundamental in the design of vehicles and structures. They influence:
Automobiles: Every time you hit the gas pedal or apply the brakes, you’re experiencing Newton’s Laws in action. The force you apply determines the acceleration or deceleration of your vehicle.
Aviation: In aviation, understanding the interplay of forces is essential for safe and efficient flight. Pilots must balance thrust, lift, drag, and gravity to keep an aircraft in equilibrium.
Maritime Travel: Ships, too, rely on Newton’s Laws. When a ship’s engine propels it forward, the reaction is a force that moves the ship in the opposite direction.
Sports and Athletics
Newton’s Laws govern the dynamics of sports and athletics:
Baseball: When a pitcher throws a fastball, the force applied to the ball determines its speed and trajectory. Hitters use this knowledge to time their swings.
Swimming: Swimmers maximize their efficiency by minimizing water resistance and using the principles of buoyancy and propulsion, all in line with Newton’s Laws.
Billiards: In billiards, players use the principles of angles and collisions to control the path of the balls.
Space Exploration and Astronomy
Newton’s Laws are integral to space exploration. They underpin:
Spacecraft: Rockets and spacecraft rely on these laws to propel them through the vacuum of space. Rocket engines function by expelling mass at high velocities, creating a reaction that propels the spacecraft in the opposite direction. Without Newton’s Third Law, space exploration as we know it would be impossible.
Historical Significance
The importance of Newton’s Laws is not limited to their practical applications. They also hold historical significance as they mark a turning point in the world of science.
A Scientific Revolution
Newton’s Laws were part of a broader scientific revolution that also included the works of Galileo, Copernicus, and Kepler. They offered a unified and mathematically rigorous framework for understanding motion and paved the way for classical physics.
Revolutionizing Astronomy
In addition to their impact on physics, Newton’s Laws played a pivotal role in the field of astronomy. His law of universal gravitation, which describes how all objects with mass are attracted to each other, led to the understanding of planetary motion and orbits. It was Newton’s mathematical genius that allowed scientists to predict the movement of celestial bodies with unparalleled accuracy.
Educational Impact
Newton’s Laws play a crucial role in science education and the development of critical thinking skills.
Importance of Teaching
These laws are fundamental to STEM (Science, Technology, Engineering, and Mathematics) education. They introduce students to the principles of physics and help them develop critical thinking skills. Educators emphasize the importance of teaching Newton’s Laws as a gateway to understanding the scientific method.
Resources for Learning
Numerous educational resources and tools are available to facilitate the learning of Newton’s Laws. These include interactive simulations, video lessons, and hands-on experiments that make physics accessible and engaging for students of all ages.
Newton’s Laws and the Bigger Picture
Newton’s Laws are not isolated principles; they are interconnected with other scientific concepts.
Gravity and Universal Gravitation
Newton’s Laws and the law of universal gravitation are closely related. Gravity, the force of attraction between masses, adheres to the principles set forth in Newton’s Laws. The orbits of celestial bodies, planetary motion, and the behavior of objects in free fall are all governed by a combination of these laws.
While the laws of thermodynamics primarily describe the flow of heat and energy, they are not separate from Newton’s Laws. For example, the behavior of gas particles in a container can be explained using the laws of motion as molecules collide and exert forces on one another.
Quantum Mechanics
Even in the realm of quantum mechanics, which explores the behavior of particles on the smallest scales, the foundations laid by Newton’s Laws still apply. While quantum mechanics introduces new principles, the motion of particles and the forces acting upon them remain consistent with Newton’s insights.
Quotes and Insights from Experts
To highlight the enduring importance of Newton’s Laws, let’s turn to the insights of renowned physicists and scientists.
Albert Einstein
Albert Einstein, the legendary physicist, once remarked, “You have this elementary particle, the electron, and it is attracted by positive charge, and repelled by negative charge. Newton would have despised that theory.” Einstein’s comment underscores the profound impact of Newton’s work and the subsequent developments in physics.
Richard Feynman
Richard Feynman, another Nobel laureate in physics, praised Newton’s contributions by stating, “In his day, the great truths which now seem so obvious were unknown… He took the known facts and extended our ideas to a remarkable degree in a precise quantitative way.” Feynman recognized the precision and depth of Newton’s work, which laid the groundwork for future scientific discoveries.
Interactive Demonstrations and Simulations
To truly grasp the significance of Newton’s Laws, there’s no substitute for hands-on experience. Interactive simulations and demonstrations can bring these laws to life, enabling learners to experiment and observe the principles of motion.
Hands-On Learning
We recommend exploring online platforms, such as PhET Interactive Simulations, where you can interact with virtual experiments that demonstrate the concepts of force, motion, and equilibrium. These tools offer an intuitive and engaging way to gain a deeper understanding of Newton’s Laws.
Conclusion
In conclusion, the importance of Newton’s Laws of Motion cannot be overstated. These fundamental principles, introduced by the genius of Sir Isaac Newton, continue to shape our understanding of the physical world. They govern everything from the way we travel to the devices we use, from the structures we build to the scientific inquiries we pursue.
The enduring significance of Newton’s Laws is a testament to their timeless relevance in science, technology, and education. As we journey further into the future, we carry with us the profound insights provided by these laws, reminding us of the boundless possibilities that lie ahead.
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We hope you’ve enjoyed this comprehensive exploration of the importance of Newton’s Laws of Motion. Your journey into the world of physics has only just begun, and we invite you to continue your quest for knowledge and understanding.
Law of conservation of energy states that energy can neither be created nor destroyed but can be transferred from one medium to another.
Additionally, the law states that in an isolated or closed system, the total amount of energy is always constant, although energy may be changed from one form to another.
Data
what is the system’s potential energy when its kinetic energy is equal to 34e?
The net energy in a system is
Net Energy (Enet) = Potential Energy (P.E) + Kinetic Energy (K.E)
Therefore, the system’s potential energy is (1/8) Kx2
Drop a question in the comment section if you have a challenge with the question: “what is the system’s potential energy when its kinetic energy is equal to 34e?”
Here are Newton’s second law practice problems to help you master the calculations involving the second law of motion.
Newton’s Second Law Practice Problems
Problem 1
An unbalanced force of 20 Newtons acts on a 4.0-kilogram mass. What acceleration does it give it?
Newton’s Second Law Practice Problems
Data
Force acting on the body, f = 20 N
Mass of the body, m = 4 kg
Unknown
Acceleration of the body, a = ?
Formula
The formula for Newton’s second law of motion is F = ma
and to find the acceleration, we need to make a subject of the formula
a = f/m
We will apply the above formula to solve the problem
Solution
Substitute your data into the formula
a = 20/4 = 5 ms-2 [ Because the unit of acceleration is ms-2 ]
Therefore, the acceleration of the body is 5 meters per second square.
Problem 2
A 900 grams stone is pushed along a tarmac by a horizontal force of 20 Newton. A frictional force of 8 newtons opposes the motion. What is the acceleration given to the stone?
Data
Mass of the stone, m = 900 g = (900/1000) kg = 0.9 kg
Horizontal force, fh = 20 N
Frictional force, fr = 8 N
Unknown
Unbalanced force, F = fh – fr = 20 – 8 = 12 N
The acceleration, a = ?
Formula
Force = mass x acceleration
Which can be written as
F = m x a
Solution
Now we can insert our data into the formula
The acceleration, a = F/m = 12/0.9 = 13.3 ms-2
Therefore, the acceleration of the stone is 13 ms-2
Problem 3
A ball of mass 0.3 kilograms moving at a velocity of 20 meters per second is suddenly hit by a force of 5 newtons for a time of 0.03 seconds. Find its new velocity of motion.
Data
Mass of the ball, m = 0.3 kg
Initial velocity, u = 20 m/s
Force, F = 5 N
Time, t = 0.03 s
Unknown
Final velocity, v = ?
Formula
Since F = ma
and a = (v-u)/t
We can see that
F = m(v-u)/t = (mv – mu)/t
We can make v subject of the formula from F = m(v-u)/t
F/m = (v-u)/t
and now
Ft/m = v – u
Therefore, the formula we need to apply is
v = (Ft/m) + u
Solution
It is now time for us to put our data into the formula
v = (5 x 0.03/0.3) + 20
Now, our final velocity, v = (0.15/0.3) + 20 = 0.5 + 20 = 20.5 m/s
Therefore, our final velocity is 20.5 meters per second
Problem 4
A body of mass 0.1 kilograms dropped from a height of 8 meters onto a hard floor and bounces back to a height of 2 meters. Calculate the change of momentum. If the body is in contact with the floor for 0.1 seconds, what is the force exerted on the body? ( Take gravitational acceleration, g = 10 ms-2).
Data
Mass of the body, m = 0.1 kg
Before hitting the floor, the height h1 = 8 m
After hitting the floor, the height of bouncing h2 = 2 m
Gravitational acceleration, g = 10 ms-2
Time, t = 0.1 s
Before hitting the floor, the initial velocity, u1 = 0
After hitting the floor, the final velocity after bouncing, v2 = 0
Unknown
Before hitting the floor, the final velocity, v1 = ?
After hitting the floor, the initial velocity after bouncing, u2 = ?
Formula
The first formula
v12 = u12 + 2gh1
The second formula
v22 = u22 – 2gh2
The final formula to apply is
F = (mv1 – (-mu2))/t = (mv1 + mu2)/t [For the body bouncing and moving away from the floor, the momentum is negative]
Solution
We can apply our data to the first formula
v12 = u12 +2gh1 = u12 +2gh1 = 0 + 2 x 10 x 8 = 160
Therefore
v12 = 160
and v1 = √160 = 12.7 m/s = 13 m/s
Applying our data to the second formula, we will have
v22 = u22 – 2gh2
0 = u22 – 2 x 10 x 2
u22 – 40 = 0
We can now collect like terms to obtain
u22 = 40
u2 = √40 = 6.325 m/s = 6 m/s
Applying the final formula, we can now find the force exerted on the body
F = (mv1 + mu2)/t
substituting our data and the rest of the values, we will get
Force, F = (mv1 + mu2)/t = m(v1 + u2)/t = 0.1(13 + 6)/0.1 = 19 N
Therefore, the force exerted on the body is 19 Newton
Problem 5: Newton’s Second Law Practice Problems
A body of mass 2 kilograms falls freely from rest, through a height of 50 meters, and comes to rest having penetrated 5.0 centimeters of sand.
a. Calculate the velocity with which the body hits the sand
b. The time taken in falling
c. The average force exerted by the sand in bringing the body to rest.
Data
Mass of the body, m = 2 kg
Initial height, h1 = 50 m
Final height, h2 = 5 cm = 5/100 m = 0.05 m
Gravitational acceleration, g = 10 ms-2
Initial velocity, u = 0
Unknown
v = ?
t = ?
retardation, a = ?
F = ?
Formula
The first formula is
v2 = u2 + 2gs
The second formula to apply is
v = u + gt
We will also apply v2 = u2 + 2as to find retardation
And then finally apply the formula for Newton’s second law of motion to solve the force exerted by the sand on the body
F = ma
Solutions
a. We can add our data into the equation v2 = u2 + 2gs
Hence,
v2 = 0 + 2 x 10 x 50 = 1000
v = √1000 = 31.623 m/s = 31 m/s
b. Substitute your data into the second formula
v = u + gt
we make t subject of the formula
v = gt [since u = 0]
Thus,
t = v/g = 31/10 = 3.1 s
c. We need to find the retardation by applying the formula v2 = u2 + 2as before we can go ahead to find the force
Insert our data into v2 = u2 + 2as
Note: After hitting the sand, our initial velocity = 31 m/s and the final velocity = 0
Hence,
0 = 312 + 2 x a x 0.05
We make a subject of the formula
a = -(961/0.1) = -9610 m/s2 [The negative sign indicates negative acceleration or retardation]
Therefore, a = 9610 m/s2
We can now comfortably apply our final formula that says Force (F) = Mass(m) x Acceleration (a)
Thus,
F = ma = 2 x 9610 = 19,220 N
Therefore, the force exerted by the sand in bringing the body to rest is 19,220 Newton
Problem 6: Newton’s Second Law Practice Problems
A towing van tows a car of mass 1,500 kilograms along a level road and accelerates at 0.8 meters per second square. What is the tension in the towing string?
If the towing string suddenly breaks when the car reaches a speed of 20 meters per second, calculate the distance covered by the car before it comes to rest on the application of a braking force of 2000 newtons.
Data
Mass of the car, m = 1,500 kg
Acceleration, a = 0.8 ms-2
Initial speed, u = 20 m/s
Final speed, v = 0 [Because the car finally came to rest, then its final velocity is zero]
Braking force = 2000 N
Unknown
The tension of the string, F = ?
Change in momentum = ?
Time taken, t = ?
Deceleration, aD = ?
Distance covered, s = ?
Formula
To find the tension of the string, we will apply
F = ma
For change in momentum, the formula will be
Change in momentum = mu – mv
The to find time taken is
t = change in momentum / tension on the string (F)
Finding deceleration, we will apply
a = (v – u) / t
And finally, the last formula to find the distance covered, we use
s = (v2 – u2) / 2a
Solution
Let us find the tension on the string by substituting our data into F = ma
F = 1,500 x 0.8 = 1,200 N
The change in momentum = mv – mu = m(v – u) = 1,500 (20 – 0) = 1,500 x 20 = 30,000 kgms-1
To calculate the time taken, we will insert our data into the formula
Time taken, t = change in momentum / tension on the string = 30,000 / 1,200 = 25 seconds
The deceleration will be
aD = (v – u) / t = (0 – 20) / 25 = – 0.8 ms-2
Now, we can go ahead to calculate the distance covered
s = (v2 – u2) / 2aD = (02 – 202) / 2 x (-0.8) = -400/-1.6 = 250 m
Therefore, the distance covered is 250 meters
Therefore, I hope these Newton’s second law practice problems have helped you understand how to solve problems related to the second law of motion.
You can share your biggest challenge in Newton’s second law practice problems so that i can help you solve them.
Newton’s Laws of Motion, formulated by the renowned physicist Sir Isaac Newton, are fundamental principles that describe the behaviour of objects in motion. These laws have been pivotal in our understanding of the physical world and have countless applications in science and engineering. In educational and hands-on settings, projects that explore Newton’s Laws of Motion not only deepen our comprehension of these principles but also foster a sense of wonder and discovery.
The following project ideas are designed to engage students, science enthusiasts, and aspiring physicists in practical experiments and demonstrations that showcase the three laws of motion:
Solar-powered water heater
Eco-friendly packaging design
Urban vertical garden
Home automation system
Virtual reality game
Renewable energy-powered vehicle
Smart healthcare wearable
Autonomous delivery drone
Sustainable fashion line
AI-driven language translation app
Indoor air quality monitor
Community recycling program
Smartphone app for mental health
Aquaponics system
Green energy-efficient home
Educational coding platform for children
Low-cost prosthetic limb
Hydroponic gardening system
AI-based traffic management
Sustainable urban transportation
Home composting solution
Automated home cleaning robot
Water purification system
Smart farming technology
Zero-waste lifestyle app
Newton’s Laws of Motion Project Ideas
13 Newton’s Laws of Motion Project Ideas with Brief Explanation
1. Building a model roller coaster to demonstrate the laws of motion.
In this project, you will use materials like marble, PVC pipes, and form tubing to build a roller coaster. The aim and objective of the experiment is to describe the impact of Newton’s first law of motion.
The law states that a body will continue to be at rest or in motion until it is acted upon by a force. It is also called “law of Inertia“.
Thus, the roller coaster will not move until a force is applied to it, and then it will start rolling its cars. The roller coaster will continue to move until another force is applied to stop its motion.
Additionally, we know that the second law of motion talks about the force being proportional to the rate of change of momentum.
Hence, the mass and the acceleration of the roller coaster play a vital role in its motion. You can use this model to explain newton’s laws of motion
2. Designing a car crash experiment to show the effects of momentum
In this project, you can use toy cars and crash-test dummies to explain Newton’s second law of motion.
You can carry out a task to explain how the force of the collision is affected by the mass and velocity of the toy cars you have used.
3. Creating a balloon rocket car to demonstrate third law
In this project, you will create a car that will work with balloons. The balloon will set the toy car in motion.
The aim of the project is to demonstrate the effect of Newton’s third law of motion. The law states that for every action, there is an equal and opposite reaction.
4. Constructing a marble run to illustrate the conservation of energy.
This project will help you to demonstrate the effect of Newton’s second law of motion. You will apply materials such as plastic pipes, marbles, and cardboard tubes to build a marble run.
5. Building a Rube Goldberg machine
The Rube Goldberg machine was designed by a cartoonist called Rube Goldberg. The essence of this design is to create a machine that performs different tasks.
And the machine cannot work until you apply a force to set it in motion. Hence, the first Newton’s law of motion comes into play.
Additionally, the machine obeys both the second and the third Newton’s law of motion. This is because we are dealing with the momentum of the machine. Also, the third law states that actions and reactions are equal and opposite.
Therefore, the Rube Goldberg machine is a perfect example of a project to demonstrate Newton’s laws of motion.
6. Designing a paper plane flight experiment to show the effects of lift and drag.
Another project to demonstrate the laws of motion is a paper plane flight. Students can design a paper plane flight to show the impact of Newton’s law of motion.
Additionally, you can also design and build a toy plane that can be controlled with a remote (more like a drone) to demonstrate the effect of laws of motion.
7. A Pendulum Experiment
You can also come up with a project that involves the motion of a pendulum bob. You can show the application of the three laws of motion when the pendulum is set in motion.
8. Creating a simple robot to demonstrate Newton’s third law of motion.
Coming up with a project on a simple robot is a very good idea to demonstrate the applications of Newton’s law of motion.
The robot can only move when a force is exerted on it. Additionally, the robot obeys the third law of motion.
9. Designing a model rocket to illustrate the principles of thrust and propulsion.
You can also build a model rocket to demonstrate how it’s set in motion. The mass of the rocket, its acceleration. You can also show the application of Newton’s third law of motion where actions and reactions are equal and opposite.
10. Building a trebuchet to demonstrate the laws of projectile motion.
You can also construct a trebuchet to demonstrate the applications of these famous laws of motion.
11. Designing an experiment to demonstrate the effects of air resistance
In this project, you can come up with an experiment to show how air resistance can limit the motion of objects.
12. Building a balance scale to illustrate the principle of equilibrium
This project will help you to determine the effect of mass and gravity. You will be able to demonstrate Newton’s first and second laws of motion.
13. Creating a simple machine, such as a lever or pulley, to demonstrate mechanical advantage.
Building a simple machine, such as a lever pulley is another way to show the application of laws of motion
Designing a physics game that teaches the laws of motion in a fun and interactive way.
Building a simple electric car to demonstrate the relationship between force and motion.
Therefore, now you can confidently explore more Newton’s laws of motion project ideas.
Examples of Newton’s Second Law of Motion in Everyday Life
Newton’s second law of motion states that the net force acting upon a body is the product of mass and acceleration. In other words, Newton’s second law of motion can be defined as the rate of change of momentum being directly proportional to the force acting upon it and it happens in the direction of the force. Here are 25 Newton’s Second Law of Motion examples:
When you push a book with twice the force, it accelerates at twice the rate.
A soccer player kicking a ball harder makes it accelerate faster.
A car accelerates when the driver presses the gas pedal.
Flicking a lighter requires a small force to create a large acceleration.
Swinging a baseball bat with more force increases the speed of the swing.
Pushing a shopping cart with more groceries requires a greater force to accelerate it.
When a rocket engine ignites, the force it generates propels the rocket into space.
When a person jumps off a diving board, they exert force on the board, causing it to accelerate.
Riding a bicycle requires pedaling to provide the force for acceleration.
A basketball player dribbling the ball exerts force to control its motion.
Pushing a lawn mower forward accelerates it due to the applied force.
Rowing a boat involves applying force to the oars, causing the boat to move.
A skydiver falling faster experiences greater air resistance due to higher acceleration.
Using a slingshot requires a pulling force to accelerate the projectile.
A weightlifter lifting a heavier weight requires more force for acceleration.
An ice skater pushes off the ice to accelerate in a specific direction.
A car’s brakes apply force to slow down and eventually stop the car.
A swimmer kicks their legs to accelerate in the water.
A person using a pogo stick bounces by applying force to the ground.
An archer’s arrow accelerates when the bowstring is pulled back.
A rocket launching into the sky generates tremendous force for acceleration.
A person using a sled pushes off with their feet to accelerate downhill.
A skateboarder pushes the skateboard to increase its speed.
A car’s engine provides the force necessary for acceleration on the road.
A roller coaster accelerates due to the force applied by the tracks and gravity.
These examples demonstrate how Newton’s Second Law of Motion relates force, mass, and acceleration in various everyday scenarios.
We apply Newton’s second law of motion to understand the motion of objects at various physical phenomena. This phenomenon ranges from the movement of a rocket up in space to a car running on the road. Here is a detailed explanation of some of the examples of Newton’s second law of motion that applies to us in our day-to-day activities:
Car on a Highway
Newton’s Second Law of Motion Examples
When a car is speeding on a highway, two physical variables in form of force, and mass would play a major role in the car. The engine of the car will force it to move on the road. Hence, the car engine is responsible for the force moving the car.
Additionally, the mass of the car will determine it is acceleration. This is because when the mass of the car is very large, the weight of the car might reduce the acceleration due to the force of gravity.
Thus, the car will accelerate at a lower speed when we compare it with an acceleration of a car with a lighter mass.
Skydiver
Newton’s Second Law of Motion Examples
When a skydiver is falling to the ground, his body will accelerate downward. The mass of the skydiver is proportional to his acceleration. The mass of the skydiver determines his acceleration. The skydiver will accelerate at a constant rate until they reach terminal velocity.
Additionally, the force acting on the skydiver will pull him to the ground. Therefore, we can easily identify that type of force as the force of gravity. Hence, the force of gravity and mass of his body plays a vital role in bringing him down to the ground.
A Hockey Puck
Newton’s Second Law of Motion Examples
Another example of newton’s second law of motion is when a hockey puck is sliding on the ice. As the puck slides on the ice, there is always a force of friction that is decreasing its acceleration. Therefore, the mass of the hockey puck plays a vital role in influencing its acceleration.
The frictional force between the hockey puck and the ice will resist and limit the acceleration of the puck. If enough force is applied to the puck, it will accelerate toward its target. We can now see that a hockey puck sliding on also obeys newton’s second law of motion.
Rocket
Newton’s Second Law of Motion Examples
When we launch a rocket into space, it accelerates toward a specific destination. As the rocket continues to move upward, the engine will continue to work. During this process, the fuel in the engine will be burnt, and the rocket will be lighter. Hence the acceleration of the rocket will continue to increase.
The above smooth operation of the rocket is only possible thanks to a decrease in mass (from the burnt fuel). As the fuel of the rocket continues to decrease, the mass of the rocket will also continue to decrease. When the mass of the rocket decreases, the rocket will be lighter and the force pushing it upward will increase. The force that is acting on the rocket is called thrust and it was influenced by the working of the engine.
When the thrust of the rocket increases, the rocket will accelerate into space.
Shopping Cart
Newton’s Second Law of Motion Examples
When you visit a supermarket, you are bound to push a shopping cart because of the items you bought. The shopping cart makes it easy for you to push all the items you have bought at once. While pushing the shopping cart, you must apply force to make it move.
Additionally, the mass of both the shopping cart and the items inside plays a role in determining the acceleration of the cart. Buying many items will increase the mass of the cart and hence demand more force to make it accelerate.
Therefore, as you push a shopping cart toward your car. You are applying newton’s second law of motion.
A Baseball Hit by a Bat
In this case, the force of the bat will hit the ball and causes it to accelerate. The mass of the ball is a very important factor in deciding its acceleration. Additionally, the greater the force, the higher the acceleration of the ball.
An Airplane
We have an airplane as another one of newton’s second law of motion examples. When an airplane is about to take off, its engine forces the plane to accelerate on the runway. This acceleration helps the plane to conquer its weight through the force of lift.
A Boat Rowing Through Water
The oars of a boat rowing through water help to propel the boat in the water. Therefore, the oars influence the force of the boat, while the mass of the boat and its passengers determines the acceleration of the boat on the water.
Stronger and more people can influence the force of the boat and make it run faster. Therefore, with the right people controlling the boat, you stand a chance to win the boat game. This practical experience affirms the application of newton’s second law of motion in real life.
A Soccer Ball
When a football player hits a ball, he applies force on the ball thereby causing it to accelerate. Additionally, the ball is made lighter by adding air into it, this helps the ball to accelerate as soon as the force of the leg is applied to it.
A Basketball
A basketball player applies force from his biceps to throw a ball into the net. As he throws the ball, its acceleration is proportional to the force he applied and the mass of the basketball.
Understanding Newton’s Second Law of Motion?
Newton’s second law of motion is the vehicle behind the motion of bodies around us. The law is one of the fundamental principles of physics. It was stated by a renowned physicist and mathematician called Isaac Newton. Newton hailed from England and lived between 1642 AD to 1727 AD. He came up with 3 laws of motion, and we shifted our focus to the second law and its examples.
A small rock with a mass of 0.20 kg is released from rest at point A which is at the top edge of a large, hemispherical bowl with a radius R = 0.5 m. Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle. And that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has a magnitude of 0.22 J.
a) Between points A to point B, how much work is done on the rock by;
to find the speed, we need to make v subject of the formula
(2 x K.E)/m = v2
v = √(2 x K.E)/m
we insert our data from the question to get
v = √[(2 x 0.76)/0.2] = √(1.52/0.2) = √7.6 = 2.757
Therefore,
v = 2.8 ms-1
Solution c
Of the three forces acting on the rock as it slides down the bowl, the gravitational force is constant. While the frictional force and the normal force between the surface of the rock and the bowl is not constant.
d) The normal force N is
N = mg + mv2/R= 0.2 x 10 + (0.2 x 2.82)/0.5= 2 + (0.2 x 7.84)/0.5 = 2 + (1.568/0.5)
