CHAPTER ONE

Physical Quantities, Units, Dimensions, Space and Time

Measurement is the process of comparing a physical quantity with an accepted standard of the same kind in order to assign a numerical value and a unit to that quantity.

If the length of a room is written as 8, the statement is incomplete. If it is written as 8 m, the number 8 gives the magnitude while the metre specifies the standard unit used.

A physical quantity is a measurable property of a body, substance, phenomenon or physical system that can be expressed by a numerical value together with an appropriate unit.

A fundamental or base quantity is a physical quantity chosen as independent and not defined in terms of other quantities within a given system of measurement.

A derived quantity is a physical quantity defined through an algebraic combination of fundamental quantities or other derived quantities.

Question: Determine the SI unit of acceleration from its definition.

Given: Acceleration is change in velocity per unit time.

Formula:

Solution:

The SI unit of velocity is

Therefore,

Answer: The SI unit of acceleration is m s^-2.

Question: Deduce the SI unit of force from the relation F=ma.

Formula:

Where: m is mass and a is acceleration.

Solution:

This derived unit is called the newton.

Answer: The SI unit of force is the newton (N).

Unit symbols are not normally pluralized and are not followed by a full stop unless they occur at the end of a sentence. Thus, 5 kg, 20 m and 15 s are standard forms.

Question: Express 7.5 km in metres.

Given:

Solution:

Answer: 7500 m.

Question: Express 0.0000042 m in micrometres.

Given:

Solution:

Therefore,

Answer: 4.2 μm.

Question: Convert 72 km/h to m/s.

Given:

Solution:

Answer: 20 m/s.

Question: Convert 25 m/s to km/h.

Solution:

Answer: 90 km/h.

Question: Convert 250 cm² to square metres.

Solution:

Since

then

Therefore,

Answer: 0.025 m².

The dimensions of a physical quantity show how that quantity depends on fundamental quantities such as mass, length and time.

Question: Determine the dimensions of force.

Formula:

Solution:

and

Therefore,

Answer: The dimensions of force are MLT^-2.

Question: Determine the dimensions of kinetic energy.

Formula:

Solution:

The numerical constant 1/2 has no dimensions. Therefore,

Answer: [K]=ML²T^-2.

Question: Pressure is defined as force per unit area. Determine its dimensions.

Formula:

Solution:

Hence,

Answer: The dimensions of pressure are ML^-1T^-2.

Every term in a physically meaningful equation that is added to or subtracted from another term must have the same dimensions.

Question: Test whether the equation

is dimensionally consistent.

Solution:

Also,

Hence,

Answer: The equation is dimensionally consistent.

Question: Determine whether

can represent a physically valid equation for velocity.

Solution:

However,

The dimensions LT^-1 and L cannot be added in a physically meaningful equation.

Answer: The equation is not dimensionally consistent.

Question: Use dimensional analysis to determine how the period of a simple pendulum depends on its length l and gravitational acceleration g.

Given:

Dimensions:

Solution:

Equating powers of L:

Equating powers of T:

Therefore,

Hence,

Thus,

Answer: Dimensional analysis predicts that T is proportional to √(l/g). It cannot determine the dimensionless constant k; for a simple pendulum undergoing small oscillations, further analysis gives k=2π.

Work and torque both have dimensions ML²T^-2, but they are physically different quantities. Dimensional equality does not imply physical equivalence.

A reference frame is a coordinate system together with a means of measuring time, relative to which the position and motion of an object are described.

Question: A motion begins at t=2.5 s and ends at t=11.8 s. Determine the elapsed time.

Formula:

Solution:

Answer: The elapsed time is 9.3 s.

Accuracy describes how close a measured value is to the accepted or true value of the quantity.

Precision describes the degree of agreement among repeated measurements or the fineness with which a measurement is reported.

Question: An experimental measurement of gravitational acceleration gives 9.70 m/s², while the accepted value for the comparison is 9.81 m/s². Calculate the absolute and percentage errors.

Solution:

Percentage error:

Answer: Absolute error = 0.11 m/s²; percentage error ≈ 1.12%.

• Physics describes natural phenomena using measurable physical quantities and mathematical relationships.
• A physical measurement consists of a numerical magnitude and an appropriate unit.
• Physical quantities may be fundamental or derived.
• The SI has seven base quantities, with length, mass and time being especially important in mechanics.
• Derived units are obtained by combining base units according to the defining relationships of physical quantities.
• SI prefixes provide convenient ways of expressing very large and very small quantities.
• Unit conversions must account correctly for powers when area and volume units are involved.
• Dimensions express the physical nature of a quantity in terms of fundamental dimensions.
• Velocity has dimensions LT^-1, acceleration LT^-2 and force MLT^-2.
• The principle of dimensional homogeneity requires terms that are added or equated to have compatible dimensions.
• Dimensional analysis can test equations and determine the possible form of some physical relationships.
• Dimensional analysis cannot generally determine dimensionless numerical constants.
• Position and motion are described relative to a reference frame.
• Elapsed time is the difference between final and initial times.
• Accuracy concerns closeness to an accepted value, whereas precision concerns consistency or fineness of measurement.

CHAPTER TWO

Scalars, Vectors and Vector Operations

A scalar is a physical quantity that is completely specified by magnitude and an appropriate unit, without the need to specify a direction.

A statement that a vehicle travels at 20 m/s describes its speed. No direction is required. Speed is therefore a scalar quantity.

A vector is a physical quantity having both magnitude and direction and obeying the mathematical rules of vector addition.

A velocity of 20 m/s east contains both magnitude and direction. If the direction changes while the magnitude remains 20 m/s, the velocity has changed even though the speed has not.

Question: A person walks 6 m east and then 8 m north. Determine the magnitude and direction of the resultant displacement.

Given:

Required: Resultant displacement.

Solution:

The two displacements are perpendicular. Therefore,

Let θ be the angle north of east.

Answer: The resultant displacement is 10 m at approximately 53.1° north of east.

Question: Two forces of magnitudes 10 N and 6 N act at an angle of 60°. Find the magnitude of their resultant.

Formula:

Solution:

Since cos60°=1/2,

Answer: The resultant force has magnitude 14 N.

Question: An object initially moves east at 12 m/s and later moves east at 20 m/s. Determine its change in velocity.

Solution:

Taking east as positive,

Answer: The change in velocity is 8 m/s east.

Question: A force of 50 N acts at 30° above the horizontal. Determine its horizontal and vertical components.

Given:

Formula:

Solution:

Answer: Horizontal component ≈ 43.3 N; vertical component = 25 N.

Question: A displacement of 20 m is directed 40° north of east. Find its components.

Solution:

Answer: Eastward component ≈ 15.32 m; northward component ≈ 12.86 m.

Question: Find the magnitude of

Solution:

Answer: |A|=5.

Question: Given

and

find A+B.

Solution:

Add corresponding components:

Answer:

Question: A force has components

Determine its magnitude and direction.

Solution:

For the direction,

Answer: The force has magnitude 13 N and is directed approximately 22.6° above the positive x-axis.

The scalar or dot product of two vectors A and B is defined by

where θ is the angle between the vectors. The result is a scalar.

Question: Given

and

calculate A·B.

Solution:

Answer: A·B=5.

Question: Two vectors have magnitudes 6 and 8 and an angle of 60° between them. Find their scalar product.

Formula:

Solution:

Answer: The scalar product is 24.

The vector or cross product of vectors A and B is a vector defined by

where n is a unit vector perpendicular to the plane containing A and B. Its direction is determined by the right-hand rule.

Question: Two vectors have magnitudes 5 and 4 and are perpendicular. Find the magnitude of their vector product.

Formula:

Solution:

Answer: The magnitude of the vector product is 20.

Question: Given

and

find A×B.

Solution:

Using distributivity,

Since

we obtain

Answer: A×B=-2k.

Question: A force of 20 N acts through a displacement of 5 m at an angle of 60° to the displacement. Find the work done.

Formula:

Solution:

Answer: The work done is 50 J.

Question: A force of 30 N is applied perpendicular to a spanner at a distance of 0.25 m from the pivot. Calculate the magnitude of the torque.

Given:

Formula:

Solution:

Answer: The torque magnitude is 7.5 N m.

Question: A particle moves from

to

Determine its displacement vector and the magnitude of the displacement.

Solution:

The magnitude is

Answer: Displacement vector = 5i-4j m; magnitude ≈ 6.40 m.

Question: Given

find dA/dt.

Solution:

Differentiate each component:

and

Therefore,

Answer:

Question: The position vector of a particle is

Determine its velocity and acceleration at time t.

Required: v(t) and a(t).

Formula:

Solution:

Differentiate the position vector:

Differentiate again:

Answer:

Question: For the particle in Worked Example 17, find its velocity, speed and acceleration at t=2 s.

Solution:

Velocity:

The speed is the magnitude of the velocity:

Acceleration:

Answer: At t=2 s, the velocity is 8i+8j m/s, the speed is approximately 11.31 m/s, and the acceleration is 4i+12j m/s².

Position, velocity and acceleration are linked by vector differentiation. Position tells where the particle is, velocity tells how its position is changing, and acceleration tells how its velocity is changing. This relationship provides the mathematical foundation for the study of kinematics in the next chapter.

• A scalar quantity has magnitude but does not require direction, while a vector quantity has both magnitude and direction.
• Distance, speed, mass, time and energy are scalars; displacement, velocity, acceleration, force and momentum are vectors.
• Vectors can be represented by directed line segments or by their Cartesian components.
• Vector addition may be performed geometrically or by adding corresponding components.
• Vector subtraction is equivalent to addition of the negative vector.
• A vector of magnitude A at angle θ to the positive x-axis has components Acosθ and Asinθ.
• The unit vectors i, j and k specify the positive Cartesian coordinate directions.
• The scalar product is A·B=ABcosθ and produces a scalar.
• The vector product has magnitude ABsinθ and produces a vector perpendicular to the plane of the original vectors.
• Work is an important application of the scalar product, while torque is an important application of the vector product.
• The displacement vector is the difference between final and initial position vectors.
• A vector function can be differentiated by differentiating its Cartesian components.
• Velocity is the time derivative of the position vector.
• Acceleration is the time derivative of velocity and the second time derivative of position.
• Vector methods provide the mathematical language required for the systematic study of motion and forces in mechanics.

CHAPTER THREE

Kinematics and Equations of Motion

Kinematics is the branch of mechanics that describes the motion of objects in terms of quantities such as position, displacement, velocity and acceleration without considering the forces causing the motion.

Distance is the total length of the actual path travelled by an object. It is a scalar quantity and is always non-negative.

Displacement is the change in position of an object from its initial position to its final position. It is a vector quantity.

Question: A student walks 12 m east and then 5 m west. Find the total distance travelled and the displacement.

Given: First motion = 12 m east; second motion = 5 m west.

Solution:

Total distance is the sum of the path lengths:

Taking east as positive, the displacement is

Answer: Distance travelled = 17 m; displacement = 7 m east.

Distance depends on the actual route followed. Displacement depends only on the initial and final positions. An object can travel a considerable distance and still have zero displacement if it returns to its starting point.

Speed is the rate at which distance is travelled.

Velocity is the rate of change of displacement with time. It is a vector quantity.

Question: A car travels 120 km east in 2 h. Find its average speed and average velocity.

Solution:

Since the displacement is 120 km east,

Answer: Average speed = 60 km/h; average velocity = 60 km/h east.

Question: A runner completes one full lap of a circular track of length 400 m in 50 s. Find the average speed and average velocity.

Solution:

After one complete lap, the runner returns to the starting point, so the displacement is zero.

Answer: Average speed = 8 m/s; average velocity = 0 m/s.

Question: The position of a particle is given by

metres. Find its instantaneous velocity at t=4 s.

Formula:

Solution:

At t=4,

Answer: The instantaneous velocity is 26 m/s.

Acceleration is the rate of change of velocity with time.

Question: The velocity of a particle changes from 10 m/s to 28 m/s in 6 s. Find its average acceleration.

Formula:

Solution:

Answer: The average acceleration is 3 m/s².

Question: The position of a particle is

Find its velocity and acceleration at t=2 s.

Solution:

Velocity:

At t=2,

Acceleration:

At t=2,

Answer: Velocity = 8 m/s; acceleration = 14 m/s².

Question: A train moves at a constant velocity of 25 m/s for 3 minutes. Find the distance travelled.

Given:

Formula:

Solution:

Answer: The train travels 4500 m or 4.5 km.

A body undergoes uniform acceleration when its velocity changes by equal amounts in equal intervals of time.

These equations are valid only when acceleration is constant during the interval under consideration.

Question: A car starts from rest and accelerates uniformly at 4 m/s² for 6 s. Find its final velocity.

Given:

Required: Final velocity v.

Formula:

Solution:

Answer: 24 m/s.

Question: A body with initial velocity 5 m/s accelerates uniformly at 3 m/s² for 4 s. Find its displacement.

Given:

Formula:

Solution:

Answer: The displacement is 44 m.

Question: A motorcycle moving at 20 m/s accelerates uniformly at 2.5 m/s² over a distance of 60 m. Find its final speed.

Given:

Formula:

Solution:

Answer: The final speed is approximately 26.46 m/s.

Question: A vehicle travelling at 30 m/s is brought to rest uniformly in 5 s. Find its acceleration and stopping distance.

Given:

Acceleration:

The negative sign indicates acceleration opposite to the chosen positive direction.

Stopping Distance:

Answer: Acceleration = -6 m/s²; stopping distance = 75 m.

Question: A body's velocity increases uniformly from 4 m/s to 16 m/s in 6 s. Use the velocity-time interpretation to find the acceleration and displacement.

Acceleration:

Displacement:

The area under the velocity-time graph is a trapezium:

Answer: Acceleration = 2 m/s²; displacement = 60 m.

A projectile is an object that, after being launched, moves under the influence of gravity alone when air resistance is neglected.

Question: A ball is projected horizontally at 10 m/s from the top of a cliff 20 m high. Neglect air resistance and take g=9.8 m/s². Find the time to reach the ground and the horizontal distance travelled.

Vertical Motion:

Horizontal Motion:

Answer: Time of flight ≈ 2.02 s; horizontal distance ≈ 20.2 m.

Question: A projectile is launched at 20 m/s at an angle of 30° above the horizontal. Take g=9.8 m/s². Find its time of flight, maximum height and horizontal range.

Given:

Time of Flight:

Maximum Height:

Range:

Answer: Time of flight ≈ 2.04 s; maximum height ≈ 5.10 m; range ≈ 35.35 m.

Question: At what angle should a projectile be launched to obtain maximum range when it lands at its launch level?

Formula:

Solution:

For fixed u and g, range is maximum when sin2θ has its maximum value, which is 1.

Therefore,

Answer: The maximum range occurs at a launch angle of 45°.

The velocity of object A relative to object B is

Question: Car A moves east at 25 m/s and Car B moves east at 15 m/s. Find the velocity of A relative to B.

Solution:

Answer: Car A moves at 10 m/s east relative to Car B.

Question: Two cars approach each other on a straight road. Car A moves east at 20 m/s, while Car B moves west at 15 m/s. Find the velocity of A relative to B.

Solution:

Take east as positive:

Answer: The magnitude of the relative velocity is 35 m/s.

Kinematics tells us how motion occurs. It does not explain why velocity changes. To understand why an object accelerates, slows down, changes direction or remains at rest, we must introduce force and Newton's laws of motion.

• Kinematics describes motion without considering its causes.
• Distance is the total path length, while displacement is the change in position.
• Speed is a scalar rate of distance travelled, whereas velocity is a vector rate of displacement.
• Instantaneous velocity is dx/dt.
• Acceleration is the rate of change of velocity and may be written as dv/dt or d²x/dt².
• For constant velocity, displacement is proportional to time.
• For constant acceleration, the principal equations are v=u+at, s=ut+(1/2)at², v²=u²+2as and s=(u+v)t/2.
• The slope of a position-time graph gives velocity.
• The slope of a velocity-time graph gives acceleration, while its area gives displacement.
• The area under an acceleration-time graph gives change in velocity.
• Two-dimensional motion may be analysed by resolving it into independent Cartesian components.
• Projectile motion combines constant horizontal velocity with constant downward acceleration due to gravity.
• For equal launch and landing levels, projectile time of flight, maximum height and range can be determined from standard equations.
• The maximum horizontal range of an ideal projectile occurs at 45°.
• Relative velocity is obtained by subtracting one object's velocity vector from another's.

CHAPTER FOUR

Newton's Laws of Motion and Applications

Dynamics is the branch of mechanics concerned with the relationship between the motion of bodies and the forces responsible for that motion.

A force is an interaction that can change the state of motion of an object or deform it. Force is a vector quantity.

A body remains at rest or continues in uniform straight-line motion unless acted upon by a non-zero resultant external force.

A book resting on a table does not have zero forces acting on it. Its weight acts downward, while the table exerts an upward normal force of equal magnitude. The resultant force is zero, so the book remains at rest.

Inertia is the tendency of an object to resist any change in its state of rest or uniform straight-line motion.

An inertial frame is a reference frame in which Newton's first law is valid. Such a frame is either at rest or moves with constant velocity relative to another inertial frame.

The resultant external force on a body is equal to the time rate of change of its momentum.

Question: A net horizontal force of 24 N acts on a 6 kg object. Find its acceleration.

Given:

Required: Acceleration.

Formula:

Solution:

Answer: The acceleration is 4 m/s² in the direction of the net force.

Question: Two horizontal forces of 50 N east and 20 N west act on a 10 kg block. Find the acceleration.

Solution:

Taking east as positive, the resultant force is

Therefore,

Answer: The block accelerates at 3 m/s² east.

Mass is a measure of the inertia of a body and the amount of matter it contains. In classical mechanics, it is treated as a scalar quantity and is measured in kilograms.

Weight is the gravitational force exerted on a body.

Question: Find the weight of a 70 kg person at a place where g=9.8 m/s².

Formula:

Solution:

Answer: The person's weight is 686 N downward.

When one body exerts a force on a second body, the second body simultaneously exerts an equal-magnitude force in the opposite direction on the first body.

When a person pushes a wall, the person exerts a force on the wall and the wall exerts an equal and opposite force on the person. The pair constitutes a third-law interaction.

A free-body diagram is a simplified representation of an isolated object showing all external forces acting on that object.

Do not draw forces that the object exerts on other bodies. Draw only the forces acting on the object being analysed. Choose coordinate axes that make the equations as simple as possible.

The normal reaction is the contact force exerted by a surface on an object, directed perpendicular to the surface.

Question: A 15 kg box rests on a horizontal floor. Find the normal reaction exerted by the floor. Take g=9.8m/s².

Solution:

The box has no vertical acceleration, so

Answer: The normal reaction is 147 N upward.

Friction is a contact force that opposes relative motion or the tendency of relative motion between surfaces in contact.

Question: A 10 kg block slides on a horizontal surface with coefficient of kinetic friction 0.20. Find the frictional force. Take g=9.8m/s².

Given:

Solution:

On a horizontal surface with no other vertical forces,

Then,

Answer: The kinetic frictional force has magnitude 19.6 N and acts opposite to the sliding motion.

Question: A 12 kg block is pulled horizontally by a force of 50 N across a rough surface. The coefficient of kinetic friction is 0.25. Find its acceleration. Take g=9.8m/s².

Given:

Vertical Direction:

Friction:

Horizontal Resultant Force:

Acceleration:

Answer: The acceleration is approximately 1.72 m/s².

Question: A 5 kg block on a smooth horizontal surface is acted on by a force of 40 N at 30° above the horizontal. Find its horizontal acceleration and the normal reaction. Take g=9.8m/s².

Resolve the Applied Force:

Horizontal Motion:

Vertical Direction:

There is no vertical acceleration, so

Answer: Horizontal acceleration ≈ 6.93 m/s²; normal reaction = 29 N.

Question: A 6 kg block slides down a smooth plane inclined at 30° to the horizontal. Find its acceleration. Take g=9.8m/s².

Force Parallel to Plane:

Using F=ma,

Cancel m:

Answer: The block accelerates down the plane at 4.9 m/s².

Question: A 10 kg block slides down a plane inclined at 25°. The coefficient of kinetic friction is 0.15. Find its acceleration. Take g=9.8m/s².

Normal Reaction:

Friction:

Down-Slope Component of Weight:

Net Force Down the Plane:

Acceleration:

Answer: The block accelerates down the plane at approximately 2.81 m/s².

Question: Two masses of 6 kg and 4 kg are connected by a light string passing over a frictionless pulley. Find the acceleration of the system and the tension in the string. Take g=9.8m/s².

Given:

The 6 kg mass moves downward and the 4 kg mass moves upward.

For the 6 kg Mass:

For the 4 kg Mass:

Add the equations:

Now use

Answer: Acceleration = 1.96 m/s²; tension = 47.04 N.

Question: Two blocks of masses 3 kg and 5 kg are connected by a light horizontal string on a smooth surface. A horizontal force of 24 N pulls the 5 kg block. Find the acceleration and the tension in the string.

Whole System:

For the 3 kg Block:

The only horizontal force on it is the tension:

Answer: Acceleration = 3 m/s²; tension = 9 N.

Question: A 60 kg person stands on a scale in an elevator accelerating upward at 2 m/s². Find the scale reading. Take g=9.8m/s².

Formula:

Solution:

Answer: The scale reads 708 N.

Question: The same 60 kg person is in an elevator accelerating downward at 2 m/s². Find the scale reading.

Formula:

Solution:

Answer: The scale reads 468 N.

Question: A 20 kg crate is pushed horizontally with a force of 100 N across a floor. A frictional force of 35 N opposes the motion. If the crate starts from rest, find its acceleration and speed after 4 s.

Resultant Force:

Acceleration:

Speed after 4 s:

Since the crate starts from rest, u=0.

Answer: Acceleration = 3.25 m/s²; speed after 4 s = 13 m/s.

Question: A 4 kg block rests on a horizontal surface. The coefficient of static friction is 0.40. What is the largest horizontal force that can be applied without causing the block to move? Take g=9.8m/s².

Normal Reaction:

Maximum Static Friction:

Static friction can balance an applied force only up to this value.

Answer: The largest horizontal force that can be applied without causing motion is 15.68 N.

Question: A 5 kg block is pulled up a smooth plane inclined at 30° by a force of 40 N parallel to the plane. Find its acceleration. Take g=9.8m/s².

Component of Weight Down the Plane:

Resultant Force Up the Plane:

Acceleration:

Answer: The block accelerates up the plane at 3.1 m/s².

Question: A horizontal force causes a 2 kg object initially at rest to accelerate at 5 m/s². Find the resultant force and the object's velocity after 6 s.

Resultant Force:

Velocity:

Answer: Resultant force = 10 N; velocity after 6 s = 30 m/s.

Newton's second law and the equations of kinematics are often used together. Newton's laws determine the acceleration from the forces, while the kinematic equations use that acceleration to determine velocity, displacement or time.

• Dynamics studies the relationship between forces and motion.
• A force is a vector interaction capable of changing motion or causing deformation.
• Newton's first law states that a body remains at rest or in uniform straight-line motion when the resultant external force is zero.
• Inertia is the tendency of a body to resist changes in its state of motion.
• An inertial frame is a reference frame in which Newton's first law holds.
• Newton's second law is fundamentally ΣF=dp/dt; for constant mass it becomes ΣF=ma.
• Mass measures inertia, while weight is the gravitational force W=mg.
• Newton's third law states that interaction forces occur in equal-magnitude, opposite-direction pairs acting on different bodies.
• Free-body diagrams identify all external forces acting on a selected body.
• The normal reaction acts perpendicular to a supporting surface.
• Static friction prevents sliding up to a limiting value, while kinetic friction acts during sliding.
• On an inclined plane, the components of weight are mg sinθ parallel to the plane and mg cosθ perpendicular to it.
• Tension transmits pulling forces through strings or ropes and can be analysed using Newton's laws.
• The normal reaction measured by a scale in an accelerating elevator represents apparent weight.
• Gravitational force is a non-contact interaction and is treated classically as an action-at-a-distance force.
• Newtonian mechanics problems are solved by identifying forces, resolving them into appropriate directions and applying ΣF=ma.

CHAPTER FIVE

Momentum, Impulse and Conservation of Linear Momentum

The linear momentum of a particle is the product of its mass and velocity.

Question: A car of mass 1200 kg moves east at 20 m/s. Determine its momentum.

Given:

Formula:

Solution:

Answer: The momentum is 2.4×10⁴ kg m/s east.

Question: A 0.15 kg ball moves west at 30 m/s. Taking east as positive, find its momentum.

Solution:

Since west is negative,

Answer: The momentum is -4.5 kg m/s, or equivalently 4.5 kg m/s west.

Question: The momentum of an object changes from 10 kg m/s east to 34 kg m/s east in 6 s. Find the average resultant force.

Formula:

Solution:

Answer: The average resultant force is 4 N east.

Impulse is the product of a force and the time interval for which it acts when the force is constant.

The impulse of the resultant force acting on a body equals the change in the body's linear momentum.

Question: A constant force of 40 N acts on an object for 0.25 s. Calculate the impulse.

Formula:

Solution:

Answer: The impulse is 10 N s.

Question: A 0.20 kg ball moving east at 15 m/s is struck and leaves westward at 25 m/s. Find the impulse on the ball.

Given: Take east as positive.

Formula:

Solution:

Answer: The impulse is 8 N s west.

Question: A 0.5 kg ball initially at rest receives an impulse of 6 N s. Find its final speed.

Formula:

Solution: Since u=0,

Answer: The final speed is 12 m/s.

For a given change in momentum, increasing the time over which the change occurs reduces the average force. This principle explains the protective function of airbags, seat belts, crash cushions, helmets and padded landing surfaces.

If the resultant external force on a system is zero, the total linear momentum of the system remains constant.

Question: A 4 kg trolley moving at 6 m/s collides with a stationary 2 kg trolley. They stick together. Find their common velocity after collision.

Given:

Formula:

Solution:

Answer: Their common velocity is 4 m/s in the original direction of the 4 kg trolley.

The centre of mass of a system is the mass-weighted average position of all the particles making up the system.

Question: A 2 kg particle is at x=1 m and a 6 kg particle is at x=5 m. Find the centre of mass.

Formula:

Solution:

Answer: The centre of mass is at x=4 m.

Question: A system has total mass 20 kg and total momentum 100 kg m/s east. Find the velocity of its centre of mass.

Formula:

Solution:

Answer: The centre of mass moves at 5 m/s east.

An elastic collision is a collision in which both total linear momentum and total kinetic energy are conserved.

An inelastic collision is one in which total momentum is conserved for an isolated system but kinetic energy is not conserved.

Question: A 3 kg body moving at 8 m/s collides with a stationary 5 kg body. They stick together. Find their final velocity and determine the kinetic energy lost.

Momentum Conservation:

Initial Kinetic Energy:

Final Kinetic Energy:

Kinetic Energy Lost:

Answer: Final velocity = 3 m/s; kinetic energy lost = 60 J.

Question: A 2 kg object moving at 6 m/s collides head-on elastically with an identical stationary object. The first object stops after the collision. Find the velocity of the second object.

Momentum Conservation:

Answer: The second object moves at 6 m/s in the original direction of the first object.

Question: A stationary object of mass 10 kg explodes into two fragments of masses 4 kg and 6 kg. The 4 kg fragment moves east at 9 m/s. Find the velocity of the 6 kg fragment.

Initial Momentum:

Take east as positive:

Answer: The 6 kg fragment moves at 6 m/s west.

Question: A rifle of mass 5 kg fires a bullet of mass 0.02 kg at 400 m/s. If the rifle and bullet were initially at rest, find the recoil velocity of the rifle.

Solution:

Take the bullet's direction as positive:

Answer: The rifle recoils at 1.6 m/s opposite to the bullet's motion.

Question: A stationary object explodes into two fragments. One fragment has momentum 12 kg m/s east and the other has momentum 5 kg m/s north. What momentum must the remaining fragment have if the original total momentum was zero?

Solution:

The first two fragments have resultant momentum

To make total momentum zero, the remaining fragment must have the opposite momentum:

Its magnitude is

Answer: The required momentum has magnitude 13 kg m/s and is directed opposite to the resultant of the other two momenta.

Question: A 1000 kg car moving at 20 m/s is brought to rest. Compare the average stopping force when it stops in 0.10 s and when it stops in 0.50 s.

Change in Momentum:

First Stopping Time:

Second Stopping Time:

Answer: The average force magnitudes are 200,000 N and 40,000 N. Increasing the stopping time from 0.10 s to 0.50 s reduces the average force by a factor of five.

• Linear momentum is defined by p=mv and is a vector quantity.
• Newton's second law may be written as ΣF=dp/dt.
• Impulse is the effect of force acting over a time interval and is equal to the change in momentum.
• For constant force, J=FΔt=Δp.
• The area under a force-time graph represents impulse.
• Total linear momentum remains constant when the resultant external force or external impulse on a system is zero.
• The total momentum of a system is the vector sum of the momenta of its particles.
• The centre of mass is the mass-weighted average position of a system.
• The total momentum of a system can be expressed as P=Mv_cm.
• Momentum is conserved in isolated elastic and inelastic collisions.
• Kinetic energy is conserved in elastic collisions but not generally in inelastic collisions.
• In a perfectly inelastic collision, the colliding bodies move together after impact.
• Recoil and explosions are explained using conservation of momentum.
• In two dimensions, momentum conservation must be applied separately to each component.

CHAPTER SIX

Work, Energy and Conservation of Energy

Energy is the capacity of a physical system to produce change or perform work. In mechanics, important forms include kinetic energy and potential energy.

The work done by a constant force is the scalar product of the force and displacement.

Question: A horizontal force of 50 N moves a box through a horizontal distance of 8 m. Find the work done by the force.

Given:

Formula:

Solution:

Answer: The work done is 400 J.

Question: A force of 100 N acts at 60° to the horizontal while an object moves horizontally through 5 m. Calculate the work done by the force.

Formula:

Solution:

Answer: The work done is 250 J.

Question: A frictional force of 20 N acts opposite to the motion of a block that moves 10 m. Find the work done by friction.

Given:

Solution:

Answer: The work done by friction is -200 J.

When an object moves horizontally at constant height, its weight acts vertically downward. Since weight is perpendicular to the horizontal displacement, gravity does zero work during that horizontal displacement.

Question: A force varies with position according to F(x)=4x newtons. Find the work done as the object moves from x=0 to x=3 m.

Formula:

Solution:

Answer: The work done is 18 J.

Kinetic energy is the energy possessed by a body because of its motion.

Question: Calculate the kinetic energy of a 1500 kg car moving at 20 m/s.

Formula:

Solution:

Answer: The kinetic energy is 3.0×10⁵ J.

Kinetic energy is proportional to the square of speed. If the speed of an object doubles while its mass remains constant, its kinetic energy becomes four times as large.

The net work done on a particle equals the change in its kinetic energy.

Question: A 4 kg object initially moving at 3 m/s has 50 J of net work done on it. Find its final speed.

Initial Kinetic Energy:

Work-Energy Theorem:

Now,

Answer: The final speed is approximately 5.83 m/s.

Question: A 1000 kg car moving at 15 m/s is brought to rest by braking forces. Determine the net work done on the car.

Formula:

Solution:

Answer: The net work done is -1.125×10⁵ J.

Potential energy is energy associated with the position or configuration of a system when conservative forces are involved.

Question: A 5 kg object is raised vertically through 8 m. Find the increase in gravitational potential energy. Take g=9.8m/s².

Formula:

Solution:

Answer: The gravitational potential energy increases by 392 J.

Question: A spring of force constant 200 N/m is compressed by 0.10 m. Find the elastic potential energy stored.

Given:

Formula:

Solution:

Answer: The spring stores 1 J of elastic potential energy.

A force is conservative if the work it does between two points is independent of the path followed and depends only on the initial and final positions.

A non-conservative force is one for which the work done generally depends on the path followed.

When only conservative forces perform work on a system, the total mechanical energy of the system remains constant.

Question: A stone is released from rest at a height of 20 m above the ground. Neglect air resistance. Find its speed just before reaching the ground. Take g=9.8m/s².

Initial Energy:

Final Energy:

By conservation of mechanical energy,

Cancel m:

Answer: The speed just before reaching the ground is approximately 19.8 m/s.

Question: A ball is projected vertically upward at 14 m/s. Neglect air resistance. Use conservation of energy to determine its maximum height above the point of projection. Take g=9.8m/s².

Initial Energy:

At maximum height the instantaneous speed is zero, so the gain in gravitational potential energy equals the initial kinetic energy:

Cancel m:

Answer: The maximum height is 10 m.

Question: A 2 kg block is released from rest and slides down a smooth track through a vertical height of 5 m. Find its speed at the bottom. Take g=9.8m/s².

Solution:

Answer: The speed at the bottom is approximately 9.90 m/s.

Question: A 5 kg block initially moving at 10 m/s slides along a horizontal rough surface and stops after 8 m. Determine the average frictional force.

Initial Kinetic Energy:

Final Kinetic Energy:

Therefore,

Since friction acts opposite to displacement,

Answer: The average frictional force has magnitude 31.25 N.

Power is the rate at which work is done or energy is transferred.

Question: A machine performs 24,000 J of work in 30 s. Find its average power.

Formula:

Solution:

Answer: The average power is 800 W.

Question: A car experiences a driving force of 2500 N while moving at a constant speed of 20 m/s. Determine the power delivered by the driving force.

Formula:

Solution:

Answer: The power delivered is 50 kW.

Question: A motor lifts a 200 kg load vertically through 15 m in 10 s at constant speed. Find the useful power output. Take g=9.8m/s².

Work Done:

Power:

Answer: The useful power output is 2940 W or 2.94 kW.

Efficiency is the ratio of useful energy output to total energy input, or equivalently useful power output to total power input.

Question: A machine receives 5000 J of energy and delivers 3750 J as useful work. Determine its efficiency.

Formula:

Solution:

Answer: The machine is 75% efficient.

Question: A 2 kg block starts from rest and is pushed by a constant horizontal force of 10 N through 5 m on a smooth horizontal surface. Find its final speed using the work-energy theorem.

Work Done:

Work-Energy Theorem:

Since the block starts from rest, K_i=0.

Answer: The final speed is approximately 7.07 m/s.

Question: A 3 kg object starts from rest at the top of a smooth slope 12 m above its lowest point. Find its speed at the bottom. Take g=9.8m/s².

Principle: Since the slope is smooth, mechanical energy is conserved.

Mass cancels:

Answer: The speed at the bottom is approximately 15.34 m/s.

Question: A 1000 kg car climbs a hill of vertical height 50 m at constant speed. Determine the increase in gravitational potential energy. If this is accomplished in 25 s, determine the minimum average power required to increase the car's gravitational potential energy. Take g=9.8m/s².

Potential Energy Increase:

Average Power:

Answer: The increase in gravitational potential energy is 4.90×10⁵ J, and the corresponding minimum average power is 19.6 kW.

Newton's laws are especially useful when forces and accelerations are required. Energy methods are often more efficient when a problem asks for speed, height, displacement or energy changes between two states. Both approaches describe the same mechanics and should give consistent results when applied correctly.

• Work is the scalar product of force and displacement: W=Fs cosθ.
• Work is positive when a force assists displacement, negative when it opposes displacement and zero when force is perpendicular to displacement.
• For a variable one-dimensional force, work is W=∫F(x)dx.
• Kinetic energy is the energy associated with motion and is given by K=(1/2)mv².
• The work-energy theorem states that net work equals the change in kinetic energy.
• Potential energy is associated with position or configuration when conservative forces are involved.
• Near the Earth's surface, the change in gravitational potential energy is ΔU=m g Δh.
• An ideal spring stores elastic potential energy U=(1/2)kx².
• Conservative forces do path-independent work and can be associated with potential-energy functions.
• When only conservative forces do work, total mechanical energy K+U remains constant.
• Non-conservative forces can change the mechanical energy of a system while total energy remains conserved when all forms of energy are considered.
• Power is the rate of doing work or transferring energy and is measured in watts.
• Instantaneous mechanical power can be expressed as P=F·v.
• Efficiency compares useful output with total input and is commonly expressed as a percentage.
• Work-energy and conservation-of-energy methods provide efficient alternatives to direct force-and-acceleration calculations in many mechanics problems.

CHAPTER NINE

Circular Motion and Moment of Inertia

Circular motion is the motion of a body or particle along a path whose distance from a fixed centre remains constant.

Question: A wheel of radius 0.50 m rotates through an angle of 8 rad. Find the distance travelled by a point on its rim.

Given:

Required: Arc length s.

Formula:

Solution:

Answer: The point travels a distance of 4 m.

The period T is the time required for one complete revolution or cycle.

The frequency f is the number of complete revolutions or cycles made per unit time.

Question: A rotating wheel completes 120 revolutions in one minute. Determine its frequency, period and angular speed.

Solution:

Convert the number of revolutions per minute to revolutions per second:

The period is

The angular speed is

Answer: Frequency = 2 Hz; period = 0.50 s; angular speed = 4π rad/s ≈ 12.57 rad/s.

Uniform circular motion is motion along a circular path at constant speed.

Constant speed does not necessarily mean zero acceleration. In uniform circular motion, speed is constant but velocity changes because its direction changes. The particle therefore possesses acceleration.

Centripetal acceleration is the inward acceleration of an object moving along a circular path. It is always directed toward the centre of the circle.

Question: A particle moves in a circle of radius 2 m at a constant speed of 6 m/s. Find its centripetal acceleration.

Given:

Formula:

Solution:

Answer: The centripetal acceleration is 18 m/s², directed toward the centre.

Question: A point on a rotating disc is 0.40 m from the axis and rotates at 10 rad/s. Determine its centripetal acceleration.

Formula:

Solution:

Answer: The centripetal acceleration is 40 m/s².

Centripetal force is the name given to the resultant force directed toward the centre of a circular path that produces centripetal acceleration.

Centripetal force is not an additional fundamental force. Depending on the physical situation, the inward force may be supplied by tension, friction, gravity, a normal reaction, or a combination of forces.

Question: A 0.50 kg stone moves in a horizontal circle of radius 1.5 m at 6 m/s. Determine the required centripetal force.

Formula:

Solution:

Answer: The required inward force is 12 N.

Question: A car moves around a level circular bend of radius 50 m. If the coefficient of static friction between the tyres and road is 0.40, estimate the maximum speed without skidding. Take g=9.8 m/s².

Formula:

Solution:

Answer: The maximum speed is approximately 14 m/s.

Question: A frictionless road of radius 80 m is banked at 15°. Find the speed for which no sideways friction is required. Take g=9.8 m/s².

Formula:

Solution:

Answer: The design speed is approximately 14.5 m/s.

Question: A 0.20 kg mass moves at 5 m/s at the bottom of a vertical circle of radius 1 m. Find the tension in the string at that instant. Take g=9.8 m/s².

Formula:

Solution:

Answer: The tension is 6.96 N.

The moment of inertia of a body about a specified axis is a measure of its resistance to angular acceleration about that axis.

Question: Two particles of masses 2 kg and 3 kg are located 0.50 m and 0.80 m respectively from a fixed axis. Find the moment of inertia of the system.

Formula:

Solution:

Answer: The moment of inertia is 2.42 kg m².

Question: Derive the moment of inertia of a uniform thin rod of mass M and length L about an axis through its centre and perpendicular to the rod.

Given: The linear mass density is

A small element of length dx has mass

Formula:

Solution:

Answer: The moment of inertia is I=ML²/12.

Question: Find the moment of inertia of a uniform solid disc of mass 6 kg and radius 0.40 m about its central axis.

Formula:

Solution:

Answer: The moment of inertia is 0.48 kg m².

The radius of gyration of a body about an axis is the distance from the axis at which the entire mass of the body could be imagined to be concentrated without changing its moment of inertia.

Question: A body of mass 8 kg has moment of inertia 2 kg m² about a certain axis. Find its radius of gyration.

Formula:

Solution:

Answer: The radius of gyration is 0.50 m.

If I_cm is the moment of inertia about an axis through the centre of mass, then the moment of inertia about a parallel axis a perpendicular distance d away is

Question: Use the parallel-axis theorem to obtain the moment of inertia of a uniform rod of length L about an axis through one end and perpendicular to the rod.

Given:

The distance from the centre to the end is

Solution:

Answer: The moment of inertia about the end is I=(1/3)ML².

Question: A uniform circular disc has moment of inertia (1/2)MR² about the axis through its centre perpendicular to its plane. Find its moment of inertia about a diameter.

Solution:

By symmetry,

Using the perpendicular-axis theorem,

Therefore,

Answer: The moment of inertia about a diameter is (1/4)MR².

Question: A solid disc of mass 10 kg and radius 0.50 m is acted upon by a constant torque of 5 N m about its central axis. Determine its angular acceleration.

Moment of Inertia:

Rotational Dynamics:

Answer: The angular acceleration is 4 rad/s².

Question: A uniform solid cylinder of mass 4 kg and radius 0.30 m rotates at 20 rad/s about its symmetry axis. Find its rotational kinetic energy.

Moment of Inertia:

Rotational Kinetic Energy:

Answer: The rotational kinetic energy is 36 J.

• Circular motion occurs when an object moves along a path of constant radius about a fixed centre.
• Arc length and angular displacement are related by s=rθ.
• Linear and angular speed are related by v=rω.
• Period and frequency satisfy T=1/f, while ω=2πf=2π/T.
• Uniform circular motion has constant speed but continuously changing velocity.
• Centripetal acceleration is directed toward the centre and has magnitude v²/r=ω²r.
• The required centripetal force is the resultant inward force, F_c=mv²/r.
• Friction, tension, gravity or normal reaction may supply the required centripetal force depending on the situation.
• Banking allows a component of the normal force to contribute to the centripetal force.
• Moment of inertia measures resistance to angular acceleration and depends on both mass and its distribution about the axis.
• For discrete particles, I=Σmr²; for continuous bodies, I=∫r²dm.
• Radius of gyration is defined through I=Mk².
• The parallel-axis theorem is I=I_cm+Md².
• For a planar body, the perpendicular-axis theorem is I_z=I_x+I_y.
• Rotational dynamics is governed by Στ=Iα.

CHAPTER TEN

Gyroscopes, Precession and Advanced Rotational Dynamics

Gyroscopic motion is the characteristic rotational behaviour of a spinning body whose angular momentum responds to an applied torque, often by changing direction rather than simply falling in the direction of the torque.

Question: A flywheel has moment of inertia 5 kg m² and rotates at 30 rad/s. Determine its angular momentum.

Formula:

Solution:

Answer: The angular momentum has magnitude 150 kg m²/s.

Question: A constant torque of 12 N m acts in the direction of the angular momentum of a rotating body for 5 s. Find the magnitude of the change in angular momentum.

Formula:

Solution:

Answer: The angular momentum changes in magnitude by 60 kg m²/s.

A gyroscope is a rapidly rotating body, commonly a wheel or disc, mounted so that its axis of rotation is able to change direction.

A stationary bicycle wheel is relatively easy to reorient. When the same wheel spins rapidly, changing the direction of its axle becomes more difficult because the wheel possesses substantial angular momentum.

Precession is the relatively slow rotation of the spin axis of a rotating body about another direction as a result of an external torque.

For a given gyroscope and geometry, increasing the spin angular speed ω decreases the precession rate Ω. A rapidly spinning gyroscope therefore precesses more slowly than the same gyroscope spinning at a lower rate, under the assumptions of steady precession.

Question: A gyroscope has mass 2 kg, moment of inertia 0.10 kg m², spin angular speed 100 rad/s, and its centre of mass is 0.20 m from the pivot. Find its steady precession angular speed when the axle is horizontal. Take g=9.8 m/s².

Given:

Formula:

Solution:

Answer: The precession angular speed is approximately 0.392 rad/s.

Question: For the gyroscope in Worked Example 3, determine the period of precession.

Formula:

Solution:

Answer: The precession period is approximately 16.0 s.

Question: A gyroscope has angular momentum 40 kg m²/s. A perpendicular torque of magnitude 8 N m produces steady precession. Determine the precession rate for the geometry in which τ=ΩL.

Formula:

Solution:

Answer: The precession angular speed is 0.20 rad/s.

Nutation is an oscillatory or nodding motion of the rotational axis that may accompany precession.

Question: A student on a rotating stool has a total moment of inertia of 6 kg m² and rotates at 2 rad/s. After drawing the arms inward, the moment of inertia becomes 3 kg m². Find the new angular speed.

Formula:

Solution:

Answer: The new angular speed is 4 rad/s.

Question: A disc with moment of inertia 2 kg m² rotates at 10 rad/s. A second non-rotating disc with moment of inertia 3 kg m² is gently dropped onto it. The discs eventually rotate together. Find their common angular speed, assuming negligible external torque.

Initial Angular Momentum:

Final Angular Momentum:

By conservation of angular momentum,

Answer: The discs rotate together at 4 rad/s.

Angular momentum can be conserved even when rotational kinetic energy is not conserved. In the two-disc example, friction between the discs converts some mechanical energy into internal energy while the total angular momentum remains constant because the external torque is negligible.

Rolling motion is the combined translational and rotational motion of a body as it moves over a surface.

For rolling without slipping, the translational speed of the centre of mass and the angular speed satisfy v_cm=Rω.

Question: A wheel of radius 0.40 m rolls without slipping at a centre-of-mass speed of 8 m/s. Find its angular speed.

Formula:

Solution:

Answer: The angular speed is 20 rad/s.

Question: A wheel rolls without slipping with its centre moving at 6 m/s. Determine the instantaneous speeds of the highest point and the point in contact with the ground.

Solution:

Answer: The highest point moves at 12 m/s, while the contact point is instantaneously at rest relative to the ground.

Question: A solid cylinder of mass 4 kg and radius 0.50 m rolls without slipping at 3 m/s. Find its total kinetic energy.

Moment of Inertia:

Angular Speed:

Translational Kinetic Energy:

Rotational Kinetic Energy:

Total:

Answer: The total kinetic energy is 27 J.

Question: A uniform solid cylinder rolls without slipping down a plane inclined at 30° to the horizontal. Find its acceleration. Take g=9.8 m/s².

For a Solid Cylinder:

Formula:

Substitution:

Answer: The acceleration is approximately 3.27 m/s².

Question: A solid sphere rolls without slipping from rest through a vertical height of 5 m. Find its speed at the bottom. Take g=9.8 m/s² and I=(2/5)MR².

Conservation of Energy:

Using I=(2/5)MR² and ω=v/R,

Answer: The sphere reaches the bottom with a speed of approximately 8.37 m/s.

Question: A flywheel has moment of inertia 20 kg m² and rotates at 50 rad/s. Find the rotational kinetic energy stored.

Formula:

Solution:

Answer: The flywheel stores 25,000 J, or 25 kJ, of rotational kinetic energy.

Question: A motor delivers a torque of 40 N m while its shaft rotates at 25 rad/s. Determine its mechanical power output.

Formula:

Solution:

Answer: The mechanical power output is 1000 W = 1.0 kW.

Gyroscopic motion is not a separate set of laws. It follows directly from Newtonian rotational mechanics. Torque changes angular momentum, conservation of angular momentum governs isolated rotating systems, and moment of inertia determines how mass distribution influences rotational response.

• A gyroscope is a rapidly spinning body whose axis is free to change direction.
• For rotation about a fixed principal axis, angular momentum is L=Iω.
• External torque changes angular momentum according to Στ_ext=dL/dt.
• A large angular momentum makes a rapidly spinning body's orientation comparatively resistant to rapid change.
• Precession is the relatively slow rotation of a spinning body's axis under an applied torque.
• For a rapidly spinning gyroscope undergoing steady precession with a horizontal axle under gravity, Ω=Mgl/(Iω).
• Nutation is an oscillatory variation in the inclination of a spinning body's axis that may accompany precession.
• When external torque is zero, total angular momentum remains constant.
• Angular momentum may be conserved even when mechanical energy is not conserved.
• Rolling motion combines translation of the centre of mass and rotation about the centre of mass.
• For pure rolling without slipping, v_cm=Rω.
• The total kinetic energy of a rolling body is the sum of translational and rotational kinetic energies.
• The acceleration of a body rolling down an incline depends on its moment of inertia and therefore on its mass distribution.
• Gyroscopic and rotational principles have applications in rotating machinery, flywheels, navigation, vehicles and many other mechanical systems.

CHAPTER ELEVEN

Gravitation and Motion Under Gravity

Gravitation is the mutual attractive interaction between bodies that possess mass.

Every two point masses attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

If the separation between two masses is doubled, the gravitational force becomes one-quarter of its original value. If the separation is tripled, the force becomes one-ninth.

Question: Two masses of 5 kg and 8 kg are separated by 2 m. Calculate the gravitational force between them. Take G=6.67×10^-11 N m²/kg².

Given:

Formula:

Solution:

Answer: The gravitational attraction is 6.67×10^-10 N.

Question: State the SI unit and dimensions of the universal gravitational constant.

Answer: SI unit: N m²/kg²; dimensions: M^-1L³T^-2.

A gravitational field is a region of space in which a mass experiences gravitational force.

The gravitational field strength at a point is the gravitational force per unit mass experienced by a small test mass placed at that point.

Question: Calculate the gravitational field strength at a distance of 7.0×10⁶ m from the centre of a planet of mass 6.0×10²⁴ kg.

Formula:

Solution:

Answer: The gravitational field strength is approximately 8.17 N/kg, directed toward the planet.

Question: Estimate the acceleration due to gravity at the Earth's surface using M_E=5.97×10²⁴ kg, R_E=6.37×10⁶ m, and G=6.67×10^-11 N m²/kg².

Formula:

Solution:

Answer: The calculated value is approximately 9.81 m/s².

Question: At what height above a planet's surface would the gravitational acceleration become one-quarter of its surface value, assuming the planet has radius R?

Given:

Formula:

Solution:

Taking positive square roots,

Answer: The height is equal to one planetary radius.

The linear relation with depth assumes uniform density. The real Earth is not of uniform density, so the actual variation is more complicated.

Question: In a uniform spherical planet, determine the gravitational acceleration at a depth equal to one-quarter of the planetary radius.

Given:

Formula:

Solution:

Answer: The gravitational acceleration is 0.75g.

The gravitational potential at a point is the gravitational potential energy per unit mass of a small test mass placed at that point.

Question: Calculate the gravitational potential at a point 4×10⁶ m from the centre of a planet of mass 2×10²⁴ kg.

Formula:

Solution:

Answer: The gravitational potential is approximately -3.34×10⁷ J/kg.

Question: A 100 kg object is at a distance of 8×10⁶ m from the centre of a planet of mass 6×10²⁴ kg. Find its gravitational potential energy.

Formula:

Solution:

Answer: The gravitational potential energy is approximately -5.00×10⁹ J.

Free fall is motion in which gravity is the only force acting on a body.

Question: A stone is dropped from rest and falls freely for 3 s. Find its speed and distance fallen. Take g=9.8 m/s².

Speed:

Distance:

Answer: Speed = 29.4 m/s downward; distance fallen = 44.1 m.

Question: A stone is dropped from a height of 78.4 m. How long does it take to reach the ground, neglecting air resistance?

Formula:

Solution:

Answer: The stone reaches the ground after 4 s.

Question: A ball is thrown vertically upward at 19.6 m/s. Find the time to reach maximum height, maximum height, and total time to return to the point of projection. Take g=9.8 m/s².

Time to Maximum Height:

Maximum Height:

Total Time:

Answer: Time to maximum height = 2 s; maximum height = 19.6 m; total time of flight = 4 s.

Question: A ball is projected vertically upward at 30 m/s. Find its velocity after 2 s. Take upward as positive and g=9.8 m/s².

Formula:

Solution:

Answer: The ball is still moving upward at 10.4 m/s.

Question: A stone is thrown vertically downward at 5 m/s from a height of 40 m. Find its speed just before reaching the ground. Take g=9.8 m/s².

Given:

Formula:

Solution:

Answer: The speed just before impact is approximately 28.44 m/s.

Astronauts in an orbiting spacecraft appear weightless because both the spacecraft and the astronauts are continuously falling under gravity together. Their orbital motion prevents them from striking the Earth's surface.

Question: A ball is thrown vertically upward from the top of a 30 m building with speed 14 m/s. Find the maximum height above the ground. Take g=9.8 m/s².

Additional Height Above the Building:

Total Height Above Ground:

Answer: The maximum height above the ground is 40 m.

Question: An object falls freely from rest. Determine its speed after falling 20 m. Take g=9.8 m/s².

Formula:

Solution:

Answer: The speed is approximately 19.8 m/s downward.

• Gravitation is the mutual attractive interaction between masses.
• Newton's law of universal gravitation is F=Gm₁m₂/r².
• The universal gravitational constant is approximately 6.67×10^-11 N m²/kg².
• Gravitational field strength is force per unit mass.
• For a point or spherical mass, g=GM/r².
• At the surface of a spherical planet, g=GM/R².
• Gravitational acceleration decreases with increasing altitude.
• For a uniform spherical planet, gravitational acceleration decreases linearly with depth below the surface.
• Taking zero potential at infinity, gravitational potential is V=-GM/r.
• Gravitational potential energy is U=-GMm/r.
• The gravitational field is related to potential through g=-∇V.
• Free fall is motion under gravity alone.
• Near the Earth's surface, free-fall motion can be treated using constant acceleration g.
• For vertical upward projection, maximum height is H=u²/(2g) and time to maximum height is u/g.
• Apparent weightlessness occurs when the normal supporting force becomes zero during free fall.

CHAPTER TWELVE

Planetary Motion, Satellites and Escape Velocity

Every planet moves in an elliptical orbit with the Sun located at one focus of the ellipse.

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Kepler's second law is a geometrical expression of angular momentum conservation for motion under a central force.

For bodies orbiting the same central mass, the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

Question: Satellite B orbits the same planet as Satellite A at four times A's orbital radius. If A's period is 2 h, find B's orbital period.

Formula:

Given:

Solution:

Answer: Satellite B has an orbital period of 16 h.

The orbital velocity of a body in a circular orbit is the tangential speed required for gravitational attraction to provide exactly the centripetal force needed to maintain that orbit.

Question: Estimate the circular orbital speed of a satellite at a distance of 7.0×10⁶ m from the centre of the Earth. Take G=6.67×10^-11 and M_E=5.97×10²⁴ kg.

Formula:

Solution:

Answer: The orbital speed is approximately 7.54 km/s.

Question: A satellite moves in a circular orbit of radius 7.0×10⁶ m around the Earth. Using GM_E≈3.986×10¹⁴ m³/s², calculate its orbital period.

Formula:

Solution:

Converting to minutes,

Answer: The period is approximately 5.83×10³ s, or 97.2 minutes.

Question: A 500 kg satellite moves in a circular orbit of radius 8.0×10⁶ m around Earth. Take GM_E=3.986×10¹⁴ m³/s². Find its total mechanical energy.

Formula:

Solution:

Answer: The total mechanical energy is approximately -1.25×10¹⁰ J.

An artificial satellite is a human-made object intentionally placed into orbit around the Earth or another astronomical body.

In orbital equations, r is measured from the centre of the Earth or other central body, not from its surface. If the altitude is h, then r=R+h.

A geostationary satellite is a satellite in a circular equatorial orbit that moves in the same rotational sense as the Earth and has an orbital period equal to the Earth's rotation period, so it appears fixed above one point on the equator.

Question: Explain why a geostationary satellite cannot orbit directly above the North Pole.

Answer: A geostationary satellite must remain above the same point on the Earth. This requires a circular orbit in the equatorial plane with the same angular speed and rotational direction as the Earth. An orbit above the poles does not remain fixed over a single surface point and is therefore not geostationary.

Escape velocity is the minimum initial speed required for an object to move away from a gravitational body and reach infinitely large distance with zero remaining speed, assuming no further propulsion and neglecting atmospheric resistance and other bodies.

Question: Estimate the escape velocity from the Earth's surface using g=9.8 m/s² and R=6.37×10⁶ m.

Formula:

Solution:

Answer: The escape velocity is approximately 11.2 km/s.

The mass of the escaping object cancels from the derivation. Therefore, in the idealized model, escape velocity does not depend on the mass of the escaping body.

Question: The circular orbital speed at a certain radius around a planet is 7.5 km/s. Find the escape speed from the same radius.

Formula:

Solution:

Answer: The escape speed is approximately 10.61 km/s.

Question: A spacecraft at a certain point has kinetic energy 8×10⁹ J and gravitational potential energy -1.1×10¹⁰ J. Determine whether it is gravitationally bound.

Formula:

Solution:

Answer: Since the total mechanical energy is negative, the spacecraft is gravitationally bound.

Question: A satellite is moved from a circular orbit of radius r to a larger circular orbit of radius 4r. Compare the new circular orbital speed with the original speed.

Formula:

Original speed:

New speed:

Answer: At four times the orbital radius, the circular orbital speed is half the original speed.

Question: A satellite in a circular orbit has period 3 h. Another satellite orbits the same planet at nine times the first satellite's orbital radius. Determine the second satellite's period.

Kepler's Third Law:

Answer: The second satellite has a period of 81 h.

Question: A satellite has circular orbital speed 8 km/s. If its orbital radius is doubled while it remains in a circular orbit around the same planet, determine its new speed.

Relationship:

Therefore,

Answer: The new circular orbital speed is approximately 5.66 km/s.

Planetary and satellite motion brings together the major ideas developed throughout mechanics: vectors, kinematics, Newton's laws, circular motion, work and energy, conservation principles and gravitation. The same fundamental laws describe both ordinary terrestrial motion and the large-scale motion of astronomical bodies.

• Kepler's first law states that planetary orbits are ellipses with the Sun at one focus.
• Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal times.
• Kepler's second law is closely connected with conservation of angular momentum.
• Kepler's third law gives T²∝a³ for bodies orbiting the same central mass.
• For circular motion under gravity, gravitational force supplies the centripetal force.
• Circular orbital speed is v=√(GM/r).
• The circular orbital period is T=2π√(r³/GM).
• For a satellite in circular orbit, kinetic energy is GMm/(2r).
• Its gravitational potential energy is -GMm/r.
• Its total mechanical energy is -GMm/(2r).
• A geostationary satellite must orbit in the equatorial plane, in the direction of Earth's rotation, with the same angular period as the Earth's rotation.
• Escape velocity from radius R is v_e=√(2GM/R).
• At the same radius, escape speed is √2 times the circular orbital speed.
• Negative total gravitational mechanical energy indicates a bound orbit.
• Zero or positive total gravitational energy corresponds to the threshold of escape or an unbound trajectory.
• Satellite mechanics has major applications in communication, navigation, meteorology, Earth observation and scientific research.