Pearson Edexcel Physics

Unit 1: Mechanics and Materials

1.3 Mechanics

Pearson Edexcel Physics

Unit 1: Mechanics and Materials

1.3 Mechanics

Candidates will be assessed on their ability to::
1. Be able to use the equations for uniformly accelerated motion in one dimension

[math]s = \frac{(u + v)t}{2} \\
v = u + at \\
s = ut + \frac{1}{2}at^2 \\
v^2 = u^2 + 2as[/math]
2. Be able to draw and interpret displacement-time, velocity-time and acceleration time graphs
3. Know the physical quantities derived from the slopes and areas of displacement time, velocity-time and acceleration-time graphs, including cases of non-uniform acceleration and understand how to use the quantities
4. Understand scalar and vector quantities and know examples of each type of quantity and recognize vector notation
5. Be able to resolve a vector into two components at right angles to each other by drawing and by calculation
6. Be able to find the resultant of two coplanar vectors at any angle to each other by drawing, and at right angles to each other by calculation
7. Understand how to make use of the independence of vertical and horizontal motion of a projectile moving freely under gravity
8. Be able to draw and interpret free-body force diagrams to represent forces on a particle or on an extended but rigid body using the concept of Centre of gravity of an extended body
9. Be able to use the equation ∑F = ma, and understand how to use this equation in situations where m is constant (Newton’s second law of motion), including Newton’s first law of motion where a = 0, objects at rest or travelling at constant velocity Use of the term ‘terminal velocity’ is expected.
10. Be able to use the equations for gravitational field strength [math]g = \frac{F}{m}[/math] and weight W = mg
11. CORE PRACTICAL 1: Determine the acceleration of a freely-falling object
12. Know and understand Newton’s third law of motion and know the properties of pairs of forces in an interaction between two bodies
13. Understand that momentum is defined as p = mv
14. Know the principle of conservation of linear momentum, understand how to relate this to Newton’s laws of motion and understand how to apply this to problems in one dimension
15. Be able to use the equation for the moment of a force = Fx, moment of force  where x is the perpendicular distance between the line of action of the force and the axis of rotation
16. Be able to use the concept of center of gravity of an extended body and apply the principle of moments to an extended body in equilibrium
17. Be able to use the equation for work∆W = F∆s , including calculations when the force is not along the line of motion
18. Be able to use the equation [math]E_k = \frac{1}{2}mv^2[/math] for the kinetic energy of a body
19. Be able to use the equation [math]\Delta E_{\text{grav}} = mg\Delta h[/math] for the difference in gravitational potential energy near the Earth’s surface
20. Know, and understand how to apply, the principle of conservation of energy including use of work done, gravitational potential energy and kinetic energy
21. Be able to use the equations relating power, time and energy transferred or work done

[math]P = \frac{E}{t} \quad \text{and} \quad P = \frac{W}{t}[/math]
22. Be able to use the equations

[math]\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}[/math]

And

[math]\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}[/math]

  • 1) The equations for uniformly accelerated motion in one dimension:

  • ⇒ Acceleration:
  • Acceleration is defined as the rate of change of velocity. Therefore, it must include the direction in which the speed is changing.
  • Quantity: (vector quantity)
  • Formula: 
  • [math]\begin{gather}
    \text{Acceleration (ms}^{-2}\text{)} = \frac{\text{Change in velocity (ms}^{-1}\text{)}}{\text{time taken to change the velocity (s)}} \\
    a = \frac{v – u}{t} \\
    a = \frac{\Delta v}{t}
    \end{gather}[/math]
  • Where u is the initial velocity and v in the final velocity.
  • The vector nature of acceleration is that if an object changes only the direction of its velocity, it is accelerating, while remaining at a constant speed. Similarly, deceleration represents a negative change in velocity, and so cloud be stated as a negative acceleration.
  • ⇒ Zero Acceleration:
  • If an object has no resultant force acting on it then it does not accelerate. This is uniform motion.
  • On its motion are very easy, as they simply involve the basic velocity equation.
  • [math] v = \frac{s}{t}[/math]
  • o The velocity is the same at the beginning and end of the motion, and need to find the displacement travelled, it is a simple case of multiplying velocity by time.
  • [math]s = v × t[/math]
  • ⇒ Constant Acceleration:
  • Some equations that allow us to work out the motion of an object that is moving with a constant acceleration. For these kinematics equations, the first step is to define the five variables used:
  • – s (displacement (m)
  • – u (initial velocity (m.s-1))
  • – v (final velocity (m.s-1))
  • – a (acceleration (m.s-2))
  • – t (time (s))
  • By re-arranging the equation that defined acceleration,
  • The first kinematics equation:
  • [math]v = u + at → 1[/math]
  • ⇒ Example:
  • If a stone is dropped off a cliff (see fig B) and takes three seconds to hit the ground, what is its speed when it does hit the ground? Identify the three things we know:
  • – Falling under gravity, so [math]a = g = 9.81 m.s^2[/math] (constant acceleration, so the kinematics equations can be used)
  • – Starts at rest, so [math]u = 0 m.s^{-1}[/math]
  • – Time to fall t = 3s
  • [math]\begin{gather}
    v = u + at \\
    v = (0) + (9.81)(3) \\
    v = 29.43 \text{ m.s}^{-1}
    \end{gather}[/math]
  • ⇒ Distance from average speed:
  • As the kinematics equations only work with uniform acceleration, the average speed during any acceleration will be halfway from the initial velocity to the final velocity. Therefore, the distance travelled is the average speed multiplied by the time:
  • [math]s = \frac{(u + v)}{2} \times t \to 2[/math]
  • Example:
  • The same stone dropping off the cliff as in the previous example, we could work out how high the cliff is.
  • Identify the three things we know:
  • – Final velocity came to be v = 29.4 [math]m.s^{-1}[/math]
  • – Starts at rest, so u = 0 [math]m.s^{-1}[/math]
  • – Time to fall t = 3s
  • [math]\begin{gather}
    s = \frac{(u + v)}{2} \times t \\
    s = \frac{0 + 29.4}{2} \times 3 \\
    s = 44.1 \text{ m}
    \end{gather}[/math]
  • Equation 1 and equation 2 combine both equations
  • [math] v = u + at[/math]
  • And
  • [math]\begin{gather}
    s = \frac{(u + v)}{2} \times t \\
    s = \frac{u + (u + at)}{2} \times t \\
    s = \left(\frac{2u}{2} + \frac{at}{2}\right) t \\
    s = ut + \frac{1}{2}at^2 \to 3
    \end{gather}[/math]
  • By equation 2
  • [math]s = \frac{(u + v)}{2} \times t \\
    t = \frac{2s}{u + v}[/math]
  • And put in equation 1
  • [math]\begin{gather}
    v = u + at \\
    v = u + a\left( \frac{2s}{u + v} \right) \\
    v(u + v) = u(u + v) + 2as \\
    vu + v^2 = u^2 + uv + 2as \\
    v^2 = u^2 + 2as \to 4
    \end{gather}[/math]
  • Example:
  • The stone’s final velocity would be
  • – Falling under gravity, so [math]a = g = 9.8 m.s^{-2}[/math]
  • – Start at rest, so u = 0 [math]m.s^{-1}[/math]
  • – Height to fall s = 44.1 m
  • [math]\begin{gather}
    v^2 = u^2 + 2as \\
    v^2 = (0)^2 + 2(9.8)(44.1) \\
    v^2 = 865.2 \\
    v = \sqrt{865.2} \\
    v = 29.4 \text{ m.s}^{-1}
    \end{gather}[/math]
Kinematics Equation Quantity not used
[math]v = u + at[/math] Distance
[math]s = \frac{(u + v)}{2} \times t [/math] Acceleration
[math]s = ut + \frac{1}{2} at^2[/math] Final Velocity
[math]v^2 = u^2 + 2as[/math] Time

  • 2) To draw and interpret displacement-time, velocity-time and acceleration time graphs:

  • 3) Know the physical quantities derived from the slopes and areas of displacement time, velocity-time and acceleration-time graphs, including cases of non-uniform acceleration and understand how to use the quantities

  • ⇒ Displacement-time graphs:
  • A boat trip on a river, we could monitor the location of the boat over the hour that it travels for and plot the displacement-time graph for these movements.
  • Depending on what information we want the graph to provide, it is often simpler to draw a distance-time graph in which the direction of movement is ignored.
  • The graphs shown in fig 1 are examples of plotting position against time, and show how a distance-time graph cannot decrease with time.
  • A displacement-time graph could have parts of it in the negative portions of the y-axis, if the movement went in the opposite direction at some points in time.
  • Figure 1 A comparison of the displacement-time graph of the boat trip up and down a river with its corresponding distance-time graph
  • The gradient of the d-t graph in figure 1 will tell us how fast the boat was moving Gradient is found from the ratio of changes in the y-axis divided by the corresponding change on the x-axis, so for a distance-time graph:
  • [math]\begin{gather}
    \text{gradient} = \frac{\text{distance}}{\text{time}} = \text{speed (m.s}^{-1}\text{)} \\
    v = \frac{d}{t}
    \end{gather}[/math]
  • For a displacement-time graph:
  • [math]\begin{gather}
    \text{gradient} = \frac{\text{displacement}}{\text{time}} = \text{velocity (m.s}^{-1}\text{)} \\
    v = \frac{\Delta s}{t}
    \end{gather}[/math]
  • ⇒ Velocity – Time graph:
  • A velocity-time graph will show the velocity of an object over time. Calculated that the velocity of the boat on the river was 0.167[math]m.s^{-1}[/math] up river for the first 15 minutes of the journey.
  • In figure 2, the line is constant at +0.167[math]m.s^{-1}[/math] for the first 15 minutes.
  • Also notice that the velocity axis includes negative values, so that the difference between travelling up river (positive y-axis values) and down river (negative y-axis values) can be represented.
  • Figure 2 Velocity – time graph of the boat trip
  • Example:
  • Between 15 and 20 minutes on the graphs, the boat slows evenly to a stop. The acceleration here can be calculated as the gradient:
  • [math]\begin{gather}
    \text{gradient} = \frac{\Delta v}{\Delta t} \\
    \text{gradient} = \frac{v – u}{t} \\
    \text{gradient} = \frac{0 – 0.167}{5 \times 60} \\
    \text{gradient} = \frac{-0.167}{300} \\
    \text{gradient} = -0.0006 \ \text{m.s}^{-2}
    \end{gather}[/math]
  •  So, the acceleration is: [math]a = -0.0006 \ \text{m.s}^{-2}[/math]
  • ⇒ Acceleration – time graph:
  • Acceleration-time graphs show how the acceleration of an object changes over time. In many instances the acceleration is zero or a constant value, in which case an acceleration-time (a-t) graph is likely to be of relatively little interest.
  • For example, the object falling in our investigation above will be accelerated by gravity throughout. Assuming it is relatively small, air resistance will be minimal, and the a-t graph of its motion would be a horizontal line at [math] a = -9.81 ms^{-2}[/math]. Compare this with your results to see how realistic it is to ignore air resistance.
  • For a larger object falling for a long period, such as a skydiver, then the acceleration will change over time as the air resistance increases with speed.
  • Figure 3 Acceleration – time graph for a skydiver

  • 4) Understand scalar and vector quantities and know examples of each type of quantity and recognize vector notation:

  • ⇒ Scalar Quantity:
  • Scalars have only magnitude (size), while vectors have both magnitude and direction.
  • – Mass: The amount of matter in an object.
  • – Distance: The total length of a path traveled.
  • – Speed: How fast an object is moving.
  • – Time: The duration of an event.
  • – Energy: The ability to do work.
  • – Temperature: The measure of hotness or coldness.
  • – Volume: The amount of space a substance occupies.
  • ⇒ Vector Quantity:
  • Vectors have both magnitude and direction. They are represented by arrows, with the length of the arrow representing the magnitude and the direction of the arrow representing the direction.
  • Figure 4 Vector quantity
  • – Displacement:
  • The change in position of an object, including its direction.
  • – Velocity:
  • The rate at which an object changes position, including its direction
  • – Acceleration:
  • The rate at which an object changes velocity.
  • – Force:
  • A push or pull on an object, including its direction.

  • 5) Be able to resolve a vector into two components at right angles to each other by drawing and by calculation

  • 6) Be able to find the resultant of two coplanar vectors at any angle to each other by drawing, and at right angles to each other by calculation

  • Vectors that act perpendicular to each other can be combined to produce a single vector that has the same effect as the two separate vectors.
  • A single vector, we can find a pair of vectors at right angles to each other that would combine to give our single original vector. This reverse process is called resolution or resolving vectors.
  • Figure 5 Resolve a vector into two components at right angles to each other
  • ⇒ Alternative resolution angles:
  • The velocity of an object, that resolve this into a pair of velocity vectors at right angles to each other.
  • The choice of direction of the right-angle pair, though, is arbitrary, and can be chosen to suit a given situation.
  • Example:
  • A submarine descending underwater close to an angled.
  • Figure 6 Submarine velocity resolved into horizontal and vertical components.
  • The speed at which the submarine is getting closer to the seafloor is perhaps what the submarine commander is most interested in.
  • This might be discovered by calculating the submarine’s velocity.
  • The right-angled pair with that will be a component perpendicular to the seafloor, and becomes a component parallel to the seabed.
  • The submarine commander will be able to know how fast he is getting to the seabed thanks to this [math]v_{perp}[/math].
  • Figure 7 Submarine velocity resolved into components parallel and perpendicular to the seabed.

  • 7) Understand how to make use of the independence of vertical and horizontal motion of a projectile moving freely under gravity

  • ⇒ Projectile motion:
  • Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile, and its path is called its trajectory.
  • Figure 8 Projectile motion
  •  Example:
  • There’s a cannonball out of the cannon. Since air resistance is disregarded in these computations, the object’s weight will be the sole force at work.
  • Therefore, gravity will act as the acceleration in all vertical motion, which will obey kinematics equations. Since there won’t be any horizontal force, there won’t be any acceleration, and [math]v = s/t[/math].
  • ⇒ Horizontal Throws:
  • An item thrown horizontally will have no vertical velocity at first. But because of gravity, its velocity will curve downward in the form of a parabola, like the stone in the figure 9
  • Figure 9 Vertical acceleration on a horizontally moving stone
  • o It will be the same time to strike the shore, t, as if the stone had just been dropped.
  • – u = 0 [math]ms^{-1}[/math],
  • – a = -9.81 [math]ms^{-2}[/math],
  • – s = -60 m is the height plummeted.
  • [math]\begin{gather}
    s = ut + \frac{1}{2}at^2 \\
    s = (0)t + \frac{1}{2}at^2 \\
    s = \frac{1}{2}at^2 \\
    t = \sqrt{\frac{2s}{a}} \\
    t = \sqrt{\frac{2(-60)}{-9.81}} \\
    t = \sqrt{12.23} \\
    t = 3.5 \ \text{s}
    \end{gather}[/math]
  • The distance travelled horizontally, d:
  • [math]\begin{gather}
    v = \frac{d}{t} \\
    d = v \times t \\
    d = 8.2 \times 3.5 \\
    d = 28.7 \ \text{m}
    \end{gather}[/math]
  • ⇒ Vertical Throws:
  • Consider tossing a pal a ball. The ball moves both forward and upward. An item thrown with a vertical upward component of motion will have a symmetrical trajectory, according to a frequent assumption in these computations.
  • The vertical velocity is briefly zero at the highest point. It will take half of the flying time to get to this position.
  • Figure 10 Considering only the vertical component of velocity.
  • In figure 10: A ball is thrown with a vertical velocity component of 5.5 ms¹
  • If someone tossed the ball vertically up and caught it again, this demonstrates once more the degree to which vertical and horizontal movements are.
  • It’s possible that a beginning velocity at an angle is mentioned; thus, in order to determine that vertical[math] = 5.5 m s^{-1}[/math] in this case, we must split the velocity vector into its horizontal and vertical components.
  • ⇒ Calculate the vertical distance:
  • First, look at the second query: the kinematics equations may be used since uniform acceleration under gravity equals [math]a = – 9.81 m s^{-2}[/math]. To determine the height, s, and we know that [math]u = 5.5 ms^{-1}[/math] and [math]v = 0 ms^{-1}[/math] at the top of the route.
  • [math]\begin{gather}
    v^2 = u^2 + 2as \\
    s = \frac{v^2 – u^2}{2a} \\
    s = \frac{(0)^2 – (5.5)^2}{2(-9.8)} \\
    s = \frac{-30.25}{-19.6} \\
    s = 1.54 \ \text{m}
    \end{gather}[/math]
  • The time of flight for the ball will be just the time taken to rise and fall vertically. We find the time to reach the highest point, and then double that value. We know u = 5.5 [math]ms^{-1}[/math]; at the top of the path, [math]v = 0 ms^{-1}[/math]; and we want to find the time, t.
  • [math]\begin{gather}
    v = u + at \\
    t = \frac{v – u}{a} \\
    t = \frac{0 – 5.5}{-9.81} \\
    t = 0.56 \ \text{s}
    \end{gather}[/math]
  • So, the overall time of flight will be 0.56 seconds doubled: total time = 1.12 s.

  • 8) Be able to draw and interpret free-body force diagrams to represent forces on a particle or on an extended but rigid body using the concept of Centre of gravity of an extended body:

  • ⇒ Free-body force diagram:
  • It may be easier to determine how an object will move if we can identify the forces operating on it. This is often accomplished by drawing a free-body force diagram, in which the object is separated and all acting forces are shown at their respective positions. Other objects and the forces operating upon them are not depicted.
  • ⇒ Adding perpendicular forces:
  • The vector sum of the effects of two forces acting perpendicularly, or at right angles, to one another will have an impact on an object.
  • To determine the outcome, we must sum the sizes while taking the instructions into account.
  • Figure 11 These two rugby players are each putting a force on their opponent. The forces are at right angles, so the overall effect would be to move him in a third direction, which we could calculate.
  • ⇒ Magnitude of the resultant force:
  • To calculate the resultant magnitude of two perpendicular forces, we can draw them, one after the other, as the two sides of a right-angled triangle and use Pythagoras’ theorem to calculate the size of the hypotenuse.
  • Figure 12 The resultant force here is calculated using Pythagoras’ theorem:
  • [math]\begin{gather}
    \text{Pythagoras’ theorem:} \quad F = \sqrt{(70)^2 + (110)^2} \\
    F = \sqrt{17000} \\
    F = 130 \ \text{N}
    \end{gather}[/math]
  • ⇒ Direction of the resultant force:
  • As forces are vectors, when we find a resultant force, it must have both magnitude and direction. For perpendicular forces (vectors), trigonometry will determine the direction.
  • Figure 13 The resultant force here is at an angle up from the
  • [math]\begin{gather}
    \theta = \tan^{-1}\left(\frac{70}{110}\right) \\
    \theta = \tan^{-1}(0.64) \\
    \theta = 32^\circ
    \end{gather}[/math]

  • 9) Be able to use the equation [math]\Sigma F = ma [/math] , and understand how to use this equation in situations where m is constant (Newton’s second law of motion), including Newton’s first law of motion where a = 0, objects at rest or travelling at constant velocity Use of the term ‘terminal velocity’ is expected.

  • ⇒ Newton’s Second Law of motion:
  • The second law asserts that an item’s acceleration depends on two factors: the mass of the object and the net force exerted on it.
  • An object’s acceleration is directly influenced by the net force applied on it, and it is inversely correlated with its mass.
  • [math]\begin{gather}
    \Sigma F = ma \\
    \text{Resultant force (N)} = \text{mass (kg)} \times \text{acceleration (m.s}^{-2}\text{)}
    \end{gather}[/math]
  • Figure 14 Newton’s second law apply on rocket
  • This relationship allows us to calculate the acceleration due to gravity (g) if we measure the force (F) accelerating a mass (m) downwards.
  • [math]\begin{gather}
    F = ma \\
    F = mg \\
    g = \frac{F}{m}
    \end{gather}[/math]
  • Acceleration due to gravity near the surface of the Earth is about 9.81 [math]ms^{-2}[/math]. An object falling in a vacuum does accelerate at this rate. However, it is unusual for objects to be dropped near the surface of the Earth in a vacuum (in nearly all such cases a physics teacher is likely to be demonstrating to a class).
  • In reality, in order to calculate an object’s actual acceleration when falling, we need to take account of all the forces acting on it, combine these to find a resultant ΣF force, and then use Newton’s second law [math]a = \frac{\Sigma F}{m}[/math] to calculate the resulting acceleration.
  • For a falling object such as a skydiver, this means we need to include the weight, the upthrust caused by the object being in the fluid air, and the viscous drag force caused by the movement.
  • The difficult part is that the viscous drag varies with speed through the fluid, and speed is constantly changing as a result of the acceleration.
  • Usually, we consider the equilibrium situation, in which the weight exactly balances the sum of upthrust and drag, which means that the falling velocity remains constant. This constant velocity is the terminal velocity.

  • 10) Be able to use the equations for gravitational field strength [math]g = \frac{F}{m}[/math]  and weight [math]W = mg[/math]

  • Gravitational field strength, which basically describes how strong gravity is at a given place in space, is the force that gravity applies to a unit mass there.
  • It’s a measurement of the acceleration brought on by gravity there. Any object at that position is subject to a stronger gravitational pull when the gravitational field intensity is higher.
  • The calculation of weight, or the gravitational force, which is commonly expressed mathematically as W, is one of the most significant uses of Newton’s second rule.
  • People sometimes fail to understand that gravity is actually an acceleration when they discuss it. An item accelerates towards the Earth’s Centre when it is dropped.
  • According to Newton’s second law, an object’s acceleration is caused by the net external force acting on it.
  • The gravitational force—that is, the item’s weight—is the sole external force acting on a falling object when air resistance is minimal.
  • Consider a mass-m object descending towards Earth. It only feels the gravitational force, also known as weight, which is described by W.
  • Figure 15 Newton’s second law apply on a weight
  • – Newton’s second law, which states that [math]F_{net} = ma[/math].
  • – We have [math]F_{net} = W[/math] since the gravitational force is the sole force acting on the item.
  • – We have [math]a = g[/math]since we know that an object’s acceleration due to gravity equals g.
  • – When these two phrases are entered into Newton’s second law, the result is
  • [math]W = mg[/math]
  • Example:
  • A mass m’s gravitational pull. Since [math]g = 9.80 m/s^2[/math] on Earth, the weight of a 1.0 kg item on Earth (ignoring for the moment the weight’s direction)
  • [math]\begin{gather}
    W = mg \\
    W = (1)(9.8) \\
    W = 9.8 \ \text{N}
    \end{gather}[/math]

  • 11) CORE PRACTICAL 1: Determine the acceleration of a freely-falling object:

  • We can measure the acceleration caused by gravity by using a device that timed an object’s descent under gravity.
  • In this experiment, we time how long it takes for an item to fall from a given height under the force of gravity, then we change the height and take another measurement.
  • Figure 16 The freefall time of an object from different heights allows us to find the acceleration due to gravity,
  • If vary the height from which the object falls, the time taken to land will vary. The kinematics equations tell us that:
  • [math]s = ut + \frac{1}{2}at^2[/math]
  • It always begins at rest, thus u = 0 throughout; gravity causes the acceleration, g; and the distance is the height at which it is launched, h. So:
  • [math]\begin{gather}
    h = \frac{1}{2}gt^2 \\
    t^2 = \frac{2h}{g}
    \end{gather}[/math]
  • Compare this equation with the equation for a straight-line graph:
  • [math]y = mx + c[/math]
  • We plot a graph of h on the x-axis and [math]t^2[/math] on the y-axis to give a straight best-fit line. The gradient of the line on this graph will be 2/g. from which we can find g.
  • We could find a value for g by taking a single measurement from this experiment and using the equation to calculate it:
  • [math]g = \frac{2h}{t^2}[/math]
  • However, a single measurement in any experiment is subject to uncertainty from both random and systematic errors. We can reduce such uncertainties significantly by taking many readings and plotting a graph, which leads to much more reliable conclusions.

  • 12) Know and understand Newton’s third law of motion and know the properties of pairs of forces in an interaction between two bodies:

  • ⇒ Newton’s third law of motion:
  • “Newton’s third law simply states that for every action there is an equal and opposite reaction.”
  • It was developed by Sir Isaac Newton in the 17th century. The four forces of flight are always forces acting on an aircraft: thrust (forward), drag (rearward), lift (up), and weight (down).
  • Figure 17 Newton’s third law of motion
  • When a skateboarder pushes off from a wall, they exert a force on the wall with their hand. At the same time, the wall exerts a force on the skateboarder’s hand.
  • This equal and opposite reaction force is what they can feel with the sense of touch, and as the skateboard has very low friction, the wall’s push on them causes acceleration away from the wall.
  • As the wall is connected to the Earth, the Earth and wall combination will accelerate in the opposite direction.

  • 13) Understand that momentum is defined as [math]p = mv[/math]

  • ⇒ Momentum:
  • “Momentum is the product of an object’s mass and its velocity,”
  • Formula:
  • [math]p = mv[/math]
  • Unit:
  • [math]kgm.s^{-1}[/math]
  •  Quantity:
  • As momentum is the product of mass (a scalar) and velocity (a vector), momentum is a vector.
  • Figure 18 Momentum
  •                 Example:
  • An athletics hammer has a mass of 7.26 kg (men’s competition standard) and can be released at speeds in excess of 25 [math]ms^{-1}[/math]. Find momentum.
  • Solution:
  • Mass = 7.26 kg
  • Velocity = 25 [math]ms^{-1}[/math]
  • Formula:
  • [math]p = mv[/math]
  • Solve:
  • [math]\begin{gather}
    p = mv \\
    p = (7.26)(25) \\
    p = 181.5 \ \text{kg m.s}^{-1}
    \end{gather}[/math]

  • 14) Know the principle of conservation of linear momentum, understand how to relate this to Newton’s laws of motion and understand how to apply this to problems in one dimension

  • The mathematical formula for Newton’s second law is F = ma
  • Actually, only when the mass stays constant does that formula hold true.
  • Newton’s second law was first expressed as follows in his 1687 book Philosophiae naturalis principia mathematica:
  • “A body’s rate of change of momentum is directly proportional to the force that is applied to it and points in the same direction as the force.”
  • [math]\begin{gather}
    F = \frac{dp}{dt} \\
    F = \frac{d(mv)}{dt}
    \end{gather}[/math]
  • Here F is the applied force, and [math]\frac{dp}{dt}[/math] is the rate of change of the momentum in the direction of the force.
  • The [math]F = \frac{d(x)}{dt}[/math] term is a mathematical expression meaning the rate of dt change of x, or how quickly x changes. However, if the quantities are not being measured over a very short timescale, we can express this using average changes.
  • ⇒ Newton’s third law:
  • Newton’s third law is intimately related to conservation of momentum. Recall that this informed us that for any force, there an opposing and equal force.
  • It is necessary to counteract a force that changes momentum in one dimension if we consider it to be a means of altering momentum ([math]F = \frac{dp}{dt}[/math]).
  • Example:
  • If an apple falls from a tree due to the Earth’s gravitational attraction, it accumulates velocity in the direction of the Earth.
  • The Earth must acquire an equal and opposite momentum in order for momentum to be conserved. This is then brought on by the apple’s equal and opposite gravitational pull-on Earth.
  • We are unable to detect the Earth’s acceleration due to its immense mass.
  • Figure 19 Conservation of momentum causes equal and opposite forces, as Newton explained in his third law of motion.

  • 15) Be able to use the equation for the moment of a force moment of force = Fx , where x is the perpendicular distance between the line of action of the force and the axis of rotation.

  • 16) Be able to use the concept of center of gravity of an extended body and apply the principle of moments to an extended body in equilibrium

  • ⇒ The moment of a force:
  • The tendency to cause rotation is called the moment of a force.
  • [math]\begin{gather}
    \text{Moment (Nm)} = \text{force (N)} \times \text{perpendicular distance from the pivot to the line of action of the force (m)} \\
    \text{Moment} = F \times x
    \end{gather}[/math]
  • ⇒ Principle of moments:
  • All the forces acting on an object and the resultant force, accounting for their directions, is zero, then the object will be in equilibrium.
  • The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.
  • Figure 20 Balanced moments create an equilibrium situation
  • ⇒ Centre of Gravity:
  • The weight of an object is caused by the gravitational attraction between the Earth and each particle contained within the object. The sum of all these tiny weight forces appears to act from a single point for any object, and this point is called the Centre of gravity.
  • For a symmetrical object, can calculate the position of its Centre of gravity, as it must lie on every line of symmetry.
  • Figure 21 The Centre of gravity of a symmetrical object lies at the intersection of all lines of symmetry.
  • The point of intersection of all lines of symmetry will be the Centre of gravity. Figure E illustrates this with two-dimensional shapes, but the idea can be extended into three dimensions. For example, the Centre of gravity of a sphere is at the sphere’s Centre.

  • 17) Be able to use the equation for work ∆W = F∆s , including calculations when the force is not along the line of motion

  • ⇒ Work:
  • The amount of work done means the amount of energy transferred, so work is measured in joules.
  • Any energy transfer may be broadly described as work done. A 15 W lightbulb, for instance, performs 150 J of work while it operates for 10 seconds, transferring 150 J of electrical energy as heat and light.
    We can determine the energy transferred and, consequently, the work completed in any scenario where we are able to compute the energy before and after.
  • ⇒ Work forced:
  • If energy is transferred mechanically by means of a force, then the amount of work done can be calculated simply:
  • Moment (Nm) = force (N) × Prependicular distance from the pivot
  • Figure 22 Work on a building site
  • Example:
  •  A brick of mass 2.2 kg is lifted vertically against its weight through a height of 1.24 m. Find work.
  • [math]\begin{gather}
    \Delta W = F \Delta s \\
    \Delta W = (mg) \times \Delta s \\
    \Delta W = (2.2 \times 9.8) \times 1.24 \\
    \Delta W = 26.8 \ \text{J}
    \end{gather}[/math]
  • Note that this is the same amount of energy as we calculated for the gravitational potential energy of this same brick undergoing the same lift.

  • 18) Be able to use the equation [math]E_k = \frac{1}{2} m v^2[/math]  for the kinetic energy of a body

  • ⇒ Kinetic energy:
  • “Kinetic energy is a property of a moving object or particle and depends not only on its motion but also on its mass.”
  • Formula:
  • [math]\begin{gather}
    \text{Kinetic energy (J)} = \frac{1}{2} \times \text{mass} \times (\text{speed})^2 \\
    E_k = \frac{1}{2} m v^2
    \end{gather}[/math]
  •  Example:
  • A large jumbo jet plane might have a cruising speed of 900 km h¹and a flight mass of 400 tonnes. What would its kinetic energy be?
  • Solution:
  • [math]\begin{gather}
    v = 900 \ \text{km h}^{-1} \\
    v = \frac{9 \times 10^5}{60 \times 60} \\
    v = 250 \ \text{m.s}^{-1} \\
    m = 400 \times 1000 \\
    m = 4 \times 10^5 \ \text{kg} \\
    E_k = \frac{1}{2} \times m \times v^2 \\
    E_k = \frac{1}{2} \times 4 \times 10^5 \times (250)^2 \\
    E_k = 1.25 \times 10^{10} \ \text{J} \\
    E_k = 12.5 \ \text{GJ}
    \end{gather}[/math]

  • 19) Be able to use the equation [math]E_{\text{grav}} = mgh[/math]  for the difference in gravitational potential energy near the Earth’s surface

  • Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field.
  • [math]\begin{gather}
    \text{Gravitational potential energy} = \text{mass (kg)} \times \text{gravitational field strength (N.k}^{-1}\text{)} \times \text{height (m)} \\
    E_{\text{grav}} = mgh
    \end{gather}[/math]
  • The gravitational potential energy is typically thought of as a change brought on by a change in height, such as the change that occurs when you raise a thing onto a shelf. In order to account for transfers to or from gravitational potential energy, this significantly modifies the equation:
  • [math]E_{\text{grav}} = mgh[/math]
  • Example:
  • A brick of mass 2.2 kg is lifted vertically through a height of 1.24 m. Find gravitational potential energy.
  • [math]\begin{gather}
    \Delta E_{\text{grav}} = mg \Delta h \\
    \Delta E_{\text{grav}} = (2.2)(9.8)(1.24) \\
    \Delta E_{\text{grav}} = 26.8 \ \text{J}
    \end{gather}[/math]
  • The strength of the gravitational field is constant. A planet’s gravitational attraction at a distance from its core is measured by its gravitational field strength.
  • This isn’t really constant since a mass’s experience of the gravitational field is inversely proportional to its squared distance from the planet’s core.
  • However, it is a reasonable estimate to claim that g is constant at 9.81 N.kg-1 near to the Earth’s surface, over tiny scales, such as the heights that humans interact with in daily life.

  • 20) Know, and understand how to apply, the principle of conservation of energy including use of work done, gravitational potential energy and kinetic energy:

  • ⇒ The principle of conservation of energy:
  • The principle of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another within a closed system. This means the total amount of energy in the system remains constant, although its form may change.
  • If an item falls to a lower height, the gravitational potential energy can be converted to kinetic energy.
    Alternatively, as kinetic energy is converted to gravitational potential energy, an object hurled upward will slow down. In either situation:
  • [math]\begin{gather}
    \Delta E_{\text{grav}} = mg \Delta h \\
    \Delta E_{\text{grav}} = \frac{1}{2} mv^2 \\
    \Delta E_{\text{grav}} = E_k
    \end{gather}[/math]
  • Depending on the situation, it can often be useful to divide out the mass that appears on both sides of this equation. This allows a convenient calculation of how fast an object will be travelling after falling a certain distance from rest:
  • [math]v = \sqrt{2g \Delta h}[/math]
  • – or how high an object could rise if projected upwards at a certain speed:
  • [math]\Delta h = \frac{v^2}{2g}[/math]
  • Example:
  • how high would water from a fountain rise if it were ejected vertically upwards from a spout at [math]13.5 m.s^{-1}[/math]?
  • [math]\begin{gather}
    \Delta h = \frac{v^2}{2g} \\
    \Delta h = \frac{(13.5)^2}{2(9.8)} \\
    \Delta h = \frac{182.25}{19.62} \\
    \Delta h = 9.29 \ \text{m}
    \end{gather}
    [/math]

  • 21) Be able to use the equations relating power, time and energy transferred or work done

  • [math]P = \frac{E}{t} \quad \text{and} \quad P = \frac{W}{t}[/math]

  • ⇒ Power:
  • “The ability or capacity to do something or act in a particular way.”
  • Formula:
  • [math]\begin{gather}
    \text{Power (W)} = \frac{\text{energy transferred (J)}}{\text{time for the energy transfer (s)}} \\
    p = \frac{E}{t} \\
    \text{Power (W)} = \frac{\text{work done (J)}}{\text{time for the work to be done (s)}} \\
    p = \frac{\Delta W}{t}
    \end{gather}[/math]
  •  Remember:
  • [math]\begin{gather}
    \text{work (J)} = \text{force (N)} \times \text{distance moved in the direction of the force (m)} \\
    \Delta W = F \Delta s
    \end{gather}[/math]
  • So,
  • [math]\begin{gather}
    \text{Power (W)} = \frac{\text{force (N)} \times \text{distance moved (m)}}{\text{time for the force to move (s)}} \\
    \text{Power (W)} = \frac{F \Delta s}{t}
    \end{gather}[/math]
  • Example
  • The power of a forklift truck lifting a 120 kg crate vertically up 5.00 m in 4.0 seconds would be calculated as:
  • [math]\begin{gather}
    \text{Power (W)} = \frac{F \Delta s}{t} \\
    \text{Power (W)} = \frac{mg \Delta s}{t} \\
    \text{Power (W)} = \frac{(120)(9.8)(5)}{4} \\
    \text{Power (W)} = 1.47 \ \text{kW}
    \end{gather}[/math]

  • 22) Be able to use the equations

  • [math]\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}[/math]

  • And

  • [math]\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}[/math]

  • ⇒ Efficiency:
  • “The ability of a machine to transfer energy usefully is called efficiency.”
  • Formula:
  • [math]\begin{gather}
    \text{efficiency} = \frac{\text{useful work done}}{\text{total energy input}} \\
    \text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}
    \end{gather}[/math]
  • If we remember that power is energy divided by time, then measuring the energy flows in a machine for a fixed amount of time means that we can write a power version of the efficiency equation:
  • [math]\begin{gather}
    \text{efficiency} = \frac{\frac{\text{useful energy output}}{\text{time}}}{\frac{\text{total energy input}}{\text{time}}} \\
    \text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}
    \end{gather}[/math]
  • Example:
  • If the forklift truck referred to above lifted the crate when supplied with electrical energy from its battery at a rate of 3000 joules per second, what is its efficiency?
  • [math]\begin{gather}
    \text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \\
    \text{efficiency} = \frac{1470}{3000} \\
    \text{efficiency} = 0.49 \\
    \text{efficiency} = 49\%
    \end{gather}[/math]
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