Physics · 1 Motion, forces and energy

1.1 Physical quantities and measurement techniques

Before you start

Units of length and how to convert them: $10$ mm $= 1$ cm and $100$ cm $= 1$ m.
The volume of a cuboid is length × width × height.
How to find the mean of several readings: add them and divide by how many there are.
Pythagoras' theorem and $\tan θ = \frac{opposite}{adjacent}$ in a right-angled triangle, from your maths lessons.

Syllabus objectives covered (7)

Describe the use of rulers and measuring cylinders to find a length or a volume
Describe how to measure a variety of time intervals using clocks and digital timers
Determine an average value for a small distance and for a short interval of time by measuring multiples (including the period of oscillation of a pendulum)
Understand that a scalar quantity has magnitude (size) only and that a vector quantity has magnitude and direction
Know that the following quantities are scalars: distance, speed, time, mass, energy and temperature
Know that the following quantities are vectors: force, weight, velocity, acceleration, momentum, electric field strength and gravitational field strength
Determine, by calculation or graphically, the resultant of two vectors at right angles, limited to forces or velocities only

Part 1 of 7

Reading a length from a ruler

A school ruler or metre rule is marked in millimetres. The smallest division is $1$ mm, so that is the finest length you can read from it. Record a ruler length to the nearest millimetre, for example $14.6$ cm, not $14.63$ cm.
Put the edge of the ruler that carries the scale right against the object. If there is a gap between the scale and the object, the mark the object seems to line up with changes as you move your head.
That shift is called parallax error. You avoid it by putting your eye directly in line with the mark you are reading, looking at right angles to the scale, not from one side.
apparatus diagram — not drawn yet
A pencil lying along a ruler marked in cm and mm, with three eyes drawn above the far end of the pencil: one looking straight down at right angles to the scale (labelled correct) and two looking at an angle from the left and right (labelled wrong), with dashed sight-lines showing each angled eye lining the pencil tip up with a different mark.
Only the eye looking at right angles to the scale reads the true mark. The angled views give parallax error.
Do not assume the end of the ruler is zero. The zero mark is not always at the very end, and the end can be worn or chipped. Line one end of the object up with a clear mark, such as $1.0$ cm, and read the mark at the other end.
The length is then the difference between the two readings. This is the same subtraction you will use for a measuring cylinder in the next section.
$length = end reading - start reading$
Write the unit with every length, and convert before you calculate. A length of $2.7$ cm is $27$ mm. Mixing cm and mm in one calculation gives an answer ten times too big or too small.

Part 2 of 7

Finding a volume with a measuring cylinder

A measuring cylinder measures the volume of a liquid. Its scale is in cm³ or in ml, and these are the same size: $1$ ml $= 1$ cm³.
Stand the cylinder on a flat, horizontal surface before you read it. If the cylinder tilts, the liquid surface is not level against the scale.
The surface of water in a cylinder is not flat. Water creeps up the glass at the edges, so the surface curves down in the middle. This curved surface is the meniscus.
Read the scale at the bottom of the meniscus, with your eye level with it. If you look down from above, you get parallax error again, for the same reason as with the ruler.
apparatus diagram — not drawn yet
A close-up of a measuring cylinder marked in cm³ with water whose surface curves down in the middle; an eye drawn level with the lowest point of the curve, a horizontal dashed sight-line from the eye to the scale, the bottom of the meniscus labelled, and a second eye above the cylinder looking down at an angle labelled wrong.
Eye level, bottom of the meniscus. Looking from above makes the level line up with the wrong mark.
A measuring cylinder can also find the volume of a solid with an irregular shape, such as a stone. Read the water level first. Then lower the stone in on a thread, so no water splashes out, until it is completely under the water. Read the level again.
The water level rises because the stone pushes aside its own volume of water. So the volume of the stone is the rise in the reading. This is called the displacement method.
$V_{stone} = V_{final} - V_{initial}$
For a solid with a regular shape you do not need the cylinder. Measure each dimension with a ruler and calculate. For a cuboid, the volume is length × width × height, in cm³ if the lengths were in cm.

Part 3 of 7

Timing an interval: clocks, stopwatches and light gates

Time is measured in seconds, s. The right timer depends on how long the interval is compared with the error you make when you start and stop it.
A long interval, such as the minutes an ice cube takes to melt or the hours a plant takes to wilt, can be timed with a clock or a watch. An error of a second or so at each end is tiny compared with the whole interval.
An interval of several seconds, such as a ball rolling down a long ramp, is timed with a stopwatch. You start and stop it with your thumb, so your reaction time, a few tenths of a second, is part of every reading.
Do not be fooled by the stopwatch display. It shows hundredths of a second, but your reaction time is much larger than $0.01$ s, so the last digit does not tell you anything real about the event.
A very short interval, such as the time a trolley takes to pass through a gap, can be shorter than your reaction time. A stopwatch cannot time that usefully. You use a digital timer that is started and stopped electronically instead.
A light gate is one way to trigger it. It sends a beam across a gap to a sensor. When an object breaks the beam, the timer starts or stops. With one gate, the timer records how long the beam is blocked. With two gates, it records the time between the object passing the first and the second.
No human reaction is involved with a light gate, which is why it is the right choice for intervals too short to time by hand.

Part 4 of 7

Measuring many to find one

Some things are too small for your instrument. A sheet of paper is about $0.1$ mm thick, which is much less than the $1$ mm smallest division on a ruler. You cannot read it directly.
The fix is to measure many identical things together and divide by how many there are. If a stack of $200$ sheets is $18$ mm thick, one sheet is $18 \div 200 = 0.090$ mm thick on average.
$thickness of one sheet = \frac{thickness of the stack}{number of sheets}$
Here is why it works. Your reading error, about $1$ mm on the ruler, happens once for the whole stack, not once for each sheet. When you divide by $200$ , you divide that error by $200$ as well.
The same trick finds the diameter of a thin wire. Wind the wire tightly around a pencil so the turns lie side by side with no gaps. Measure the length of the coil and divide by the number of turns.
It also works for time. A pendulum swings too quickly for one swing to be timed well by hand. One complete oscillation is the bob going from one side, over to the other side, and back to where it started. The time for one complete oscillation is the period.
Time a number of complete oscillations, such as $20$ , with a stopwatch. Then divide the total time by $20$ . Your reaction-time error happens once at the start and once at the end, so dividing by $20$ makes its effect on the period $20$ times smaller.
Put a fixed marker, such as a pin in a lump of modelling clay, directly below the bob when it hangs at rest. Start and stop the stopwatch as the bob passes the marker, moving in the same direction each time.
apparatus diagram — not drawn yet
A clamp stand with a string and a small metal bob hanging from it, the bob drawn at both ends of its swing with a curved double-headed arrow between them, and a vertical pin marker on the bench directly below the rest position of the bob, labelled 'marker at centre of swing'.
Time the swings as the bob passes the marker, where it is moving fastest.
The marker goes at the centre of the swing because the bob moves fastest there, so the moment it passes is sharp. At the ends of the swing the bob slows down and stops for an instant, so it is hard to judge exactly when it gets there.
Count from zero. Say 'zero' as you start the stopwatch, then 'one' the next time the bob passes the marker in the same direction, and stop the watch on 'twenty'. The trap here is saying 'one' as you start: you then time only $19$ oscillations.
Repeat the whole timing at least once more and take the mean of the totals. Then divide the mean by the number of oscillations. A reading far from the others is anomalous: time it again rather than include it in the mean.

Part 5 of 7

Scalars and vectors

A scalar quantity has magnitude (size) only. A mass of $3.0$ kg is completely described by its number and unit. Asking 'which way?' makes no sense.
A vector quantity has magnitude and direction. A force of $20$ N is only half the story. A force of $20$ N vertically downwards is the whole of it, because the same force pushing sideways would have a different effect.
These are the scalars you must know: distance, speed, time, mass, energy and temperature.
These are the vectors you must know: force, weight, velocity, acceleration, momentum, electric field strength and gravitational field strength.
Learn the lists as pairs that are easy to mix up. Speed is a scalar, but velocity is a vector, because velocity is speed in a given direction. Mass is a scalar, but weight is a vector, because weight is a force: the pull of gravity, acting towards the centre of the Earth.
Momentum is a vector because it is mass × velocity, and the velocity carries a direction. Acceleration is a vector because it is the change in velocity per unit time, and velocity has a direction.
Both field strengths are vectors because they tell you about the force something feels in the field, and a force has a direction. Near the Earth, gravitational field strength points down, towards the centre of the Earth. The direction of an electric field is the direction of the force on a positive charge.
A minus sign does not make a quantity a vector. A temperature of $- 5$ °C has a minus sign, but it points nowhere: the minus only means below zero on the Celsius scale. Temperature is still a scalar.
The difference matters when you add. Scalars add by ordinary arithmetic: $2.0$ kg and $3.0$ kg of sand always make $5.0$ kg. Two forces of $3.0$ N can give a resultant of anything from $0$ N to $6.0$ N, depending on their directions.

Part 6 of 7

The resultant of two vectors at right angles: by calculation

The resultant is the single vector that has the same effect as the two vectors acting together. On this syllabus the two vectors are always forces or always velocities, and the method is the same for both.
Draw the two vectors tip to tail: the second one starts where the first one ends. When they are at right angles, they form the two shorter sides of a right-angled triangle. The resultant runs from the tail of the first to the tip of the second, so it is the hypotenuse.
forces diagram — not drawn yet
A right-angled vector triangle: a horizontal arrow labelled 7.0 N pointing right, then from its tip a vertical arrow labelled 4.0 N pointing up, then a thicker arrow labelled 'resultant' from the tail of the 7.0 N arrow to the tip of the 4.0 N arrow, with a right-angle mark between the two forces and the angle θ marked at the tail between the 7.0 N arrow and the resultant.
Tip to tail, the two forces make the shorter sides; the resultant is the hypotenuse, and θ is measured from the first force.
So the size of the resultant comes from Pythagoras. Square each vector, add the squares, then take the square root.
$R = \sqrt{F_{1}^{2} + F_{2}^{2}}$
Check your answer makes sense. The resultant is always bigger than either of the two vectors but smaller than their sum. Forces of $7.0$ N and $4.0$ N at right angles give $8.1$ N, which lies between $7.0$ N and $11$ N.
The resultant is a vector, so you also need its direction. Give it as an angle $θ$ from one of the two original vectors. With $θ$ at the tail of the first vector, the second vector is opposite $θ$ and the first vector is adjacent to it.
$\tan θ = \frac{F_{2}}{F_{1}}$
An angle on its own is not a direction. Say what it is measured from and which way: ' $30 °$ north of east', or ' $30 °$ to the $7.0$ N force, towards the $4.0$ N force'.
Paper 4 gives you no formula sheet. Pythagoras and $\tan$ come from your own memory, so practise writing both from scratch.

Part 7 of 7

The resultant of two vectors at right angles: by scale drawing

Choose a scale and write it next to your drawing, for example ' $1$ cm represents $1.0$ N'. Pick a scale that makes the drawing large. A $1$ mm error in measuring a $13$ cm line matters far less than a $1$ mm error in a $1.3$ cm line.
Draw the first vector as an arrow of the right length pointing in its direction. From its tip, draw the second vector at right angles, using a protractor or set square, again to scale.
Draw the resultant from the tail of the first arrow to the tip of the second. Measure its length with a ruler and use the scale to turn centimetres back into newtons or metres per second.
Measure the angle between the first arrow and the resultant with a protractor. That gives you the direction, which you state in the same way as for the calculation.
There is a second way to draw it. Draw both arrows starting from the same point and complete the rectangle. The diagonal from that starting point is the resultant. It is the same line as the hypotenuse of the triangle, so it gives the same answer.
A scale drawing is only as good as your drawing: use a sharp pencil, measure to the nearest millimetre and the nearest degree. If you have time, check it with Pythagoras.

Worked examples

Example 1. A student winds a thin copper wire tightly around a pencil so that $30$ turns lie side by side with no gaps. She holds a ruler against the coil. The left edge of the coil is at the $3.0$ cm mark and the right edge is at the $5.7$ cm mark. Determine the diameter of the wire in mm.

Show the working

She did not start at zero, so the length of the coil is the difference between the two readings.
$5.7 - 3.0 = 2.7 cm$
The question wants mm, so convert now, before dividing: $2.7$ cm $= 27$ mm.
The coil is $30$ wire diameters lying side by side, so one diameter is the coil length divided by $30$ .
$d = \frac{27}{30} = 0.90 mm$
Notice why the coil was needed. One diameter, $0.90$ mm, is smaller than the ruler's $1$ mm division. A $1$ mm error in reading the coil is shared over $30$ turns, so it becomes only about $0.03$ mm in the diameter.

Answer: $0.90$ mm

Example 2. A student uses a stopwatch to time $20$ complete oscillations of a pendulum. She does this three times and gets $32.4$ s, $32.1$ s and $32.7$ s. Determine the period of the pendulum.

Show the working

First look at the readings. They agree to within $0.6$ s, so none is anomalous and all three go into the mean.
Find the mean time for $20$ oscillations: add the three readings and divide by $3$ , because there are three repeats.
$\frac{32.4 + 32.1 + 32.7}{3} = \frac{97.2}{3} = 32.4 s$
That is the time for $20$ oscillations. The period is the time for one, so divide by $20$ .
$T = \frac{32.4}{20} = 1.62 s$
You would get the same answer by dividing each reading by $20$ first and then taking the mean. The order does not matter, but do not forget either step.

Answer: $1.62$ s

Example 3. Two forces act at the same point on a box. One is $7.0$ N due east and the other is $4.0$ N due north. Calculate the size and direction of the resultant force.

Show the working

The forces are at right angles, so they form the two shorter sides of a right-angled triangle and the resultant is the hypotenuse. Use Pythagoras.
$R = \sqrt{{7.0}^{2} + {4.0}^{2}} = \sqrt{49 + 16} = \sqrt{65} = 8.1 N$
Check: $8.1$ N is bigger than $7.0$ N and smaller than $7.0 + 4.0 = 11$ N, as it should be.
For the direction, put the angle $θ$ at the tail of the east force. The north force is opposite $θ$ ; the east force is adjacent.
$\tan θ = \frac{4.0}{7.0} \Rightarrow θ = 29.7 ° \approx 30 °$
State what the angle is measured from. It is measured from east, turning towards north.

Answer: $8.1$ N at $30 °$ north of east.

Example 4. A swimmer swims due north across a river at $1.2$ m/s relative to the water. The river flows due east at $0.50$ m/s. Determine the swimmer's resultant velocity by scale drawing.

Show the working

Choose a scale that makes a large drawing: $1$ cm represents $0.10$ m/s. Write it on the drawing.
Draw the swimmer's velocity as an arrow $1.2 \div 0.10 = 12.0$ cm long, pointing north (up the page).
From the tip of that arrow, draw the river's velocity at right angles: $0.50 \div 0.10 = 5.0$ cm long, pointing east.
Draw the resultant from the start of the first arrow to the tip of the second. It measures $13.0$ cm, so the resultant speed is $13.0 \times 0.10 = 1.3$ m/s.
Measure the angle at the start between the north arrow and the resultant with a protractor. It is about $23 °$ .
Check with Pythagoras.
$\sqrt{{1.2}^{2} + {0.50}^{2}} = \sqrt{1.69} = 1.3 m/s$

Answer: $1.3$ m/s at about $23 °$ east of north (downstream from straight across).

Where people go wrong

Mistake: Adding two forces at right angles like ordinary numbers: $7.0$ N $+ 4.0$ N $= 11$ N.

Instead: Use Pythagoras: $\sqrt{{7.0}^{2} + {4.0}^{2}} = 8.1$ N.

Forces are vectors. Only vectors pointing the same way add as plain numbers. At right angles, neither force helps the other along its own direction, so the resultant is less than the sum.

Mistake: Giving the size of a resultant and stopping there.

Instead: Give the direction too: an angle and what it is measured from, such as ' $30 °$ north of east'.

A resultant force or velocity is a vector. A size alone describes only half of it.

Mistake: Calling temperature a vector because it can be negative.

Instead: Temperature is a scalar. The minus sign in $- 5$ °C means below zero on the scale, not a direction.

A vector needs a direction in space. A minus sign only shows direction for a vector along a line; for a scalar it just means less than zero.

Mistake: Saying 'one' as the stopwatch starts, stopping on 'twenty', then dividing by $20$ .

Instead: Count 'zero' as you start the watch, so that stopping on 'twenty' really means $20$ oscillations.

Counting from one times only $19$ oscillations. Dividing that time by $20$ gives a period that is too small.

Mistake: Reading a measuring cylinder from above, or at the edge of the water where it climbs the glass.

Instead: Stand the cylinder on a level bench and read the bottom of the meniscus with your eye level with it.

The edges of the water surface are higher than the middle, and looking from above lines the surface up with the wrong mark (parallax error).

Check yourself

You are reading the volume of water in a measuring cylinder. Where should your eye be, and which part of the water surface do you read?
Answer
Your eye should be level with the water surface, and you read the bottom of the meniscus. The cylinder should be standing on a flat, horizontal surface.
A measuring cylinder contains water at the $35$ cm³ mark. A key is lowered in on a thread until it is completely under the water, and the level rises to the $41$ cm³ mark. What is the volume of the key?
Answer
$41 - 35 = 6$ cm³. The key pushes aside its own volume of water.
Suggest a suitable timer for (a) the time a candle takes to burn down, and (b) the time a trolley takes to pass through a narrow gap. Give a reason for (b).
Answer
(a) A clock or a watch. (b) A light gate connected to a digital timer, because the interval is so short that your reaction time with a stopwatch would be about as large as the interval itself.
A stack of $200$ identical sheets of paper is $18$ mm thick. Determine the thickness of one sheet, and explain why you would not measure one sheet directly with a ruler.
Answer
$18 \div 200 = 0.090$ mm. One sheet is much thinner than the ruler's $1$ mm smallest division, so it cannot be read directly. Measuring the stack shares the reading error over $200$ sheets.
A student says 'one' as she starts the stopwatch when the pendulum bob passes the marker, and stops the watch when she says 'twenty'. How many oscillations has she timed, and what happens to her value for the period if she divides by $20$ ?
Answer
She has timed $19$ oscillations. Dividing by $20$ gives a period that is too small.
Sort these into scalars and vectors: speed, weight, energy, acceleration, temperature, momentum, time, gravitational field strength.
Answer
Scalars: speed, energy, temperature, time. Vectors: weight, acceleration, momentum, gravitational field strength.
Explain why velocity is a vector but speed is not.
Answer
A vector has magnitude and direction; a scalar has magnitude only. Velocity is speed in a given direction, so it has both. Speed tells you only how fast, not which way.
Forces of $12$ N and $5.0$ N act at right angles on the same point. Calculate the size of the resultant and its angle to the $12$ N force.
Answer
$R = \sqrt{12^{2} + {5.0}^{2}} = \sqrt{169} = 13$ N. $\tan θ = 5.0 / 12$ , so $θ = 23 °$ to the $12$ N force, towards the $5.0$ N force.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student investigates a pendulum made from a small metal bob on a string hanging from a clamp. She has a stopwatch.

(a) Describe how the student uses the stopwatch to find an accurate value for the period of the pendulum. 3 marks No calculator

Mark scheme

B1 times a stated number of complete oscillations, at least 10 (e.g. 20)
B1 divides the total time by the number of oscillations
B1 any one of: repeats the timing and calculates the mean / starts and stops the stopwatch as the bob passes a fixed marker at the centre of the swing / counts zero as the stopwatch is started

(b) The student times $20$ oscillations three times. Her readings are $28.4$ s, $28.8$ s and $28.5$ s. Calculate the period of the pendulum. 3 marks

Mark scheme

C1 mean of the three readings = 28.6 s (accept 28.57 s)
C1 mean divided by 20, e.g. 28.57 / 20; or each reading divided by 20 (1.42, 1.44, 1.425) and the mean of these taken
A1 1.43 s (accept 1.4 s or 1.428 s); unit required

(c) Explain why timing $20$ oscillations gives a more accurate value for the period than timing one oscillation. 2 marks No calculator

Mark scheme

B1 the error from starting and stopping the stopwatch (reaction time) is about the same whether 1 or 20 oscillations are timed
B1 dividing by 20 makes the error in the period 20 times smaller / the error is a smaller fraction of the time measured, owtte

Question 2. A student finds the volume of a small stone using a measuring cylinder containing water. Before the stone is added, the water level reads $24$ cm³. She lowers the stone on a thread until it is completely under the water, and the level then reads $31$ cm³.

(a) State two things the student does to make each reading of the measuring cylinder accurate. 2 marks No calculator

Mark scheme

B1 first of any two from: (i) eye level with the water surface / line of sight at right angles to the scale, to avoid parallax, owtte; (ii) reads the bottom of the meniscus; (iii) cylinder stands on a flat, horizontal surface
B1 a second, different point from (i)–(iii)

(b) Calculate the volume of the stone. 2 marks No calculator

Mark scheme

C1 31 − 24 seen
A1 7 cm³ (accept 7 ml); unit required

Question 3. Two tugs pull a barge across a harbour. Tug A pulls with a force of $15$ kN due east. Tug B pulls with a force of $8.0$ kN due north. Both forces act at the same point on the barge.

(a) State the difference between a scalar quantity and a vector quantity. 1 mark No calculator

Mark scheme

B1 a vector has magnitude (size) and direction and a scalar has magnitude only; both halves needed

(b) Identify the two vector quantities in this list: energy, force, mass, speed, temperature, velocity. 2 marks No calculator

Mark scheme

B1 force
B1 velocity

(c) Determine the size of the resultant of the two forces. You may use a calculation or a scale drawing. 3 marks

Mark scheme

C1 R² = 15² + 8.0², or a scale drawing with the two forces at right angles to a stated scale (tip to tail, or both from one point with the rectangle completed)
C1 289 seen, or the resultant drawn from the tail of the first force to the tip of the second / the diagonal of the rectangle from the common starting point
A1 17 kN (accept 16.5 to 17.5 kN from a scale drawing); unit required

(d) Determine the angle between the resultant force and the direction due east. 2 marks

Mark scheme

C1 tan θ = 8.0 / 15, or sin θ = 8.0 / 17, or cos θ = 15 / 17 (ecf from (c)), or the angle measured on a correct scale drawing
A1 28° (accept 27° to 29°)

How this is examined

Finding the size and direction of the resultant of two forces or velocities at right angles: a Determine worth about three marks for the size and two for the direction, where you may choose between Pythagoras and a scale drawing The two values squared, added and square-rooted (or the two arrows drawn tip to tail at right angles to a stated scale); the size with its unit; then $\tan θ$ = opposite / adjacent (or the angle measured with a protractor), stated from a named direction such as 'north of east' or 'to the $15$ kN force'.
Describing how to get an accurate period from a pendulum: a three-mark Describe of the timing method Three separate points: 'time a stated number of complete oscillations, such as $20$ '; 'divide the total time by that number'; and one refinement, such as 'repeat and take the mean' or 'start and stop the watch as the bob passes a marker at the centre of the swing'.
Turning repeat timings of many oscillations into one period: a two- or three-mark Calculate from a set of repeat readings The mean of the repeat readings (add them, divide by the number of repeats), then that mean divided by the number of oscillations, then the period with the unit s.
Reading a ruler or a measuring cylinder without parallax, and finding a volume by displacement: a two-mark State of precautions, or a short Calculate of the volume of an irregular solid 'Eye level with the bottom of the meniscus' and 'cylinder on a flat, horizontal surface' for a cylinder; 'eye in line with the mark, at right angles to the scale' for a ruler; for displacement, final reading minus initial reading, with the unit cm³.
Saying what makes a quantity a scalar or a vector, and sorting quantities: a one-mark State of the difference, then an Identify from a list of quantities 'A vector has magnitude and direction; a scalar has magnitude only': both halves are needed. For the list, use the syllabus lists exactly: weight and velocity are vectors, mass and speed are scalars.
Why timing many oscillations, or using a light gate, is better than one stopwatch reading: a two-mark Explain 'The reaction-time error is about the same however long you time for' and 'dividing by $20$ makes its effect on one period $20$ times smaller'. For a very short interval: 'a light gate and digital timer start and stop automatically, so there is no reaction time'.

Words to know

Parallax error: An error in a reading caused by looking at a scale from an angle instead of at right angles to it.
Meniscus: The curved surface of a liquid in a narrow container. For water it curves down in the middle, and you read the bottom of the curve.
Displacement method: Finding the volume of an irregular solid from the rise in water level when it is put completely under water in a measuring cylinder.
Light gate: A sensor with a beam across a gap that starts or stops a digital timer when an object breaks the beam.
Oscillation: One complete swing of a pendulum: from one side, to the other side, and back to the start.
Period: The time for one complete oscillation, measured in s.
Scalar: A quantity that has magnitude (size) only, such as mass or speed.
Vector: A quantity that has magnitude and direction, such as force or velocity.
Resultant: The single vector that has the same effect as two or more vectors acting together.

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