Answers · Resultant force

Scalars, vectors and resultants — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. FQ13 · Practice

Options as numbered on the pupil sheet: 1 — scalar: size only; 2 — vector: size and direction.

Model answer. Speed, time, mass and energy have a size only, so they are scalars: “in which direction?” makes no sense for them. Velocity is speed in a given direction, so it is a vector. Weight is a force, the pull of the Earth, and it points down, so it is a vector. Gravitational field strength, 9.8 N/kg on the Earth, also points down, towards the centre of the Earth, so it is a vector too.

A full answer includes

I matched speed with scalar.
I matched velocity with vector.
I matched time with scalar.
I matched weight with vector.
I matched mass with scalar.
I matched gravitational field strength with vector.
I matched energy with scalar.
I did not match weight with scalar, as if it were the same as mass.

Also accept

speed – scalar; velocity – vector; time – scalar; weight – vector; mass – scalar; gravitational field strength – vector; energy – scalar.

Watch for

“Weight is a scalar, like mass.” — Weight is a force, the pull of the Earth, and it points down. Mass is the amount of matter in an object, with no direction.
“Speed is a vector, because a moving object always goes in some direction.” — Speed tells only how fast. Velocity is speed in a given direction, so velocity is the vector and speed is the scalar.
“Energy is a vector, because energy is transferred from one place to another.” — Energy has a size only, measured in joules. It cannot point east or north.

2. FQ14 · Practice

Model answer. A scalar has a size (magnitude) only; a vector has a size and a direction. Mass is a scalar: 2.0 kg is complete without a direction. Force is a vector: a pull of 5.0 N is not complete until you say which way it pulls. Both kinds have units, and both can be measured or calculated. A minus sign does not decide it either: a temperature of −3 °C is still a scalar.

A full answer includes

I chose “A scalar has size only; a vector has size and direction”.
I gave a correct scalar, such as mass, speed or time.
I gave a correct vector, such as force, weight or velocity.
I did not choose the statement about positive and negative values.

Also accept

a: a scalar has size only, a vector has size and direction; for example mass (scalar) and force (vector).

Watch for

“b: A scalar is always positive, and a vector can be negative.” — A temperature of −3 °C is negative, but it is a scalar. The difference is direction: a vector points somewhere, a scalar does not.
“c: A scalar is measured, and a vector can only be calculated.” — A force meter measures a force, which is a vector, and you can calculate a mass, which is a scalar. How you get the value does not decide it.
“d: A scalar has a unit, and a vector is a number with no unit.” — Force is a vector, measured in newtons. Velocity is a vector, measured in m/s. Vectors have units just like scalars.

3. FQ15 · Practice

Model answer. Row B. X and Y both move at 30 m/s, so they have the same speed: speed is a scalar, with a size only. Their directions are opposite, so their velocities are different: velocity is a vector, with a size and a direction. X and Z both move north, but at 30 m/s and 25 m/s: their speeds differ, so their velocities differ too. Two velocities are the same only when both the size and the direction are the same.

A full answer includes

I chose row B.
X and Y: the same speed, 30 m/s, but opposite directions, so different velocities.
X and Z: they move the same way, but at different speeds.
A velocity has a size and a direction, so a different size is a different velocity, even in the same direction.
I did not choose row C: two velocities in the same direction differ if their sizes differ.

Also accept

B: X and Y same speed, different velocity; X and Z different speed and different velocity.

Watch for

“Row A: X and Y have the same velocity, because both move at 30 m/s.” — Velocity includes the direction. 30 m/s north and 30 m/s south are different velocities.
“Row C: X and Z have the same velocity, because they move in the same direction.” — A velocity has a size as well as a direction. 30 m/s north and 25 m/s north have different sizes, so they are different velocities.
“Row D: X and Y have the same velocity, because their speeds are the same.” — Speed and velocity are not the same thing. The trains move in opposite directions, so their velocities differ.

4. FQ16 · Practice

Model answer. The two pulls are at right angles, so the pulls and the resultant make a right-angled triangle, with the resultant as its longest side. R² = 0.15² + 0.10² = 0.0225 + 0.0100 = 0.0325, so R = √0.0325 = 0.18 N. For the direction, the side opposite the angle at east is the north pull: tan θ = 0.10 ÷ 0.15 = 0.67, so θ = 34° north of east. The resultant points closer to east, the bigger pull.

Checked automatically

Size of the resultant force: 0.1803 N (within ±0.002 N)
- Common wrong answer 0.25 : You added the two pulls. They are at right angles, so use Pythagoras’ theorem: √(0.15² + 0.10²).
- Common wrong answer 0.05 : You subtracted the two pulls. That works only for opposite forces along one line. Use √(0.15² + 0.10²).
- Common wrong answer 0.0325 : That is R², 0.15² + 0.10². Now take the square root.
- Common wrong answer 0.125 : You found the average of the two pulls. Use Pythagoras’ theorem: √(0.15² + 0.10²).
Angle north of east: 33.69 ° (within ±1 °)
- Common wrong answer 56 : That angle is measured from north. From east, the opposite side is the north pull: tan θ = 0.10 ÷ 0.15.
- Common wrong answer 0.67 : That is tan θ. Now find the angle whose tangent is 0.67.
- Common wrong answer 45 : The resultant is not halfway. It points closer to the bigger pull: tan θ = 0.10 ÷ 0.15.

A full answer includes

Method: R² = 0.15² + 0.10² = 0.0225 + 0.0100 = 0.0325.
Answer: R = √0.0325 = 0.18 N.
Method: tan θ = 0.10 ÷ 0.15 = 0.67.
Answer: θ = 34° north of east.
I did not add the pulls to get 0.25 N.

Also accept

0.18 N, at 34° north of east.

Watch for

“The resultant is 0.25 N, because 0.15 N + 0.10 N = 0.25 N.” — Forces add like that only along the same line. At right angles, the resultant is less than the sum: use Pythagoras’ theorem.
“The resultant points halfway between east and north, at 45°.” — It points closer to the bigger pull. East has 0.15 N and north only 0.10 N, so the resultant is 34° from east.
“The angle is 56° north of east.” — 56° is the angle from north. Measured from east, the opposite side is the north pull: tan θ = 0.10 ÷ 0.15.

5. FQ17 · Practice

Options as numbered on the pupil sheet: 1 — From the tip of that arrow, draw an arrow 2.0 cm long towards the north; 2 — Measure the resultant’s length and its angle from the east line, and change the length back into kilonewtons with the scale; 3 — Choose a scale, such as 1 cm = 10 kN; 4 — Draw the resultant from the start of the first arrow to the tip of the second; 5 — Draw an arrow 5.0 cm long towards the east.

Model answer. First choose a scale, such as 1 cm = 10 kN, so that the drawing fits the page. Draw the first force to scale: 5.0 cm towards the east. From its tip, draw the second force to scale at right angles: 2.0 cm towards the north. The resultant is the arrow from the start of the first arrow to the tip of the second. Measure it: about 5.4 cm, which is 54 kN with the scale. A protractor at the start gives about 22°, so the resultant is about 54 kN at 22° north of east.

A full answer includes

Step 1: choose a scale, such as 1 cm = 10 kN.
Step 2: draw an arrow 5.0 cm long towards the east.
Step 3: from the tip of that arrow, draw an arrow 2.0 cm long towards the north.
Step 4: draw the resultant from the start of the first arrow to the tip of the second.
Step 5: measure the resultant’s length and its angle from the east line, and change the length back into kilonewtons with the scale.
I did not draw both arrows from the same point and join their tips.

Also accept

scale; 5.0 cm east; 2.0 cm north from its tip; resultant from start to last tip; measure the length and the angle, then use the scale (about 54 kN at 22° north of east).

Watch for

“Draw both arrows from the same starting point, then measure the gap between their tips.” — The line between the two tips is not the resultant. Put the arrows tip to tail, and draw the resultant from the start of the first to the tip of the second.
“The scale can be chosen after the drawing.” — You need the scale to know how long to draw each arrow. Choose it first, so the drawing fits the page.

6. FQ18 · Practice

Model answer. The velocity through the water, straight across, and the velocity of the current, downstream, are at right angles. Their resultant, 1.9 m/s, is the longest side of the triangle. So 1.9² = v² + 0.70², and v² = 1.9² − 0.70² = 3.61 − 0.49 = 3.12. The speed through the water is v = √3.12 = 1.8 m/s (1.77 m/s). Check: √(1.77² + 0.70²) = 1.9 m/s.

Checked automatically

Speed through the water: 1.766 m/s (within ±0.035 m/s)
- Common wrong answer 1.2 : You subtracted the speeds. They are at right angles, so subtract their squares: √(1.9² − 0.70²).
- Common wrong answer 2.6 : You added the speeds. The resultant, 1.9 m/s, is the longest side: √(1.9² − 0.70²).
- Common wrong answer 2.0 : You added the squares. The resultant is the longest side, so take the square of the current away: √(1.9² − 0.70²).
- Common wrong answer 3.12 : That is v², 1.9² − 0.70². Now take the square root.
- Common wrong answer 1.9 : That is the resultant over the river bed. The speed through the water is the other side: √(1.9² − 0.70²).

A full answer includes

Method: (speed through the water)² = 1.9² − 0.70².
Working: 3.61 − 0.49 = 3.12.
Answer: √3.12 = 1.8 m/s.
I did not add the squares: the resultant, 1.9 m/s, is the longest side.

Also accept

1.8 m/s (√(1.9² − 0.70²) = 1.77 m/s).

Watch for

“The speed through the water is 2.6 m/s, because 1.9 + 0.70 = 2.6.” — The velocities are at right angles, so they do not simply add or take away. The resultant is the longest side: v² = 1.9² − 0.70².
“v² = 1.9² + 0.70², as in every Pythagoras question.” — Add the squares only to find the longest side. Here the longest side, 1.9 m/s, is given, so take the square of the current away from it.
“The speed through the water is 1.9 m/s, the GPS reading.” — The GPS shows the resultant over the river bed. The current carries the kayak too, so its speed through the water is less.

7. FQ19 · Practice

Model answer. Momentum, electric field strength, acceleration and force are vectors: each one has a direction, even when the question does not give it. The distance of 2.0 km, the temperature of −10 °C and the energy of 500 J are scalars, with a size only. The minus sign in −10 °C means the temperature is below zero on the Celsius scale. It does not point anywhere, so it does not make the temperature a vector.

A full answer includes

I chose the momentum of a moving trolley.
I chose an electric field strength.
I chose an acceleration of 2.0 m/s².
I chose a force of 5.0 N.
I did not choose the distance, the temperature or the energy: they are scalars.
I explained that the minus sign in −10 °C means below zero on a scale, not a direction in space.

Also accept

b, c, e, g: momentum, electric field strength, acceleration and force; −10 °C is below zero on a scale, not a direction.

Watch for

“The temperature of −10 °C is a vector, because it has a minus sign.” — A minus sign on a temperature shows a value below zero on the scale. A vector has a direction in space, which a temperature never has.
“Momentum is a scalar, like energy.” — Momentum is on the list of vectors. It points the same way as the velocity of the moving object.
“A force of 5.0 N or an acceleration of 2.0 m/s² is a scalar, because no direction is given.” — A force and an acceleration always have a direction, even if the question leaves it out. Both are vectors; “5.0 N” alone is an incomplete answer.
“A distance of 2.0 km is a vector, because you travel it in some direction.” — Distance tells you only how far, with no direction, so it is a scalar. A velocity or a force would need a direction to be complete; a distance does not.
“An energy of 500 J is a vector, because energy is transferred from place to place.” — Energy has a size only, measured in joules. It cannot point east or north, so it is a scalar.

8. FQ20 · Practice

Model answer. (a) With 1 cm = 50 N, the 600 N force needs 600 ÷ 50 = 12.0 cm, and the 350 N force needs 350 ÷ 50 = 7.0 cm. Hana draws the arrows tip to tail: 7.0 cm towards the east and then, from its tip, 12.0 cm towards the north; the resultant joins the start to the last tip. (b) The resultant arrow is 13.9 cm long, so it stands for 13.9 × 50 = 695 N, about 690 N. It points 60° north of east, closer to north, because the north line pulls harder. A calculation agrees: √(350² + 600²) = 695 N.

Checked automatically

(a) Length of the arrow for 600 N: 12.0 cm (within ±0.1 cm)
- Common wrong answer 30000 : You multiplied by the scale. Each centimetre stands for 50 N, so divide.
- Common wrong answer 6 : You used 1 cm = 100 N. With 1 cm = 50 N, each centimetre stands for less, so the arrow is longer.
- Common wrong answer 7 : That is the arrow for the 350 N force. The question asks for the 600 N force.
- Common wrong answer 600 : That is the force. Change it into a length with the scale.
(b) Size of the resultant force: 695 N (within ±10 N)
- Common wrong answer 950 : You added the two forces. At right angles they do not simply add: read the size from the 13.9 cm arrow.
- Common wrong answer 0.278 : You divided by the scale. Each centimetre stands for 50 N, so multiply.
- Common wrong answer 13.9 : That is the length of the arrow in centimetres. Change it back into newtons with the scale.
- Common wrong answer 250 : You took one force from the other. Forces at right angles do not subtract: read the 13.9 cm arrow.

A full answer includes

(a) Method: 600 N ÷ 50 N per cm.
(a) Answer: 12.0 cm.
(b) Method: 13.9 cm × 50 N per cm.
(b) Answer: 695 N, about 690 N.
(b) Direction: 60° north of east.
I did not add the two forces to get 950 N.

Also accept

(a) 12.0 cm; (b) about 695 N, at 60° north of east.

Watch for

“(b) The resultant is 950 N, because 350 N + 600 N = 950 N.” — The lines pull at right angles, so the forces do not simply add. In the drawing, the resultant is shorter than the two arrows put end to end.
“(b) The direction is 60° east of north.” — The 60° is measured from the east arrow, so the direction is 60° north of east: closer to north, the bigger pull.

9. FQ21 · Practice

Model answer. The velocity through the air, 60 m/s north, and the velocity of the wind, 15 m/s east, are at right angles. (a) v² = 60² + 15² = 3600 + 225 = 3825, so v = √3825 = 61.8… m/s, which is 62 m/s to 2 significant figures. (b) Measured from north, the side opposite the angle is the wind: tan θ = 15 ÷ 60 = 0.25, so θ = 14° east of north. The aircraft points north but moves over the ground at 62 m/s, 14° east of north.

Checked automatically

(a) Size of the resultant velocity: 62 m/s (2 significant figures)
- Common wrong answer 75 : You added the two velocities. They are at right angles, so use Pythagoras’ theorem: √(60² + 15²).
- Common wrong answer 45 : You subtracted. The wind blows at right angles to the aircraft’s heading, not against it. Use √(60² + 15²).
- Common wrong answer 3825 : That is v², 60² + 15². Now take the square root.
- Common wrong answer 60 : That is the velocity through the air only. The wind carries the aircraft east at the same time.
- Common wrong answer 58 : You took the squares away. Add them to find the resultant, the longest side: √(60² + 15²).
(b) Angle east of north: 14 ° (2 significant figures)
- Common wrong answer 76 : That angle is measured from east. From north, the side opposite the angle is the wind: tan θ = 15 ÷ 60.
- Common wrong answer 0.25 : That is tan θ. Now find the angle whose tangent is 0.25.

A full answer includes

(a) Method: v² = 60² + 15² = 3600 + 225 = 3825.
(a) Answer: v = √3825 = 62 m/s, to 2 significant figures.
(b) Method: tan θ = 15 ÷ 60 = 0.25.
(b) Answer: θ = 14° east of north.
I did not add the velocities to get 75 m/s.

Also accept

(a) 62 m/s; (b) 14° east of north.

Watch for

“The aircraft moves north at 60 m/s, because that is the way it points.” — The wind carries the air, and the aircraft with it, towards the east. Its velocity over the ground is the resultant of the two.
“(a) The resultant is 75 m/s, because 60 + 15 = 75.” — The velocities are at right angles, so the resultant is less than their sum: √(60² + 15²).
“(b) The angle is 76°.” — 76° is measured from east. From north, the side opposite the angle is the wind: tan θ = 15 ÷ 60.

10. FT2 · Apply

Model answer. (a) The velocity through the air, 2.0 m/s up, and the velocity of the wind, 5.0 m/s east, are at right angles. v² = 2.0² + 5.0² = 4.0 + 25 = 29, so v = √29 = 5.4 m/s. From the vertical, the side opposite the angle is the wind: tan θ = 5.0 ÷ 2.0 = 2.5, so θ = 68° from the vertical, towards the east. (b) The wind carries the balloon sideways, not up, so only the 2.0 m/s upwards counts: 300 m ÷ 2.0 m/s = 150 s. (c) In 150 s the wind carries it 5.0 m/s × 150 s = 750 m east. (d) The same time, 150 s. The wind acts at right angles to the upward velocity, so a stronger wind carries the balloon further east, 8.0 m/s × 150 s = 1200 m, but does not change how fast it rises. (e) Velocity is a vector: it has a size and a direction. Speed is a scalar, with a size only. Two velocities add as plain numbers only when they point the same way. Leo adds them as if they were speeds, scalars with size only. These two are at right angles, so the resultant is √(2.0² + 5.0²) = 5.4 m/s, less than 7.0 m/s.

A full answer includes

(a) Method: v² = 2.0² + 5.0² = 4.0 + 25 = 29.
(a) Answer: v = √29 = 5.4 m/s.
(a) Method: tan θ = 5.0 ÷ 2.0 = 2.5.
(a) Answer: θ = 68° from the vertical.
(b) Only the 2.0 m/s upwards counts: 300 m ÷ 2.0 m/s = 150 s.
(b) I did not divide 300 m by 5.4 m/s.
(c) 5.0 m/s × 150 s = 750 m east.
(d) The same time, 150 s.
(d) I explained that the wind is at right angles to the upward velocity, so it does not change the rise.
(e) Velocity is a vector, with size and direction.
(e) Speed is a scalar, with size only.
(e) Leo added the velocities as if they were scalars.
(e) At right angles, velocities combine by Pythagoras’ theorem, to 5.4 m/s.

Also accept

(a) 5.4 m/s, 68° from the vertical; (b) 150 s; (c) 750 m east; (d) the same time, the wind is at right angles to the rise; (e) velocities are vectors, so at right angles they combine by Pythagoras’ theorem, not by adding.

Watch for

“(b) It takes 300 m ÷ 5.4 m/s = 56 s.” — 5.4 m/s is along the slanting path. Only the upward 2.0 m/s carries the balloon up the 300 m: 300 ÷ 2.0 = 150 s.
“(d) A stronger wind makes the balloon take longer to rise.” — The wind pushes the balloon sideways, at right angles to its rise. The upward velocity is still 2.0 m/s, so the time is still 150 s.
“(a) The angle is 22° from the vertical.” — 22° is measured from the ground, the horizontal. From the vertical, the side opposite the angle is the wind: tan θ = 5.0 ÷ 2.0, so 68°.