Answer all the questions in the spaces provided. (a) Distinguish between scalars and vectors. (b) Underline all the vector quantities in the list below. acceleration kinetic energy momentum power weight (c) A force of $7.5\ \text{N}$ acts at $40^{\circ}$ to the horizontal, as shown in Fig. 1.1. Calculate the component of the force that acts (i) horizontally, horizontal component = \_\_\_\_\_\_ $\text{N}$ (ii) vertically. vertical component = \_\_\_\_\_\_ $\text{N}$ (d) Two…

(a) Scalar: magnitude only. Vector: magnitude and direction. · (b) acceleration, momentum, weight · (c)(i) 5.7 N · (c)(ii) 4.8 N · (d) T1 = 5.7 N, T2 = 4.8 N

Physics 9702/23 — October/November 2011

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

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Topics Dynamics · Physical Quantities and Units · Forces, Density and Pressure · Kinematics · Work, Energy and Power · Deformation of Solids · +4 more

Q1Physical Quantities and UnitsForces, Density and PressureFree sample

Answer all the questions in the spaces provided.

(a)

Distinguish between scalars and vectors.

DifficultyEasy

Worked solution

Answer

A scalar has magnitude only.

A vector has magnitude and direction.

Final answer

Scalar: magnitude only. Vector: magnitude and direction.

Detailed explanation

Background Concept

Physical quantities can be classified by whether they require direction information.

A scalar quantity is fully described by a magnitude (a single number with a unit).
A vector quantity requires both magnitude and direction to be fully described.

Understanding the Question

You are asked to distinguish (i.e. clearly state the difference between) scalars and vectors.

Approach

Give the defining feature of each type:

scalar: magnitude only
vector: magnitude and direction

Step-by-Step Reasoning

Write what information is needed to specify a scalar: only size (with units).
Write what extra information is needed for a vector: a direction as well as size.

Key Takeaways

Scalars: one value (with unit).
Vectors: value (with unit) plus direction.

Common Mistakes

Saying “vectors have direction” but forgetting to mention they also have magnitude.
Giving examples only, without stating the defining difference (often loses the mark if the question asks to distinguish).

Things to Be Careful About

Words like “speed” vs “velocity” can be confusing: speed is scalar, velocity is vector because velocity includes direction.

Techniques used

state the definition of a scalar quantitystate the definition of a vector quantity

(b)

Underline all the vector quantities in the list below.

acceleration kinetic energy momentum power weight

DifficultyEasy

Worked solution

Answer

Vector quantities: acceleration, momentum, weight.

Final answer

acceleration, momentum, weight

Detailed explanation

Background Concept

A vector requires direction; a scalar does not.

Typical examples:

Vectors: displacement, velocity, acceleration, force, momentum, weight.
Scalars: mass, time, energy, power, temperature.

Understanding the Question

From the list

acceleration
kinetic energy
momentum
power
weight

you must select only those that need direction as well as magnitude.

Approach

Go through each quantity and decide: does it need direction?

Step-by-Step Reasoning

Acceleration: change of velocity per unit time, so it has direction (vector).
Kinetic energy: energy depends on speed, not direction (scalar).
Momentum: $\vec{p} = m\vec{v}$ , so it has the same direction as velocity (vector).
Power: rate of energy transfer, no direction (scalar).
Weight: a force (gravitational force), acts vertically downward (vector).

Key Takeaways

If a quantity is a type of force or depends on a vector like velocity, it is usually a vector.
Energies and rates like power are typically scalars.

Common Mistakes

Choosing kinetic energy as a vector because it relates to velocity (it depends on $v^2$ , so direction cancels).
Forgetting weight is a force and therefore a vector.

Things to Be Careful About

“Momentum” is a vector even though you may often only calculate its magnitude in 1D problems.

Techniques used

classify quantities as scalar or vector based on whether direction is requireduse standard physics definitions of the listed quantities

(c)

A force of $7.5\ \text{N}$ acts at $40^{\circ}$ to the horizontal, as shown in Fig. 1.1.

Fig. 1.1

Calculate the component of the force that acts

(i)

horizontally,

horizontal component = ______ $\text{N}$

DifficultyMedium-Easy

Worked solution

Working

Horizontal component

F_x = 7.5\cos 40^{\circ}

F_x = 5.7\ \text{N}

Answer

$5.7\ \text{N}$

Final answer

5.7 N

Detailed explanation

Background Concept

Any vector can be split into perpendicular components (usually horizontal and vertical). If a force $F$ makes an angle $\theta$ to the horizontal, then:

F_x = F\cos\theta

F_y = F\sin\theta

The cosine corresponds to the adjacent side to the angle, and sine to the opposite side.

Understanding the Question

A force of $7.5\ \text{N}$ acts at $40^{\circ}$ above the horizontal. You are asked only for the horizontal component (the part acting along the horizontal direction).

diagram-c-resolution

Approach

Use the component formula with cosine because the given angle is from the horizontal.

Step-by-Step Reasoning

Identify $F = 7.5\ \text{N}$ and $\theta = 40^{\circ}$ measured from the horizontal.
Horizontal component:

F_x = F\cos\theta = 7.5\cos 40^{\circ}

Calculate:

F_x \approx 5.7\ \text{N}

Key Takeaways

Angle to the horizontal: horizontal uses $\cos$ , vertical uses $\sin$ .
Components are always smaller than the original vector (unless angle is $0^{\circ}$ or $90^{\circ}$ ).

Common Mistakes

Swapping sine and cosine (giving $7.5\sin 40^{\circ}$ for the horizontal component).
Forgetting the unit N.

Things to Be Careful About

Check what the angle is measured from. If the angle were to the vertical, the trig functions would swap.
Rounding: give a sensible number of significant figures (typically 2 s.f. here).

Techniques used

resolve a vector into perpendicular componentsuse trigonometry to find the adjacent component

(ii)

vertically.

vertical component = ______ $\text{N}$

DifficultyMedium-Easy

Worked solution

Working

Vertical component

F_y = 7.5\sin 40^{\circ}

F_y = 4.8\ \text{N}

Answer

$4.8\ \text{N}$

Final answer

4.8 N

Detailed explanation

Background Concept

When a force $F$ is at an angle $\theta$ to the horizontal, the vertical component is the side opposite $\theta$ in the right triangle, so it is given by sine:

F_y = F\sin\theta

Understanding the Question

The same force of $7.5\ \text{N}$ acts at $40^{\circ}$ above the horizontal, and now you need the vertical component.

diagram-c-resolution

Approach

Use $F_y = F\sin\theta$ because the given angle is to the horizontal.

Step-by-Step Reasoning

Take $F = 7.5\ \text{N}$ , $\theta = 40^{\circ}$ .
Vertical component:

F_y = 7.5\sin 40^{\circ}

Evaluate:

F_y \approx 4.8\ \text{N}

Key Takeaways

With angle to the horizontal, vertical component uses sine.
Components depend on the choice of axes; here axes are horizontal/vertical.

Common Mistakes

Using cosine for the vertical component when the angle is given to the horizontal.
Giving the correct number but with wrong unit or missing unit.

Things to Be Careful About

Ensure the angle used in the calculator is in degrees, not radians.
Do not round too early if you are using the value in later calculations.

Techniques used

resolve a vector into perpendicular componentsuse trigonometry to find the opposite component

(d)

Two strings support a load of weight $7.5\ \text{N}$ , as shown in Fig. 1.2.

Fig. 1.2

One string has a tension $T_1$ and is at an angle $50^{\circ}$ to the horizontal. The other string has a tension $T_2$ and is at an angle $40^{\circ}$ to the horizontal. The object is in equilibrium.
Determine the values of $T_1$ and $T_2$ by using a vector triangle or by resolving forces.

$T_1$ = ______ $\text{N}$
$T_2$ = ______ $\text{N}$

DifficultyMedium

Worked solution

Working

In equilibrium, take rightward and upward as positive.

Horizontal:

T_1\cos 50^{\circ} = T_2\cos 40^{\circ}

Vertical:

T_1\sin 50^{\circ} + T_2\sin 40^{\circ} = 7.5

From horizontal,

T_1 = T_2\frac{\cos 40^{\circ}}{\cos 50^{\circ}}

Substitute into vertical:

T_2\frac{\cos 40^{\circ}}{\cos 50^{\circ}}\sin 50^{\circ} + T_2\sin 40^{\circ} = 7.5

T_2 = 4.82\ \text{N},\quad T_1 = 5.74\ \text{N}

Answer

$T_1 = 5.7\ \text{N}$

$T_2 = 4.8\ \text{N}$

Final answer

T1 = 5.7 N, T2 = 4.8 N

Detailed explanation

Background Concept

For a body in equilibrium, the resultant force is zero. In 2D this means the sum of components in each perpendicular direction is zero:

\sum F_x = 0,\qquad \sum F_y = 0

When forces are at angles, we resolve each force into horizontal and vertical components using sine and cosine with the angle given.

Understanding the Question

A load of weight $7.5\ \text{N}$ is supported by two strings:

tension $T_1$ at $50^{\circ}$ to the horizontal
tension $T_2$ at $40^{\circ}$ to the horizontal

The load is at rest, so forces balance. You must determine $T_1$ and $T_2$ .

diagram-d-fbd

Approach

Draw a free-body diagram of the load showing three forces: $T_1$ , $T_2$ , and weight $7.5\ \text{N}$ downward.
Resolve the tensions into horizontal and vertical components.
Apply equilibrium:
- horizontal components cancel
- vertical components add to $7.5\ \text{N}$
Solve the two simultaneous equations for $T_1$ and $T_2$ .

Step-by-Step Reasoning

Resolve horizontally (take right as positive). One tension pulls left, the other pulls right, so their horizontal components must be equal in magnitude:

T_1\cos 50^{\circ} = T_2\cos 40^{\circ}

Resolve vertically (up positive). Both tensions have upward components, balancing the weight:

T_1\sin 50^{\circ} + T_2\sin 40^{\circ} = 7.5

Use the horizontal equation to express $T_1$ in terms of $T_2$ :

T_1 = T_2\frac{\cos 40^{\circ}}{\cos 50^{\circ}}

Substitute into the vertical equation:

T_2\frac{\cos 40^{\circ}}{\cos 50^{\circ}}\sin 50^{\circ} + T_2\sin 40^{\circ} = 7.5

Solve for $T_2$ (numerically this gives $T_2 \approx 4.82\ \text{N}$ ), then substitute back to get $T_1 \approx 5.74\ \text{N}$ .
Quote to appropriate significant figures:

T_1 = 5.7\ \text{N},\qquad T_2 = 4.8\ \text{N}

Key Takeaways

In equilibrium, resolve forces and apply $\sum F_x = 0$ and $\sum F_y = 0$ .
Always relate sine/cosine to the angle given (here angles are to the horizontal).

Common Mistakes

Using $\sin$ and $\cos$ the wrong way round (common when unsure what the angle is measured from).
Setting vertical components equal (they do not cancel; they add to balance the weight).
Forgetting there are two independent equations needed to find two unknown tensions.

Things to Be Careful About

Keep a consistent sign convention (e.g. right/up positive) before writing equations.
Ensure calculator is in degrees.
Quote units (N) and sensible significant figures.

Techniques used

draw a free-body diagram for forces in equilibriumresolve each tension into horizontal and vertical componentsapply equilibrium conditions for resultant force in perpendicular directionssolve simultaneous equations

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