Fig. 1.1 shows the directions of four forces acting on a racing car as it travels in a horizontal straight line. (a) Draw a line from each box on the left to the correct description of each force. (b) The table shows the sizes of the forces acting on the car at one time. force A / $\text{N}$ force B / $\text{N}$ force C / $\text{N}$ force D / $\text{N}$ --- --- --- --- 8000 1000 8000 600 The gravitational field strength $g$ is $10\ \text{N / kg}$. Calculate (i) the mass of…

Physics 5054/21 — May/June 2017

Cambridge O-Level · Theory · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Mass, Weight and Density · Electric Circuits · Electromagnetic Induction and Transformers · Forces · Kinematics · Pressure · +14 more

Q1ForcesMass, Weight and DensityKinematicsFree sample

Fig. 1.1 shows the directions of four forces acting on a racing car as it travels in a horizontal straight line.

Fig. 1.1

(a)

Draw a line from each box on the left to the correct description of each force.

Fig. 1.2

DifficultyEasy

Worked solution

Answer

force A: contact or normal reaction force
force B: driving force
force C: force of gravity
force D: air resistance and friction

Final answer

A–contact or normal reaction force, B–driving force, C–force of gravity, D–air resistance and friction

Detailed explanation

Walkthrough

The diagram shows a racing car moving horizontally. We identify each force by its direction and physical cause:

Force A points vertically upwards. This is the force from the ground pushing up on the car, known as the contact force or normal reaction force.
Force B points horizontally forward (to the right). This is the force propelling the car forward, generated by the wheels pushing against the road. This is the driving force.
Force C points vertically downwards. This is the pull of the Earth on the car, which is the force of gravity (or weight).
Force D points horizontally backward (to the left). This opposes the motion and is caused by the air pushing against the car and friction in the moving parts. This is air resistance and friction.

Key Takeaways

A free-body diagram shows all forces acting on an object.
Vertical forces on a horizontal surface are typically weight (down) and normal reaction (up).
Horizontal forces are typically driving force (forward) and resistive forces like air resistance/friction (backward).

Common Mistakes

Confusing the driving force with air resistance (getting the direction wrong).
Calling the upward force 'lift' (unless specified, it is the normal reaction/contact force).
Using 'weight' for force A (weight always acts downwards).

Things to Be Careful About

Ensure the matching is exact to the wording in the boxes: 'contact or normal reaction force', 'driving force', 'force of gravity', 'air resistance and friction'.

Techniques used

identify the forces acting on a moving vehiclematch force arrows to their physical descriptions

(b)

The table shows the sizes of the forces acting on the car at one time.

force A / $\text{N}$	force B / $\text{N}$	force C / $\text{N}$	force D / $\text{N}$
8000	1000	8000	600

The gravitational field strength $g$ is $10\ \text{N / kg}$ .

Calculate

(i)

the mass of the car,

mass = ______

DifficultyEasy

Worked solution

Working

Force C is the weight of the car (force of gravity).
$W = 8000 \text{ N}$
$g = 10 \text{ N/kg}$

$m = \frac{W}{g} = \frac{8000}{10} = 800 \text{ kg}$

Answer

mass = 800 kg

Final answer

800 kg

Detailed explanation

Walkthrough

Identify the weight: Force C acts downwards and is labelled 'force of gravity' in part (a). Its magnitude is given as 8000 N.
Use the relationship between weight, mass, and gravitational field strength: $W = mg$ .
Rearrange to solve for mass: $m = W / g$ .
Substitute the values: $m = 8000 / 10 = 800$ kg.

Key Takeaways

Weight is a force ( $W = mg$ ).
Gravitational field strength $g$ is approximately $10 \text{ N/kg}$ on Earth (or as given in the question).
Mass is measured in kg, weight in N.

Common Mistakes

Forgetting to divide by $g$ .
Using the wrong force value (e.g., using the driving force 1000 N).

Things to Be Careful About

The question gives $g = 10 \text{ N/kg}$ , use this value, not $9.8$ .

Techniques used

calculate mass from weight and gravitational field strengthuse the equation W = mg

(ii)

the resultant force on the car,

resultant force = ______

DifficultyEasy

Worked solution

Working

Vertical forces: Force A ( $8000 \text{ N}$ up) and Force C ( $8000 \text{ N}$ down) are equal and opposite, so they cancel.

Horizontal forces: Force B ( $1000 \text{ N}$ right) and Force D ( $600 \text{ N}$ left).

$F_{\text{resultant}} = 1000 - 600 = 400 \text{ N}$

Answer

resultant force = 400 N

Final answer

400 N

Detailed explanation

Walkthrough

Check vertical forces: Force A (up) is 8000 N, Force C (down) is 8000 N. Resultant vertical force is 0.
Calculate horizontal resultant: Force B (forward) is 1000 N, Force D (backward) is 600 N.
Resultant force = Forward force - Backward force = $1000 - 600 = 400$ N.
The direction is forward (to the right), but the question asks for the size (magnitude).

Key Takeaways

Resultant force is the vector sum of all forces.
Forces in opposite directions are subtracted.
Balanced vertical forces mean no vertical acceleration.

Common Mistakes

Adding the horizontal forces ( $1000 + 600$ ) instead of subtracting.
Including the vertical forces in the calculation (they cancel out).

Things to Be Careful About

Resultant force is a vector, but here only the magnitude is asked for (size).

Techniques used

find the resultant of horizontal forcessubtract opposing forces

(iii)

the acceleration of the car.

acceleration = ______

DifficultyMedium-Easy

Worked solution

Working

$F = ma$
$a = \frac{F}{m}$

Using $F = 400 \text{ N}$ (from part ii) and $m = 800 \text{ kg}$ (from part i):

$a = \frac{400}{800} = 0.50 \text{ m/s}^2$

Answer

acceleration = 0.50 m/s $^2$

Final answer

0.50 m/s $^2$

Detailed explanation

Walkthrough

Use Newton's second law: $F = ma$ , where $F$ is the resultant force and $m$ is the mass.
Rearrange for acceleration: $a = F / m$ .
Substitute values: $F = 400$ N, $m = 800$ kg.
Calculate: $a = 400 / 800 = 0.5$ m/s $^2$ .

Key Takeaways

$F = ma$ links force, mass, and acceleration.
Acceleration is in the direction of the resultant force.

Common Mistakes

Using the wrong force value (e.g., using 1000 N instead of the resultant 400 N).
Arithmetic errors in division.

Things to Be Careful About

Significant figures: 0.5 or 0.50 are both acceptable, but 0.50 m/s $^2$ is precise.

Techniques used

apply Newton's second law F = macalculate acceleration from resultant force and mass

(c)

At another time, the car is travelling at speed $u$ . It then accelerates for $5.0\ \text{s}$ with an acceleration of $1.6\ \text{m / s}^2$ , and reaches a speed of $20\ \text{m / s}$ .

Calculate the value of $u$ .

$u$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given:

Acceleration $a = 1.6 \text{ m/s}^2$
Time $t = 5.0 \text{ s}$
Final speed $v = 20 \text{ m/s}$
Initial speed = $u$

Change in speed = acceleration $\times$ time:
$\Delta v = a \times t = 1.6 \times 5.0 = 8.0 \text{ m/s}$

$v = u + \Delta v$
$20 = u + 8.0$
$u = 20 - 8.0 = 12 \text{ m/s}$

Answer

$u$ = 12 m/s

Final answer

12 m/s

Detailed explanation

Walkthrough

Identify the knowns: $a = 1.6 \text{ m/s}^2$ , $t = 5.0 \text{ s}$ , $v = 20 \text{ m/s}$ . Unknown is $u$ .
Use the equation for acceleration: $a = (v - u) / t$ , or rearrange as change in velocity $\Delta v = at$ .
Calculate the change in velocity: $1.6 \times 5.0 = 8.0 \text{ m/s}$ .
Since the car is accelerating, the final speed is higher than the initial speed: $v = u + 8.0$ .
Solve for $u$ : $u = 20 - 8.0 = 12 \text{ m/s}$ .

Key Takeaways

Acceleration is the rate of change of velocity: $a = (v - u) / t$ .
Change in velocity = $a \times t$ .

Common Mistakes

Adding 8.0 to 20 instead of subtracting (finding final speed instead of initial).
Using the wrong formula (e.g., $s = ut + 0.5at^2$ when time and speeds are given).

Things to Be Careful About

Ensure units are consistent (m/s, m/s $^2$ , s).
The question asks for $u$ , the initial speed.

Techniques used

use the kinematic equation v = u + atcalculate change in velocity from acceleration and time

The rest of this paper

10 more questions

Q2Pressure · Turning Effect of Forces4M
Q3Transfer of Thermal Energy6M
Q4Lenses and Dispersion5M
Q5Electric Circuits5M
Q6Practical Electricity7M
Q7Magnetic Effect of a Current and the d.c. Motor · Simple Magnetism and Magnetic Fields · Electromagnetic Induction and Transformers6M
Q8Radioactivity5M
Q9Energy, Work and Power · Mass, Weight and Density · Thermal Properties of Matter · Electromagnetic Induction and Transformers15M
Q10Reflection and Refraction of Light · General Properties of Waves · Sound15M
Q11Electric Circuits · Current, Voltage and Resistance · [Legacy] Temperature and Thermometry15M

Loading the full paper…