ESAT Mock Physics: Questions & Worked Solutions

Question 1

1 mark

A neutral plastic rod is rubbed with a dry cloth. During the rubbing process,

5.0 \times 10^{12}

electrons are transferred from the cloth to the rod. What is the resulting charge on the cloth? (The elementary charge is

e = 1.6 \times 10^{-19}\,\text{C}

).

A. $-8.0 \times 10^{-7}\,\text{C}$
B. $+8.0 \times 10^{-7}\,\text{C}$
C. $-3.1 \times 10^{-32}\,\text{C}$
D. $+3.1 \times 10^{-32}\,\text{C}$
E. $0\,\text{C}$

Answer: B

Worked solution

Step 1: Identify the movement of charge. Electrons are negatively charged. Since electrons move from the cloth to the rod, the cloth loses negative charge and becomes positively charged.
Step 2: Calculate the magnitude of the charge transferred. The magnitude

Q

is the number of electrons

n

multiplied by the elementary charge

e

.

Q = n \times e = (5.0 \times 10^{12}) \times (1.6 \times 10^{-19}\,\text{C})

Q = 8.0 \times 10^{-7}\,\text{C}

Step 3: Determine the sign. As the cloth lost electrons, its charge is

+8.0 \times 10^{-7}\,\text{C}

.

Question 2

1 mark

Three identical metal spheres, X, Y, and Z, are mounted on insulating stands. Sphere X has an initial charge of

+12\,\text{nC}

. Spheres Y and Z are initially neutral (

0\,\text{nC}

). The following sequence of operations is performed:
1. Sphere X is brought into contact with sphere Y and then separated.
2. Sphere Y is then brought into contact with sphere Z and then separated.
What is the final charge on sphere Y?

A. $0\,\text{nC}$
B. $3.0\,\text{nC}$
C. $4.0\,\text{nC}$
D. $6.0\,\text{nC}$
E. $12\,\text{nC}$

Answer: B

Worked solution

Step 1: Contact between X and Y. The total charge is

+12\,\text{nC} + 0 = +12\,\text{nC}

. Since the spheres are identical, the charge divides equally. Each sphere (X and Y) now has a charge of

12 / 2 = +6.0\,\text{nC}

.
Step 2: Contact between Y and Z. The current charge on Y is

+6.0\,\text{nC}

and Z is

0\,\text{nC}

. The total charge for this pair is

+6.0\,\text{nC} + 0 = +6.0\,\text{nC}

.
Step 3: Equal distribution. When separated, the charge divides equally between Y and Z. Each sphere (Y and Z) now has a charge of

6.0 / 2 = +3.0\,\text{nC}

.
Thus, the final charge on sphere Y is

3.0\,\text{nC}

.

Question 3

1 mark

A aircraft is flying through a storm cloud and becomes electrostaticallly charged due to friction with the air. Before the aircraft is refuelled on the ground, a conducting wire is connected between the aircraft's fuselage and the Earth. Which statement correctly describes the purpose and mechanism of this process?

A.To transfer protons from the ground to the aircraft to balance the charge.
B.To allow excess electrons to flow to or from the Earth, neutralizing the aircraft.
C.To insulate the aircraft from the fuel pump to prevent current flow.
D.To increase the potential difference between the aircraft and the fuel to speed up the flow.
E.To generate a spark in a controlled environment away from the fuel tank.

Answer: B

Worked solution

Charging by friction (triboelectric effect) occurs as the aircraft moves through the air, leading to a build-up of electrons or a deficiency of them. This creates a high potential difference between the aircraft and the ground. If the aircraft were refuelled while charged, a spark could jump between the fuel nozzle and the tank, igniting the fuel vapours. Connecting the aircraft to the Earth (earthing) provides a low-resistance path for electrons. If the aircraft is negatively charged, electrons flow to the Earth; if it is positively charged, electrons flow from the Earth to the aircraft. In both cases, the aircraft becomes neutral, eliminating the risk of a spark. Protons are bound in the nuclei of the metal atoms and do not flow through wires.

Question 4

1 mark

Two long, straight wires are parallel to the

z

-axis in a Cartesian coordinate system. Wire 1 passes through the

xy

-plane at coordinates

(-d, 0)

and Wire 2 passes through the

xy

-plane at

(d, 0)

. Both wires carry an identical current

I

directed into the page (in the

-z

direction). What is the direction of the net magnetic field at the point

(0, d)

in the

xy

-plane?

A.Toward the positive $x$ -axis
B.Toward the negative $x$ -axis
C.Toward the positive $y$ -axis
D.Toward the negative $y$ -axis
E.The net magnetic field is zero

Answer: A

Worked solution

To find the net field at

P(0, d)

, we use the right-hand grip rule for each wire.
1. **Wire 1** is at

(-d, 0)

. The vector from the wire to

P

is

(d, d)

, which is at a

45^\circ

angle in the first quadrant relative to the wire. Since the current is into the page, the field lines are clockwise. A clockwise tangent at the

45^\circ

position (top-right of the wire) points down and to the right. Its components are proportional to

(+1, -1)

.
2. **Wire 2** is at

(d, 0)

. The vector from the wire to

P

is

(-d, d)

, which is at a

135^\circ

angle (top-left relative to the wire). A clockwise tangent at this position points up and to the right. Its components are proportional to

(+1, +1)

.
3. **Superposition**: The vertical components (

-1

and

+1

) are of equal magnitude because the distances

\sqrt{d^2+d^2}

and currents are identical. These cancel out. The horizontal components are both positive and add together.
Therefore, the net magnetic field is directed along the positive

x

-axis.

Question 5

1 mark

A uniform magnetic field

B

is applied at right angles to a straight copper wire of length

L

and diameter

d

. When a potential difference

V

is applied across the ends of the wire, a magnetic force

F

is exerted on it. If the wire is replaced by another copper wire of the same length

L

but with diameter

2d

, and the same potential difference

V

is applied, what is the new magnetic force exerted on the wire? (Assume the resistivity of the copper and the magnetic field remain constant).

A. $\frac{1}{4}F$
B. $\frac{1}{2}F$
C. $F$
D. $2F$
E. $4F$

Answer: E

Worked solution

The magnetic force is given by

F = BIL

. The current

I

is determined by Ohm's Law,

I = V/R

, where

R

is the resistance of the wire. Resistance is given by

R = \rho \frac{L}{A}

, where

A

is the cross-sectional area. Since

A = \pi (d/2)^2

, we have

R \propto 1/d^2

. If the diameter

d

is doubled to

2d

, the cross-sectional area

A

increases by a factor of

2^2 = 4

. Consequently, the resistance

R

decreases by a factor of

4

(since

R \propto 1/A

). If

R

is quartered and the potential difference

V

is constant, the current

I = V/R

must increase by a factor of

4

. Substituting this back into the force equation

F = BIL

, since

B

and

L

are constant and

I

has quadrupled, the new force is

4F

.

Question 6

1 mark

A coil consists of

200

turns of wire. A magnetic field passing through the coil changes such that the magnetic flux increases at a constant rate, inducing an electromotive force (emf) of

4.0\,\text{V}

. If the coil is replaced with one containing

100

turns of wire, and the magnetic flux now increases at four times the original rate, what is the new induced emf?

A. $0.5\,\text{V}$
B. $2.0\,\text{V}$
C. $4.0\,\text{V}$
D. $8.0\,\text{V}$
E. $16.0\,\text{V}$

Answer: D

Worked solution

The magnitude of the induced emf (

V

) is given by Faraday's Law:

V = N \frac{\Delta \Phi}{\Delta t}

, where

N

is the number of turns and

\frac{\Delta \Phi}{\Delta t}

is the rate of change of magnetic flux. Initially,

V_1 = 200 \times R = 4.0\,\text{V}

(where

R

is the original rate). In the new scenario, the number of turns is

N_2 = 100

(which is

0.5 \times 200

) and the rate is

R_2 = 4R

. Therefore, the new emf is

V_2 = 100 \times 4R = 400R

. Since

200R = 4.0\,\text{V}

, then

400R = 2 \times 4.0\,\text{V} = 8.0\,\text{V}

.

Question 7

1 mark

A simple ac generator consists of a coil rotating in a uniform magnetic field. The generator produces an output voltage with a peak value of

V

and a frequency of

f

. If the speed of rotation of the coil is tripled, what are the new peak voltage and the new frequency of the output?

A.Peak voltage is $V$ ; frequency is $3f$
B.Peak voltage is $3V$ ; frequency is $f$
C.Peak voltage is $3V$ ; frequency is $3f$
D.Peak voltage is $9V$ ; frequency is $3f$
E.Peak voltage is $3V$ ; frequency is $9f$

Answer: C

Worked solution

For a simple ac generator, the induced emf at any time is

V(t) = BAN\omega \sin(\omega t)

, where

\omega

is the angular velocity of rotation. The peak voltage is

V_{peak} = BAN\omega

. The frequency of the output is

f = \frac{\omega}{2\pi}

. If the speed of rotation is tripled, the new angular velocity is

3\omega

. This directly triples the frequency (

3f

) and also triples the peak voltage (

3V_{peak}

).

Question 8

1 mark

A student moves a bar magnet into and out of a stationary coil of wire. The coil is connected to a sensitive ammeter. When the student pushes the North pole of the magnet quickly into the coil, the ammeter shows a momentary deflection of

3\,\text{units}

to the right. Which of the following actions would result in a momentary deflection of more than

3\,\text{units}

to the left?

A.Pulling the North pole out of the coil slowly.
B.Pushing the South pole into the coil more quickly.
C.Pulling the South pole out of the coil more quickly.
D.Holding the North pole stationary inside the coil.
E.Pushing the North pole into the coil more slowly.

Answer: B

Worked solution

According to Lenz's Law, the direction of the induced current opposes the change in magnetic flux. If pushing a North pole in causes a deflection to the right, then: 1. Pulling a North pole out causes a deflection to the left. 2. Pushing a South pole in causes a deflection to the left. 3. Pulling a South pole out causes a deflection to the right. To obtain a deflection of 'more than

3\,\text{units}

', the rate of change of flux must be greater than in the original case, meaning the magnet must move more quickly. Therefore, pushing the South pole in more quickly will result in a larger deflection to the left.

Question 9

1 mark

A car starts from rest and accelerates at a constant rate of

4.0\,\text{m s}^{-2}

for

5.0\,\text{s}

. It then travels at a constant velocity for a further

15.0\,\text{s}

.

What is the average speed of the car for the entire

20.0\,\text{s}

journey?

A. $10.0\,\text{m s}^{-1}$
B. $15.0\,\text{m s}^{-1}$
C. $16.0\,\text{m s}^{-1}$
D. $17.5\,\text{m s}^{-1}$
E. $20.0\,\text{m s}^{-1}$

Answer: D

Worked solution

First, calculate the movement in Phase 1 (acceleration):
- Initial velocity

u = 0\,\text{m s}^{-1}

- Acceleration

a = 4.0\,\text{m s}^{-2}

- Time

t_1 = 5.0\,\text{s}

- Final velocity

v = u + at_1 = 0 + (4.0 \times 5.0) = 20\,\text{m s}^{-1}

- Distance

s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(4.0)(5.0)^2 = 2 \times 25 = 50\,\text{m}

Next, calculate the movement in Phase 2 (constant velocity):
- Velocity

v = 20\,\text{m s}^{-1}

- Time

t_2 = 15.0\,\text{s}

- Distance

s_2 = v \times t_2 = 20 \times 15.0 = 300\,\text{m}

Finally, calculate the average speed:
- Total distance

s_{total} = s_1 + s_2 = 50 + 300 = 350\,\text{m}

- Total time

t_{total} = 5.0 + 15.0 = 20.0\,\text{s}

- Average speed

v_{avg} = \frac{s_{total}}{t_{total}} = \frac{350}{20.0} = 17.5\,\text{m s}^{-1}

The correct option is D.

Question 10

1 mark

A driver is travelling at a constant speed of

30\,\text{m s}^{-1}

on a straight road. They see a hazard and apply the brakes after a reaction time of

0.50\,\text{s}

. The brakes provide a constant deceleration of

6.0\,\text{m s}^{-2}

until the car comes to a complete stop.

What is the total distance travelled by the car from the moment the driver sees the hazard until the car stops?

A. $15\,\text{m}$
B. $75\,\text{m}$
C. $90\,\text{m}$
D. $105\,\text{m}$
E. $165\,\text{m}$

Answer: C

Worked solution

The total stopping distance is the sum of the thinking distance and the braking distance.

1. Thinking distance (

s_t

):
During the reaction time, the car travels at a constant speed.

s_t = v \times t_{reaction} = 30\,\text{m s}^{-1} \times 0.50\,\text{s} = 15\,\text{m}

.

2. Braking distance (

s_b

):
Using the equation

v^2 - u^2 = 2as

, where the final velocity

v = 0

, initial velocity

u = 30\,\text{m s}^{-1}

, and acceleration

a = -6.0\,\text{m s}^{-2}

:

0^2 - (30)^2 = 2 \times (-6.0) \times s_b

-900 = -12 \times s_b

s_b = \frac{900}{12} = 75\,\text{m}

.

3. Total distance:

s_{total} = s_t + s_b = 15 + 75 = 90\,\text{m}

.

The correct option is C.

Question 11

1 mark

Two boxes, A and B, are stacked in a lift as shown. Box A, of mass

5.0\text{ kg}

, sits on top of Box B, of mass

10.0\text{ kg}

. Box B is in contact with the floor of the lift. A light rope attached to Box A exerts a constant upward tension of

30\text{ N}

on Box A. The lift is accelerating downwards at

2.0\text{ m s}^{-2}

.

What is the magnitude of the normal contact force exerted by Box B on the floor of the lift?
(gravitational field strength

g = 10\text{ N kg}^{-1}

)

A.60 N
B.90 N
C.120 N
D.150 N
E.180 N

Answer: B

Worked solution

To find the normal contact force

R

between Box B and the lift floor, we can treat Box A and Box B as a single system of total mass

M = m_A + m_B = 5.0 + 10.0 = 15.0\text{ kg}

.

Identify the vertical forces acting on the combined system:
1. Total weight

W = Mg = 15.0 \times 10 = 150\text{ N}

acting downwards.
2. Tension

T = 30\text{ N}

acting upwards on Box A.
3. Normal contact force

R

acting upwards on Box B from the floor.

The system is accelerating downwards at

a = 2.0\text{ m s}^{-2}

. Taking the downward direction as positive, we apply Newton’s Second Law (

F_{\text{net}} = Ma

):

W - T - R = Ma

Substitute the known values:

150 - 30 - R = 15.0 \times 2.0

120 - R = 30

R = 120 - 30 = 90\text{ N}

.

Alternatively, considering the boxes separately:
For Box A:

m_A g - T - R_{AB} = m_A a \Rightarrow 50 - 30 - R_{AB} = 5 \times 2 \Rightarrow R_{AB} = 10\text{ N}

(force from B on A).
For Box B:

m_B g + R_{AB} - R = m_B a \Rightarrow 100 + 10 - R = 10 \times 2 \Rightarrow 110 - R = 20 \Rightarrow R = 90\text{ N}

.

Question 12

1 mark

Three blocks, X, Y, and Z, are connected by light inextensible strings and are being pulled vertically upwards. The masses are

m_X = 2.0\text{ kg}

,

m_Y = 3.0\text{ kg}

, and

m_Z = 5.0\text{ kg}

. A pulling force

F = 160\text{ N}

is applied to block X. Each block experiences a constant air resistance (drag) force of

4.0\text{ N}

.

What is the tension in the string connecting block Y and block Z?
(gravitational field strength

g = 10\text{ N kg}^{-1}

)

A.54 N
B.74 N
C.78 N
D.84 N
E.106 N

Answer: C

Worked solution

Step 1: Find the acceleration

a

of the whole system.
Total mass

M = 2.0 + 3.0 + 5.0 = 10.0\text{ kg}

.
Total weight

W = Mg = 10.0 \times 10 = 100\text{ N}

(downwards).
Total air resistance

D = 3 \times 4.0 = 12\text{ N}

(downwards, as the blocks move upwards).
Applied force

F = 160\text{ N}

(upwards).

F_{\text{net}} = F - W - D = 160 - 100 - 12 = 48\text{ N}

.

a = \frac{F_{\text{net}}}{M} = \frac{48}{10.0} = 4.8\text{ m s}^{-2}

.

Step 2: Find the tension

T

in the string connecting Y and Z by looking at block Z.
Forces on block Z:
1. Tension

T

(upwards).
2. Weight

W_Z = m_Z g = 5.0 \times 10 = 50\text{ N}

(downwards).
3. Air resistance

D_Z = 4.0\text{ N}

(downwards).

Apply Newton’s Second Law to block Z (taking up as positive):

T - W_Z - D_Z = m_Z a

T - 50 - 4.0 = 5.0 \times 4.8

T - 54 = 24.0

T = 78\text{ N}

.

Question 13

1 mark

A hot air balloon of total mass

1200\text{ kg}

is descending vertically. The burner is adjusted to provide a constant upward upthrust of

10,500\text{ N}

. At a certain instant, the balloon is observed to be accelerating downwards at

0.5\text{ m s}^{-2}

.

What is the magnitude of the air resistance (drag) acting on the balloon at this instant?
(gravitational field strength

g = 10\text{ N kg}^{-1}

)

A.600 N
B.900 N
C.1500 N
D.2100 N
E.2700 N

Answer: B

Worked solution

First, identify all vertical forces acting on the balloon and their directions:
1. Weight

W = mg = 1200 \times 10 = 12,000\text{ N}

acting downwards.
2. Upthrust

U = 10,500\text{ N}

acting upwards.
3. Air resistance

D

acting upwards (since the balloon is descending).

The balloon is accelerating downwards at

a = 0.5\text{ m s}^{-2}

. We use Newton’s Second Law, taking the downward direction as positive:

F_{\text{net}} = W - U - D = ma

Substitute the known values:

12,000 - 10,500 - D = 1200 \times 0.5

1,500 - D = 600

Rearrange to solve for

D

:

D = 1,500 - 600 = 900\text{ N}

.

Question 14

1 mark

Two light springs,

S_1

and

S_2

, have spring constants

k

and

2k

respectively. A weight

W

is supported by these springs in two different configurations. In Arrangement 1, the springs are connected in parallel. In Arrangement 2, the springs are connected in series. What is the ratio of the total elastic potential energy stored in the springs in Arrangement 2 to the total elastic potential energy stored in Arrangement 1?
(Assume all extensions are within the limit of proportionality for both springs.)

A. $\frac{2}{9}$
B. $\frac{1}{4}$
C.4
D. $\frac{9}{2}$
E.9

Answer: D

Worked solution

First, we determine the effective spring constant for each arrangement.

In Arrangement 1 (parallel), the effective spring constant

k_P

is the sum of the individual constants:

k_P = k + 2k = 3k

The energy stored in a spring system supporting a weight

W

is given by

E = \frac{F^2}{2k_{eff}}

. For Arrangement 1:

E_1 = \frac{W^2}{2(3k)} = \frac{W^2}{6k}

In Arrangement 2 (series), the effective spring constant

k_S

is found using the reciprocal sum:

\frac{1}{k_S} = \frac{1}{k} + \frac{1}{2k} = \frac{3}{2k} \implies k_S = \frac{2k}{3}

The energy stored for Arrangement 2 is:

E_2 = \frac{W^2}{2(2k/3)} = \frac{3W^2}{4k}

Finally, we find the ratio

\frac{E_2}{E_1}

:

\frac{E_2}{E_1} = \frac{3W^2 / 4k}{W^2 / 6k} = \frac{3}{4} \times 6 = \frac{18}{4} = 4.5 = \frac{9}{2}

The correct answer is D.

Question 15

1 mark

A specialized elastic component is designed to have a variable stiffness. For extensions between

0

and

0.10\,\text{m}

, it obeys Hooke's law with a spring constant of

200\,\text{N m}^{-1}

. For extensions greater than

0.10\,\text{m}

, its stiffness increases such that the additional force required per unit of additional extension is

600\,\text{N m}^{-1}

.

What is the total work done to stretch this component from an extension of

0

to a total extension of

0.20\,\text{m}

?
(Assume the component does not exceed its elastic limit.)

A. $4.0\,\text{J}$
B. $5.0\,\text{J}$
C. $6.0\,\text{J}$
D. $8.0\,\text{J}$
E. $12.0\,\text{J}$

Answer: C

Worked solution

We can calculate the work done by finding the area under the force-extension graph in two stages.

Stage 1: Extension from

x = 0

to

x = 0.10\,\text{m}

.
The force increases linearly from

0

to

F_1 = k_1 x_1 = 200 \times 0.10 = 20\,\text{N}

.
The work done

W_1

is the area of the triangle:

W_1 = \frac{1}{2} F_1 x_1 = \frac{1}{2} \times 20 \times 0.10 = 1.0\,\text{J}

.

Stage 2: Extension from

x = 0.10\,\text{m}

to

x = 0.20\,\text{m}

.
The starting force is

20\,\text{N}

. The additional extension is

\Delta x = 0.10\,\text{m}

. The stiffness is now

k_2 = 600\,\text{N m}^{-1}

.
The force at

x = 0.20\,\text{m}

is

F_2 = 20 + (600 \times 0.10) = 20 + 60 = 80\,\text{N}

.
The work done

W_2

in this stage is the area of the trapezium:

W_2 = \frac{1}{2} (F_1 + F_2) \Delta x = \frac{1}{2} (20 + 80) \times 0.10 = 50 \times 0.10 = 5.0\,\text{J}

.

Total work done

W = W_1 + W_2 = 1.0 + 5.0 = 6.0\,\text{J}

.

The correct answer is C.

Question 16

1 mark

A solid metal sphere of mass

2.0\,\text{kg}

is dropped from a stationary helicopter at a high altitude. At a certain point during its fall, the sphere has a downward acceleration of

4.0\,\text{m s}^{-2}

. What is the magnitude of the air resistance acting on the sphere at this instant?
(The gravitational field strength

g

is

10\,\text{N kg}^{-1}

.)

A. $2.0\,\text{N}$
B. $8.0\,\text{N}$
C. $12\,\text{N}$
D. $20\,\text{N}$
E. $28\,\text{N}$

Answer: C

Worked solution

First, calculate the weight

W

of the sphere using

W = mg

:

W = 2.0\,\text{kg} \times 10\,\text{N kg}^{-1} = 20\,\text{N}

.
The net force

F_{net}

acting on the sphere can be found using Newton's Second Law,

F_{net} = ma

, where

a

is the downward acceleration:

F_{net} = 2.0\,\text{kg} \times 4.0\,\text{m s}^{-2} = 8.0\,\text{N}

(downwards).
The forces acting on the sphere are its weight

W

(downwards) and air resistance

R

(upwards). The net force is the difference between these:

F_{net} = W - R

8.0 = 20 - R

R = 20 - 8.0 = 12\,\text{N}

.
The magnitude of the air resistance is

12\,\text{N}

.

Question 17

1 mark

A scientific probe has a mass of

120\,\text{kg}

on Earth. It is sent to a planet where the gravitational field strength is

4.0\,\text{N kg}^{-1}

. The probe is lowered from a hovering spacecraft onto the planet's surface by a cable at a constant vertical speed of

2.0\,\text{m s}^{-1}

. What is the tension in the cable while the probe is being lowered?
(The gravitational field strength

g

on Earth is

10\,\text{N kg}^{-1}

.)

A. $120\,\text{N}$
B. $240\,\text{N}$
C. $480\,\text{N}$
D. $1200\,\text{N}$
E. $1680\,\text{N}$

Answer: C

Worked solution

First, identify the mass of the probe. Mass is a constant property and does not change with location; therefore, the mass on the planet is

120\,\text{kg}

.
Calculate the weight of the probe on the planet (

W_p

):

W_p = m \times g_{planet} = 120\,\text{kg} \times 4.0\,\text{N kg}^{-1} = 480\,\text{N}

.
The probe is being lowered at a constant speed, which means its acceleration

a = 0

. According to Newton's First Law (or Second Law with

a=0

), the net force acting on the probe must be zero.
The upward tension

T

must balance the downward weight

W_p

:

T - W_p = 0 \implies T = W_p = 480\,\text{N}

.

Question 18

1 mark

An object of mass

4.0\,\text{kg}

is travelling at

10\,\text{m s}^{-1}

in a straight line on a smooth horizontal surface when it explodes into two fragments, P and Q, of mass

1.0\,\text{kg}

and

3.0\,\text{kg}

respectively. Immediately after the explosion, fragment P is moving at

25\,\text{m s}^{-1}

in the original direction of motion of the object. What is the velocity of fragment Q and the increase in the total kinetic energy of the system?

A.velocity of Q: $5.0\,\text{m s}^{-1}$ in original direction; energy increase: $150\,\text{J}$
B.velocity of Q: $5.0\,\text{m s}^{-1}$ in original direction; energy increase: $350\,\text{J}$
C.velocity of Q: $5.0\,\text{m s}^{-1}$ in opposite direction; energy increase: $150\,\text{J}$
D.velocity of Q: $15\,\text{m s}^{-1}$ in original direction; energy increase: $150\,\text{J}$
E.velocity of Q: $15\,\text{m s}^{-1}$ in opposite direction; energy increase: $350\,\text{J}$

Answer: A

Worked solution

First, we use the law of conservation of momentum to find the final velocity of fragment Q (

v_Q

). Let the original direction be positive.
Initial momentum:

p_i = m_{total} u = 4.0 \times 10 = 40\,\text{kg m s}^{-1}

.
Final momentum:

p_f = m_P v_P + m_Q v_Q = (1.0 \times 25) + (3.0 \times v_Q)

.
By conservation:

40 = 25 + 3v_Q \implies 15 = 3v_Q \implies v_Q = 5.0\,\text{m s}^{-1}

.
Since the result is positive, it is in the original direction.

Next, calculate the change in kinetic energy (

KE

):

KE_{initial} = \frac{1}{2} m_{total} u^2 = \frac{1}{2} (4.0) (10)^2 = 200\,\text{J}

.

KE_{final} = \frac{1}{2} m_P v_P^2 + \frac{1}{2} m_Q v_Q^2 = \frac{1}{2} (1.0) (25)^2 + \frac{1}{2} (3.0) (5.0)^2

KE_{final} = 312.5 + 37.5 = 350\,\text{J}

.
Increase in

KE = 350 - 200 = 150\,\text{J}

.
The correct answer is A.

Question 19

1 mark

Two particles, X and Y, both of mass

m

, are moving towards each other in a vacuum. Particle X has speed

u

and particle Y has speed

\frac{u}{2}

. The particles collide head-on. Immediately after the collision, particle X is at rest. What percentage of the initial total kinetic energy of the system is lost during the collision?

A.20%
B.40%
C.50%
D.75%
E.80%

Answer: E

Worked solution

Let the direction of X be positive.
Initial momentum:

p_i = mu + m(-\frac{u}{2}) = \frac{1}{2}mu

.
Initial kinetic energy:

KE_i = \frac{1}{2}mu^2 + \frac{1}{2}m(-\frac{u}{2})^2 = \frac{1}{2}mu^2 + \frac{1}{8}mu^2 = \frac{5}{8}mu^2 = 0.625mu^2

.

After collision, X is at rest (

v_x = 0

). Let

v_y

be the velocity of Y.
Final momentum:

p_f = m(0) + mv_y = mv_y

.
By conservation of momentum:

mv_y = \frac{1}{2}mu \implies v_y = \frac{u}{2}

.
Final kinetic energy:

KE_f = \frac{1}{2}m(0)^2 + \frac{1}{2}m(\frac{u}{2})^2 = \frac{1}{8}mu^2 = 0.125mu^2

.

Kinetic energy lost:

\Delta KE = KE_i - KE_f = 0.625mu^2 - 0.125mu^2 = 0.5mu^2

.
Percentage lost:

\frac{0.5}{0.625} \times 100\% = \frac{500}{625} \times 100\% = \frac{4}{5} \times 100\% = 80\%

.
The correct answer is E.

Question 20

1 mark

A constant resultant force of

4.0\text{ N}

acts on a stationary object of mass

4.0\text{ kg}

as it moves in a straight line along a frictionless horizontal surface. The force acts in the direction of the object's motion for a distance of

8.0\text{ m}

.

What is the final speed of the object?

A. $2.0\text{ m s}^{-1}$
B. $2.8\text{ m s}^{-1}$
C. $4.0\text{ m s}^{-1}$
D. $8.0\text{ m s}^{-1}$
E. $16.0\text{ m s}^{-1}$

Answer: C

Worked solution

First, calculate the work done by the resultant force using

W = F \times d

:

W = 4.0\text{ N} \times 8.0\text{ m} = 32\text{ J}

.

According to the work-energy principle, the work done on the object is equal to its change in kinetic energy. Since the object starts from rest (

KE_{\text{initial}} = 0

):

KE_{\text{final}} = 32\text{ J}

.

Now, use the kinetic energy formula

KE = \frac{1}{2}mv^2

to solve for the final speed

v

:

32 = \frac{1}{2}(4.0)v^2

32 = 2.0v^2

v^2 = 16

v = 4.0\text{ m s}^{-1}

.

The final answer is C.

Question 21

1 mark

A wall consists of two layers of material,

X

and

Y

, in perfect thermal contact. Layer

X

has a thickness

d

and thermal conductivity

k

. Layer

Y

has a thickness

3d

and thermal conductivity

2k

. The outer surface of layer

X

is maintained at a constant temperature of

100\,\text{°C}

, and the outer surface of layer

Y

is maintained at

20\,\text{°C}

. Under steady-state conditions, what is the temperature at the interface between the two layers?

A. $32\,\text{°C}$
B. $48\,\text{°C}$
C. $52\,\text{°C}$
D. $60\,\text{°C}$
E. $68\,\text{°C}$

Answer: E

Worked solution

In steady-state conduction, the rate of thermal energy transfer

P

must be the same through both layers. The formula for the rate of conduction is

P = \frac{kA\Delta T}{x}

, where

k

is thermal conductivity,

A

is area,

\Delta T

is the temperature difference, and

x

is the thickness.

Let

T_i

be the temperature at the interface. For layer

X

, the rate is:

P_X = \frac{k A (100 - T_i)}{d}

For layer

Y

, the rate is:

P_Y = \frac{(2k) A (T_i - 20)}{3d}

Setting

P_X = P_Y

and cancelling common factors

k, A, d

:

\frac{100 - T_i}{1} = \frac{2(T_i - 20)}{3}

Multiply both sides by 3:

3(100 - T_i) = 2(T_i - 20)

300 - 3T_i = 2T_i - 40

340 = 5T_i

T_i = \frac{340}{5} = 68\,\text{°C}

.

The final answer is E.

Question 22

1 mark

A well-insulated copper rod has a length of

80\,\text{cm}

and a cross-sectional area of

2.0\,\text{cm}^2

. One end is held in a steam bath at

100\,\text{°C}

and the other end is embedded in a large block of ice at

0\,\text{°C}

.

Given the following constants:
- Thermal conductivity of copper:

400\,\text{W}\,\text{m}^{-1}\,\text{K}^{-1}

- Specific latent heat of fusion of ice:

3.3 \times 10^5\,\text{J}\,\text{kg}^{-1}

What mass of ice melts in

11

minutes?

A. $0.33\,\text{g}$
B. $2.0\,\text{g}$
C. $20\,\text{g}$
D. $200\,\text{g}$
E. $2000\,\text{g}$

Answer: C

Worked solution

Step 1: Convert all units to SI.
- Length

l = 0.80\,\text{m}

- Area

A = 2.0 \times (10^{-2}\,\text{m})^2 = 2.0 \times 10^{-4}\,\text{m}^2

- Temperature difference

\Delta T = 100 - 0 = 100\,\text{K}

- Time

t = 11 \times 60 = 660\,\text{s}

Step 2: Calculate the rate of heat transfer

P

(power):

P = \frac{kA\Delta T}{l} = \frac{400 \times 2.0 \times 10^{-4} \times 100}{0.80}

P = \frac{40,000 \times 2.0 \times 10^{-4}}{0.80} = \frac{8.0}{0.80} = 10\,\text{W}

Step 3: Calculate the total energy

Q

transferred:

Q = P \times t = 10 \times 660 = 6600\,\text{J}

Step 4: Calculate the mass of ice melted

m

using

Q = mL_f

:

m = \frac{Q}{L_f} = \frac{6600}{3.3 \times 10^5}

m = \frac{6600}{330,000} = \frac{66}{3300} = \frac{2}{100} = 0.02\,\text{kg}

0.02\,\text{kg} = 20\,\text{g}

.

Question 23

1 mark

A rectangular glass tank is filled with a fluid. A small heating element is placed at the bottom-left corner of the tank, and a cooling block is placed at the top-right corner. Both are switched on simultaneously. Assuming the fluid's density decreases as its temperature increases, which of the following best describes the resulting steady-state convection current and the physical changes driving it?

A.A clockwise circulation is established because fluid density decreases at the heater and increases at the cooling block.
B.A counter-clockwise circulation is established because fluid density decreases at the heater and increases at the cooling block.
C.A clockwise circulation is established because fluid density increases at the heater and decreases at the cooling block.
D.A counter-clockwise circulation is established because fluid density increases at the heater and decreases at the cooling block.
E.No circulation occurs because the density changes at the two corners counteract each other horizontally.

Answer: A

Worked solution

Convection is driven by changes in density due to temperature variations. 1. At the heating element (bottom-left), the fluid temperature increases, causing it to expand and its density to decrease. This lower-density fluid rises vertically. 2. At the cooling block (top-right), the fluid temperature decreases, causing it to contract and its density to increase. This higher-density fluid sinks vertically. 3. To conserve mass, the rising fluid at the left must move across the top towards the right, and the sinking fluid at the right must move across the bottom towards the left. 4. This creates a continuous clockwise loop (up on the left, right at the top, down on the right, left at the bottom). Option A correctly identifies the clockwise direction and the correct density-temperature relationship.

Question 24

1 mark

Pure water has a maximum density at a temperature of approximately

4^\circ\text{C}

. A deep lake is initially at a uniform temperature of

0^\circ\text{C}

throughout. If the sun begins to warm the surface of the lake, which statement best describes the convective mixing that occurs as the surface temperature increases from

0^\circ\text{C}

to

10^\circ\text{C}

?

A.Convection occurs continuously throughout the process because warmer water is always less dense and rises.
B.Convection occurs only while the surface water is between $0^\circ\text{C}$ and $4^\circ\text{C}$ .
C.Convection occurs only while the surface water is between $4^\circ\text{C}$ and $10^\circ\text{C}$ .
D.No convection occurs because the heat source is at the top of the fluid.
E.Convection occurs only once the surface temperature exceeds $8^\circ\text{C}$ to overcome the initial density of the $0^\circ\text{C}$ water.

Answer: B

Worked solution

1. Convection occurs when a fluid layer becomes denser than the layer beneath it, causing it to sink.
2. Between

0^\circ\text{C}

and

4^\circ\text{C}

, the density of water increases as temperature increases. Therefore, as the sun warms the surface from

0^\circ\text{C}

, the surface water becomes denser than the

0^\circ\text{C}

water below it and sinks. This drives convection.
3. At

4^\circ\text{C}

, water reaches its maximum density.
4. As the surface warms from

4^\circ\text{C}

to

10^\circ\text{C}

, its density decreases. This warmer water is now less dense than the

4^\circ\text{C}

water that has already sunk to the bottom.
5. Since the less dense water is at the top (near the heat source), it remains there (stratification), and convection ceases. Mixing would then only occur through the much slower process of conduction. Thus, convection only occurs between

0^\circ\text{C}

and

4^\circ\text{C}

.

Question 25

1 mark

An electric heater rated at

P = 50\,\text{W}

is used to heat

0.50\,\text{kg}

of a liquid in a vessel. The vessel itself has a heat capacity of

100\,\text{J}\,^\circ\text{C}^{-1}

, and the liquid has a specific heat capacity of

1800\,\text{J}\,\text{kg}^{-1}\,^\circ\text{C}^{-1}

. The liquid is initially at a temperature of

20\,^\circ\text{C}

. Only

80\%

of the energy supplied by the heater is transferred to the liquid and the vessel. How long does it take for the liquid to reach a temperature of

50\,^\circ\text{C}

?

A.540\, $\text{s}$
B.600\, $\text{s}$
C.675\, $\text{s}$
D.750\, $\text{s}$
E.1000\, $\text{s}$

Answer: D

Worked solution

First, calculate the temperature change required:

\Delta T = 50\,^\circ\text{C} - 20\,^\circ\text{C} = 30\,^\circ\text{C}

.

The total thermal energy

Q

required to heat both the liquid and the vessel is given by:

Q = (m_{\text{liquid}} \times c_{\text{liquid}} \times \Delta T) + (C_{\text{vessel}} \times \Delta T)

Q = (0.50\,\text{kg} \times 1800\,\text{J}\,\text{kg}^{-1}\,^\circ\text{C}^{-1} \times 30\,^\circ\text{C}) + (100\,\text{J}\,^\circ\text{C}^{-1} \times 30\,^\circ\text{C})

Q = 27000\,\text{J} + 3000\,\text{J} = 30000\,\text{J}

.

The heater provides

50\,\text{W}

, but only

80\%

is effective. The useful power

P_{\text{useful}}

is:

P_{\text{useful}} = 0.80 \times 50\,\text{W} = 40\,\text{W}

.

The time

t

required is:

t = \frac{Q}{P_{\text{useful}}} = \frac{30000\,\text{J}}{40\,\text{W}} = 750\,\text{s}

.

Question 26

1 mark

Two solid spheres, X and Y, are made of different materials. Sphere Y has a radius twice that of sphere X (

r_Y = 2r_X

). The density of the material in sphere Y is half that of the material in sphere X (

\rho_Y = \frac{1}{2}\rho_X

). The specific heat capacity of the material in sphere Y is three times that of the material in sphere X (

c_Y = 3c_X

). Both spheres are initially at the same temperature and are supplied with the same amount of thermal energy

Q

. What is the ratio of the temperature change of sphere X to the temperature change of sphere Y,

\frac{\Delta T_X}{\Delta T_Y}

?

A.3
B.6
C.12
D.24
E.48

Answer: C

Worked solution

The volume of a sphere is proportional to the cube of its radius (

V \propto r^3

). Thus:

V_Y = V_X \times (\frac{r_Y}{r_X})^3 = V_X \times 2^3 = 8V_X

.

Mass is density multiplied by volume (

m = \rho V

). Comparing the masses:

m_Y = \rho_Y V_Y = (\frac{1}{2}\rho_X) \times (8V_X) = 4\rho_X V_X = 4m_X

.

The energy supplied is

Q = mc\Delta T

. Rearranging for temperature change:

\Delta T = \frac{Q}{mc}

.
The ratio is:

\frac{\Delta T_X}{\Delta T_Y} = \frac{Q / (m_X c_X)}{Q / (m_Y c_Y)} = \frac{m_Y c_Y}{m_X c_X}

.

Substituting the known ratios

m_Y = 4m_X

and

c_Y = 3c_X

:

\frac{\Delta T_X}{\Delta T_Y} = \frac{(4m_X) \times (3c_X)}{m_X c_X} = 12

.

Question 27

1 mark

A

400\,\text{g}

block of metal at

150\,^\circ\text{C}

is placed into

600\,\text{g}

of oil contained in a calorimeter. The oil and the calorimeter are initially at

20\,^\circ\text{C}

. The calorimeter is made of the same metal as the block and has a mass of

200\,\text{g}

. The final equilibrium temperature of the system is

40\,^\circ\text{C}

. Assuming no heat is lost to the surroundings, what is the ratio of the specific heat capacity of the metal (

c_m

) to the specific heat capacity of the oil (

c_o

)?

A.0.18
B.0.25
C.0.27
D.0.30
E.0.50

Answer: D

Worked solution

Let

c_m

be the specific heat capacity of the metal and

c_o

be the specific heat capacity of the oil.
Heat lost by the metal block:

Q_{\text{lost}} = m_{\text{block}} c_m (T_{\text{initial, block}} - T_{\text{final}})

Q_{\text{lost}} = 0.4\,\text{kg} \times c_m \times (150 - 40) = 44c_m

.

Heat gained by the oil and the calorimeter:

Q_{\text{gained}} = m_{\text{oil}} c_o (T_{\text{final}} - T_{\text{initial, oil}}) + m_{\text{cal}} c_m (T_{\text{final}} - T_{\text{initial, cal}})

Q_{\text{gained}} = 0.6\,\text{kg} \times c_o \times (40 - 20) + 0.2\,\text{kg} \times c_m \times (40 - 20)

Q_{\text{gained}} = 12c_o + 4c_m

.

By conservation of energy,

Q_{\text{lost}} = Q_{\text{gained}}

:

44c_m = 12c_o + 4c_m

40c_m = 12c_o

\frac{c_m}{c_o} = \frac{12}{40} = 0.30

.