Physic Tutorials

# Solutions to CIE Physics May/June And Oct/Nov P1 Questions

Below are few of the soluions to CIE physics 2015, 2016 and 2018 may/june Paper 1 questions.

Question 1

What is a unit for stress?

Solution

Stress = Force/Area = (Mass x acceleration)/(length x length) = Kgms^-2/m^2 = kgm^-1s^-2

Question 2

Physical quantities can be classed as vectors or as scalars. Which pair of quantities consists of two vectors?

A kinetic energy and force

B momentum and time

C velocity and electric field strength

D weight and temperature

Velocity and Electric field strength are both vector quantities.  C is the correct answer

Question 3

Two dogs pull a sledge along an icy track, as shown. Dog X pulls with a force of 200 N at an angle of 65° to the front edge of the sledge. Dog Y pulls with a force of 120 N at an angle of 45° to the front edge of the sledge. What is the resultant forward force on the sledge exerted by the two dogs?

A 80 N    B 170 N      C 270 N        D 320 N

Dog X with force 200N at an angle of 65

Let resolve into vertical and horizontal component (resolution of forces)

Horizontal component = 200 sin 65 = 181.2615

Vertical component = 200 cos 65 = 84.5236

Dog Y with force 120N at an angle of 45 to the front of the sledge

Vertical component = 120 cos 45 = 84.8528

Horizontal component = 120 sin 45 = 84.8528

Sum of the vertical forces = Vx – Vy ( Vx is in opposite direction to Vy, the reason for the negative sign) =  84.5236 – 84.8528 = -0.3292 N

Sum of the horizontal forces = Hx + Hy (same direction) = 181.2615 + 84.8528 = 266.1143 N

Question 4

The graph shown is that of resistance R of the thermistor (not current) against temperature T. Solution

Ohm’s law: V = IR

Current I = V / R. So, the current is inversely proportional to the resistance R. The ammeter measures the current, not the resistance in a circuit.

The resistance falls rapidly with temperature at low temperatures, so the current will rise rapidly at low temperatures. Therefore, we need a larger region of the scale to represent a fixed amount of change in current at low temperatures. If the values on the scales are close to each other, the change in readings may not be noticeable.

The resistance curve becomes flatter at higher temperatures, so the current will change more slowly with temperature. Therefore, to represent this region of large temperature, a relatively smaller region of the scale is required. So, the values of the scale can be closer to each other.

Question 7

The graph shows the variation with mass of the weight of objects on a particular planet. What is the value of the acceleration of free fall on the planet?

Solution

acceleration = gradient of the graph = dw/dm (taking the gradient at any point will give the same answer)

(3.2 – 0) /(2 – 0) = 3.2/2 = 1.6 ms-2

A 0.63 m s–2     B 1.6 m s–2       C 3.2 m s–2       D 9.8 m s–2

Question 10

A horizontal metal bar PQ of length 50.0 cm is hinged at end P. The diagram shows the metal bar viewed from above. Two forces of 16.0 N and 5.0 N are in the horizontal plane and act on end Q as shown in the diagram.

What is the total moment about P due to the two forces?

Solution

Since forc 16N is acting at an angle which 30 then there is a need to do resolve the force. In this case to the vertical component.

Fv = 16 sin 30 = 8N (acting upward)

Note 5N is acting downward

The total force = 8N – 5N = 3N. The reason for the subtraction is because they acted in opposite direction

total moment about P = 3 x 50/100 = 1.5Nm

A 1.5 N m    B 4.4 N m      C 6.5 N m        D 9.4 N m

Question 13

Liquids X and Y are stored in large open tanks. Liquid X has a density of 800 kg m–3 and liquid Y has a density of 1200 kg m–3.

At which depths are the pressures equal? Solution

If pressurea are equal, then

Px = Py

hdgx =hdgy

acceleration of free fall is the same for the two liquid

800*hx = 1200*hy

2hx = 3hy

hx =1.5hy

this means that if hy = 10, then hx = 15

Question 29

A charged particle is in the electric field between two horizontal metal plates connected to a battery, as shown. There is a force F on the particle due to the electric field. The separation of the plates is doubled.

What is the new force on the particle?

A F/4      B F/2        C F      D 2F

Solution

Electric field between two parallel plates is an example of uniform electric field, meaning electric field strength is the same at any point in the field.

E = v/d = f/q

f = vq/d, q = fd/v

f1d1/v = f2d2/v

since the potential difference is the same

f1d1 = f2d2

d2 = 2* d1

f2 = f1/2, this means the force will be halved

Question 39

How many down quarks are in a nucleus of hydrogen-3, 3H1 ?

A 2    B 3     C 4      D 5

Solution

The the nuclues of hydrogen-3, there are 1 proton and 2 neutron

Proton has (UUD) i.e. two up quarks and one down quarks

Neutron has (UDD) i.e. one up quarks and two down quarks

therefore, nucleus of hydrogen -3 = 1*(UUD) + 2*(UDD) = UUUUDDDDD

So there are five down quarks

## Solutions to Cambridge A level (CIE) Physics Oct/Nov 2016 P1

The solutions below are guidlines to some topics I felt students want to see their solutions in cambridge A level Oct/Nov 2016 P11. You can comment below to ask any question you may find difficult in Oct/Nov 2016 P11, 12, and 13.

Question 1

The force F between two point charges q1 and q2, a distance r apart, is given by the equation

where k is a constant. What are the SI base units of k ? {Cambridge A level oct/nov 2016, ques 2, p11}

Solution

The SI base unit of force is Kgms-2

Charge q1 is As

Charge q2 is As

distance r is m

SI base unit of k = (Kgms-2 x m) / (As x As) = Kgm2s-4A-2 (B is the correct option)

Question 2

A student uses a cathode-ray oscilloscope (c.r.o.) to measure the period of a signal. She sets the time-base of the c.r.o. to 5 ms cm–1 and observes the trace illustrated below. The trace has a length of 10.0 cm. What is the period of the signal? {Cambridge A level oct/nov 2016, ques 5, p11}

Solution

From the graph there are 3.5 oscillations

The distance to cover one oscillations = 10/3.5 cm

Since the time-base of the c.r.o is 5 ms cm–1

The period of the signal = (10/3.5) x 5 = 14.3 ms = 1.4 x 10-2 s (D is the correct option)

Question 3

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into the air and onto a track that is vertically 2.0 m lower. The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.

What is the horizontal velocity v of the cyclist at the end of the higher track? {Cambridge A level oct/nov 2016, ques 6, p11}

Solution

There are important points to note in this question:

The horizontal velocity v is used to calculate the horizontal distance

The time to reach the maximum height is the time to travel the horizontal distance

At maximum height u = 0

Using H = ut + 1/2gt2

2 = 0 + 1/2×9.81xt2

(t=0.6395s)

Horizontal distance = horizontal velocity(v) x time(t)

6 = 0.6395v

V = 9.4ms-2 (B is the correct option)

Question 4

A car is travelling at constant velocity. At time t = 0, the driver of the car sees an obstacle in the

road and then brakes to a halt. The graph shows the variation with t of the velocity of the car. How far does the car travel in the 5.0 s after the driver sees the obstacle? {Cambridge A level oct/nov 2016, ques 8, p11}

Solution

The distance travelled by the car = 20 x 0.8 + ½ x 20 x(5 – 0.8) =16 +42 = 58m (C is the correct option)

Question 5

A car has mass m. A person needs to push the car with force F in order to give the car acceleration a. The person needs to push the car with force 2F in order to give the car acceleration 3a.

Which expression gives the constant resistive force opposing the motion of the car? {Cambridge A level oct/nov 2016, ques 11, p11}

Solution

Resultant force = applied force – resistive force

Ma = F- R

R = F- ma —-i

3ma = 2F – R

R = 2F – 3ma —–ii

Substitute for R in eq i

2F – 3ma = F – ma

F = 2ma

Therefore, R = 2ma – ma = ma

Resistive force = ma (A is the correct option)

Question 6

A car travels at a constant speed of 25 m s–1 up a slope. The wheels driven by the engine exert a forward force of 3000 N. There is a drag force due to air resistance and friction of 2100 N. The weight of the car has a component down the slope of 900 N.

What is the rate at which thermal energy is dissipated? {Cambridge A level oct/nov 2016, ques 20, p12}

Solution

Rate at which thermal energy is dissipated = power loss

Power = force x velocity

Rate at which thermal energy is dissipated = drag force x velocity = 25 x 2100 = 5.3 x 104 W (C is the correct option)

Question 7

Two parallel circular metal plates X and Y, each of diameter 18 cm, have a separation of 9.0 cm. A potential difference of 9.0 V is applied between them. Point P is 6.0 cm from the surface of plate X and 3.0 cm from the surface of plate Y.

What is the electric field strength at P? {Cambridge A level oct/nov 2016, ques 30, p12}

Solution

The kind of field in this this is a uniform electric field. Therefore, at any point in the field the electric field strength is constant.

Electric field strength = potential difference / distance between the plate = 9 / 0.09

Electric field strength at P = 100 Nc-1 (B is the correct option)

### Solutions to Cambridge A level (CIE) June 2015 and 2016 physics 9702

I took some questions out of 2015 and 2016 May/June Cambridge A level physics paper 11 and provid solutions to them based on their difficulty level. However, if you have any questions that you need additional explanation in 2015 and 2016 past papers, kindly comment below.

Question 1

The average kinetic energy E of a gas molecule is given by the equation E = 3/2 kT where T is the absolute (kelvin) temperature.
What are the SI base units of k ?
A kg–1m–1s2K          B kg–1m–2s2K         C kg m s–2K–1          D kg m2s–2K–1

Solution

The S.I base unit of Kinetic energy = Kgm2s-2

For temperature = K

K = E/T = kgm2s-2 / k

K = kgm2s-2k-1

D is the correct option

Question 2

A uranium-238 nucleus, 238U92 , undergoes nuclear decays to form uranium-234, 234U92 .Which series of decays could give this result?
A emission of four β-particles
B emission of four γ-rays
C emission of one α-particle and two β-particles
D emission of two α-particles and eight β-particles

Solution

alpha emission  = 4He2

beta emission = 0e-1

For nucleon number and proton number to be conserved Uranium-238 must undergo one α-particle and two β-particles

C is the correct option

Question 3

A cell of e.m.f. 2.0 V and negligible internal resistance is connected to a network of resistors as shown. What is the current I ?
A 0.25 A             B 0.33 A                  C 0.50 A                   D 1.5 A

Solutiothe two 4 ohms are in parallel

effective resistance is 1/R = 1/4 + 1/4

1/R = 2/4

R = 2 ohms

This 2 ohms is in series with the second 2 ohms

R = 2 + 2 = 4 ohms

This 4 ohms is in parallel with the 2 ohms above, therefore the same E.m.f flows through them

For the effective resistance of 4 ohms

E = IR

2 = I*4

I = 1/2 A

2 ohms and the two 4 ohms are in series, the same current will flow through them

The current that will flow through the 4 ohms = half of 1/2 = 1/4 = 0.25 A

A is the correct option

Question 4

A charged oil drop of mass m, with n excess electrons, is held stationary in the uniform electric field between two horizontal plates separated by a distance d. The voltage between the plates is V, the elementary charge is e and the acceleration of free fall is g.
What is the value of n ?

A eV/mgd    B mgd/ev       C meV/gd      D gd/mev

Solution

Electric field intensity = F/ne = V/d

F = mg

mg/ne = V/d

mgd = neV

n = mgd /eV

Question 5

A sound wave has a speed of 330 m s–1 and a frequency of 50 Hz. What is a possible distance between two points on the wave that have a phase difference of 60°?
A 0.03 m               B 1.1 m                    C 2.2 m                          D 6.6 m

Solution

V = fλ

330 = 50*λ

λ = 330/50 = 6.6 m

phase angle = 2πx/λ

60 = π/3

π/3 = 2πx/λ

λ = 6x

6x = 6.6

x = 1.1 m

Question 6

A fisherman lifts a fish of mass 250 g from rest through a vertical height of 1.8 m. The fish gains a speed of 1.1 m s–1.
What is the energy gained by the fish?
A 0.15 J          B 4.3 J           C 4.4 J            D 4.6 J

Solution

P.E = mgh = 0.25*1.8*9.81 = 4.415 J

K.E = 1/2 mv2 = 1/2 * 0.25 *1.1^2 = 0.15 J

Energy gained by the fish = P.E +K.E = 4.4 + 0.15 = 4.6 J

## Solutions to some questions in Cambridge A level May/June 2016 physics P11

Question 1

A car accelerates uniformly from velocity u to velocity v in time t. On the graph, which area equals the distance travelled by the car in time t ?
A NPTU + PQST      B NPW V + VRSU        C NPW V + WRST          D PST + PQS

Solution

The distance covered by the car is NPSU which is a trapezium. The area under it will give the distance covered. In the trapezium there are three full boxes. The only option that gives three full boxes is option B.

Question 2

Two cars X and Y are positioned as shown at time t = 0. They are travelling in the same direction. X is 50 m behind Y and has a constant velocity of 30 m s–1. Y has a constant velocity of 20 m s–1. What is the value of t when X is level with Y?
A 1.0 s                 B 1.7 s                       C 2.5 s                             D 5.0 s

Solution

The time X will be at the same dstance with car Y is t

the distance covered for car X is

50 + s = 30t

the distance covered for car Y is

s = 20t

50 + 20t = 30t

50 = 10t

t = 5s

D is the correct option

Question 3

Two spheres approach each other along the same straight line. Their speeds are u1 and u2 before they collide. After the collision, the spheres separate with speeds v1 and v2 in the directions shown below. The collision is perfectly elastic. Which equation must be correct?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1

Solution

relative speed of approach = relative speed of separation for perfectly elastic collision

u1 + u2 = v2 – v1

Question 4

The diagram shows a man standing on a platform that is attached to a flexible pipe. Water is pumped through the pipe so that the man and platform remain at a constant height. The resultant vertical force on the platform is zero. The combined mass of the man and platform is 96 kg. The mass of water that is discharged vertically downwards from the platform each second is 40 kg. What is the speed of the water leaving the platform?
A 2.4 m s–1         B 6.9 m s–1          C 24 m s–1              D 47 m s–1

Solution

The force exerted by the man and the platform = the vertical downward force of the water

96*9.81 = m * v/t

96 * 9.81 = m/t * v

v = 96*9.81 / 40 = 23.544 = 24ms-1

C is the correct option

Question 5

A solid metal cylinder stands on a horizontal surface, as shown. The cylinder has length x and cross-sectional area A. The cylinder exerts a pressure p on the surface. The acceleration of free fall is g. Which expression gives the density of the metal of the cylinder?
A gx/p                                             B p/gx                              C gx/pA                    D pA/gx

Solution

pressure = force / area

area = A

P = F / A

F = PA

mass = force/acceleration of free fall

mass = PA / g

density = mass / volume

volume = A *x = Ax

density =( PA/g)/Ax = P /gx

B is the correct option

Question 6

A trailer of weight 30 kN is attached to a cab at X, as shown in the diagram. What is the upward force exerted at X by the cab on the trailer?
A 3 kN                        B 15 kN                           C 30 kN                                D 60 kN

Solution

F * 20 = 30 * 10

F = 300 / 20 = 15 KN

Question 7

Some gas in a cylinder is supplied with thermal energy q. The gas does useful work in expanding at constant pressure p from volume V0 to volume VF, as shown. Which expression gives the efficiency of this change?

A pV0 / q                    B. Vf / Voq                   C. p(Vf – V0 ) / q          D.(Vf – V0 )/V0q

Solution

Efficiency = workoutput / workinput

workinput = q

workoutput = p ( Vf – V0)

efficiency = p ( Vf – V0) /q

C is the correct option

Question 8

Two copper wires of equal length are connected in parallel. A potential difference is applied across the ends of this parallel arrangement. Wire S has a diameter of 3.0 mm. Wire T has a diameter of 1.5 mm.

What is the value of the ratio current in T/ current in S?

A. 1/4  B. 1/2  C. 2  D. 4

Solution

Resistance = (resistivity x length)/Area

Area = (pi x square of diameter)/4

length and resistivity is constant

meaning, Rs(ds)^2 = RT(dT)^2

RS/RT = (dS)^2/(dT)^2 = (3×3)/(1.5×1.5) = 4

Since the wires are connected in parallel, it means that the potential difference across each wire will be the same.

V = IR

IsRs = ItRt

It/Is = Rs / Rt = 4

i.e. the ratio current in T/ current in S = 4

1. Charles says: