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# Solutions to Cambridge A level June 2015 and 2016 physics

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

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