A fairground ride consists of four carriages connected to a central vertical pole, as shown in the following view from above.

Question 39

A fairground ride consists of four carriages connected to a central vertical pole, as shown in the following view from above.

A motor rotates the central pole about its axis. This results in the four carriages each moving along a circular path.

The distance from the middle of each carriage to the centre of the pole is 3.20 m. When they are moving, each carriage experiences an air resistance force of 85.0 N. Assume that there are no other significant resistive forces.

Which torque does the motor need to apply to the pole to keep the system rotating at constant maximum speed?

A 5.44 N m                 B 272 N m                  C 544 N m                  D 1090 N m

Reference: Past Exam Paper – March 2019 Paper 12 Q12

Solution:

Answer: D.

Consider one carriage.

Distance from the middle = 3.20 m

Air resistance experienced by 1 carriage = 85.0 N

To keep a carriage rotating at constant speed, a force equal to the air resistance must be applied.

Torque = F × d

Torque to keep one carriage rotating = 85 × 3.2 = 272 Nm

This is the torque on one carriage. The system consists of 4 carriages.

Total torque = 4 × 272 = 1088 Nm

In one of the first experiments to demonstrate the Doppler effect, a train was filled with trumpeters all playing a note of frequency 440 Hz.

Question 23

In one of the first experiments to demonstrate the Doppler effect, a train was filled with trumpeters all playing a note of frequency 440 Hz. The difference in observed frequency of the note as the train directly approached a stationary observer was 22 Hz. The speed of sound was 340 m s^-1.

At which speed was the train moving?

A 15.4 m s^-1                     B 16.2 m s^-1                     C 17.0 m s^-1                     D 17.9 m s^-1

Reference: Past Exam Paper – June 2019 Paper 12 Q26

Solution:

Answer: B.

Frequency of sound, fs = 440 Hz

Speed of sound, v = 340 m s^-1

Difference in observed frequency = 22 Hz

Speed of train = v_s

For Doppler effect: f_O = fs v / (v ± v_s)

Since the train is approaching, the wavefronts are compressed, causing the observed wavelength to be smaller. Thus, the observed frequency is greater. In the equation, we use the minus sign.

Observed frequency f_O = 440 + 22 = 462 Hz

 f_O = fs v / (v – v_s)

462 = 440×340 / (340 – v_s)

340 – v_s = 440×340 / 462

340 – v_s = 323.8

Speed v_s = 340 – 323.8 = 16.2 m s^-1

A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s-1 in a vacuum.

Question 12

A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 10⁵ m s^-1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10^-2 T, as shown in Fig. 9.1.

Fig. 9.1

The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper.

A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields.

(a) State and explain the direction of the electric field. [2]

(b) Calculate the magnitude of the electric field strength.

Explain your working. [3]

(c) The electric field is now removed so that the positively-charged particle follows a curved path in the magnetic field. This path is an arc of a circle of radius 4.0 cm.

Calculate, for the particle, the ratio q/m. [3]

(d) The particle has a charge of 3e where e is the elementary charge.

(i) Use your answer in (c) to determine the mass, in u, of the particle. [2]

(ii) The particle is the nucleus of an atom. State the number of protons and the number

[Total: 11]

Reference: Past Exam Paper – March 2016 Paper 42 Q9

Solution:

(a) The direction of the force due to the electric field should be opposite to the force due to the magnetic field. So the electric field is up the page.

{From Fleming’s left hand rule, the direction of the force on the particle is downwards.  For the particle to pas undeviated, the force due to the electric field should be up. The direction of an electric field gives the direction of the force on a positive charge. So, the electric field should be upwards.}

(b)

force due to electric field = force due to magnetic field

or

Eq = Bqv

{Eliminating q, }

E = Bv                                                

E = 9.7×10^–2 × 1.6×10⁵

E = 1.6 (1.55) × 10⁴ V m^–1

(c)

{Force due to magnetic field provides the centripetal force.

Bqv = mv² / r }

q / m = v / Br                                                              

{The radius of the circle should be in metres.}

q / m = 1.6 × 10⁵ / (9.7×10^–2 × 4.0×10^–2)                  

q / m = 4.1 (4.12) × 10⁷ C kg^–1                                  

(d)

(i)

{q = 3e = 3 × 1.60×10^–19 C

q / m = 4.12 × 10⁷

m = q / 4.12 × 10⁷}

m = (3 × 1.60×10^–19) / (4.12×10⁷)     

{1 u = 1.66 × 10^–27 kg

To obtain the mass in u, we divide by 1.66 × 10^–2.}

m = 1.16×10^–26 / 1.66×10^–27

m = 7(.0) u       

(ii) 3 protons, 4 neutrons

{1 u is the mass of 1 unit / particle. 7 u means that there are 7 particles.

From the question, the charge of the particle is 3e – that is, it contains 3 protons. The rest are neutrons.}