• Physics 9702 Doubts | Help Page 148

Question 729: [Dynamics > Momentum]

Molecule of mass m travelling horizontally with velocity u hits a vertical wall at right-angles to its velocity. It then rebounds horizontally with the same speed.

What is its change in momentum?

A zero                         B mu                           C – mu                                    D – 2mu

Reference: Past Exam Paper – June 2011 Paper 11 Q10

Solution 729:

Answer: D.

Final momentum = Initial momentum + Change in momentum

Momentum is a vector quantity, so we need to consider both its magnitude and direction.

Take the direction of motion towards the vertical wall as the positive direction. Before hitting the vertical wall, the particle is moving towards the wall (in the positive direction) and after the collision, the particle rebounds away from the wall (in the negative direction).

Initial momentum = + mu

Since the particle moves with the same speed after the collision,

Final momentum = – mu

Final momentum = Initial momentum + Change in momentum

– mu = + mu + Change in momentum

Change in momentum = – mu – mu = – 2mu

Question 730: [Thermodynamics > Kinetic theory of Gases]

(a) State basic assumptions of the kinetic theory of gases.

(b) Use equations for pressure of an ideal gas to deduce that the average translational kinetic energy <E_K> of a molecule of an ideal gas is given by the expression

<E_K> = (3/2) (R / N_A) T

where R is the molar gas constant, N_A is the Avogadro constant and T is the thermodynamic temperature of the gas.

(c) A deuterium nucleus ²₁H and a proton collide. A nuclear reaction occurs, represented by the equation

²₁H                   +                      ¹₁p                   - - - >               ³₂He                +          γ

(i) State and explain whether reaction represents nuclear fission or nuclear fusion.

(ii) For the reaction to occur, minimum total kinetic energy of the deuterium nucleus and the proton is 2.4 × 10^–14 J.

Assuming that sample of a mixture of deuterium nuclei and protons behaves as an ideal gas, calculate temperature of the sample for this reaction to occur.

(iii) Suggest why assumption made in (ii) may not be valid.

Reference: Past Exam Paper – November 2010 Paper 43 Q2

Solution 730:

(a) Choose any 4:

The atoms / molecules / particles behave as elastic (identical) spheres

The volume of the atoms / molecules is negligible compared to volume of the containing vessel

Time of collision is negligible to the time between collisions

No forces of attraction or repulsion exist between the atoms / molecules

The atoms / molecules / particles are in (continuous) random motion

(b)

{The following equations may be used to describe ideal gases:}

pV = 1/3 Nm<c²>                          and pV = nRT      or pV = NkT

{where n is the number of moles and N is the number of molecules}

{Equating the equations for pV gives:}

1/3 Nm<c²> = nRT                        or = NkT 

{Kinetic energy of 1 molecule is given by: <E_K> = ½ m<c²>. We need to replace the m<c²> in this equation with an appropriate substitution (which can be obtained from the equations of the ideal gases above) so that it becomes in the form given in the question.}

EITHER

Consider 1/3 Nm<c²> = nRT

m<c²> = 3 nRT / N

Since n = N / N_A, m<c²> = 3 (N / N_A) RT / N = 3RT / N_A

OR

Consider

1/3 Nm<c²> = NkT

m<c²> = 3kT

Since k = R / N_A, m<c²> = 3 (R / N_A) T = 3RT / N_A

<E_K> = ½ m<c²> = ½ (3RT / N_A) = (3/2) (R / N_A) T

(c)

(i) The reaction represents EITHER a build-up of nucleus from light nuclei OR a build-up of heavy nucleus from nuclei.

So, it is a fusion reaction.

(ii)

{Kinetic energy <E_K> = (3/2) (R / N_A) T}

The proton and deuterium nucleus will have equal kinetic energies {the above equation for <E_K> does not depend on mass}.

1.2 × 10^–14 = 3/2 × [8.31 / (6.02 × 10²³)] × T

Temperature T = 5.8 × 10⁸ K

(iii)

EITHER Inter-molecular / atomic / nuclear forces exist

OR The proton and deuterium nucleus are positively charged / repel

Question 731: [Measurement > C.R.O]

The Y-input terminals of a cathode-ray oscilloscope (c.r.o.) are connected to supply of amplitude 5.0 V and frequency 50 Hz. Time-base is set at 10 ms per division and the Y-gain at 5.0 V per division.

Which trace is obtained?

Reference: Past Exam Paper – June 2013 Paper 11 Q6

Solution 731:

Answer: D.

The supply is of amplitude 5.0 V and frequency 50 Hz.

Period T = 1 / f = 1/ 50 = 0.02s = 20ms

Time-base setting: 10ms per division

The period is 20ms. So, a period of the signal should occupy (20 / 10 =) 2 division horizontally. [A and C are incorrect]

Y-gain setting: 5.0V per division

The amplitude of the signal is 5.0 V. So, the maximum / minimum value of the signal should be at 1 division vertically from the central horizontal line. [B is incorrect]

Question 732: [Electric field]

Two parallel metal plates have potential difference between them of 12 V. Distance between the plates is 1.0 mm.

What are the electric field strength between plates and the work done on a charge of +3.9 μC to move the charge from the negative plate to the positive plate?

electric field strength / N C^–1              work done / J

A                     12                                            4.7 × 10^–5

B                     12                                            47

C                     12 000                                     4.7 × 10^–5

D                     12 000                                     47

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q32

Solution 732:

Answer: C.

Electric field strength E = V / d = 12 / 0.001 = 12 000 N C^-1

Electric force = Eq

Work done = Force × distance = Eq × d = 12 000 × (3.9×10^-6) × 0.001 = 4.68 × 10^-5 J

OR Work done = Vq = 4.68 × 10^-5 J