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## 2021 A-level Physics (9749) Paper 3 Suggested Solutions

All solutions here are suggested. Concept First will hold no liability for any errors.

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# Answer Key to 2021 A Level Physics Paper 3:

[Maximum Marks: 9]

(ai)

a_{\parallel -\textup{slope}}=gsin\theta=(9.81)(\sin 40^{\circ})=6.3057=6.31\textup{ ms}^{-2}

(aii)

v^{2}=u^{2}+2asv^{2}=0^{2}+2(6.3057)(\frac{56}{100})

Hence, v=2.6575=2.66\textup{ ms}^{-1} (to 3 s.f.)

(bi)

F_{c}=\frac{mv^{2}}{r}=\frac{(\frac{72}{1000})(1.5)^{2}}{\frac{12}{100}}=1.35\textup{ N} (to 3 s.f.)

(bii)

F_{\textup{net}}=F_{c}=mg+N

N=F_{c}-mg=1.35-(\frac{72}{1000})(9.81)=0.64368=0.644 \textup{ N} (to 3 s.f.)

force = 0.644 N

direction: vertically downwards

[Maximum Marks: 7]

(a)

Gravitational potential is defined as the work done by an external agent to bring unit mass from infinity to a point without acceleration. By convention, potential at infinity is taken to be zero. Moreover, gravitational force is attractive. Therefore, negative work is done by the external agent in moving unit mass from infinity to that point. Hence, potential is a negative quantity.

(bi)

r_{p}=3.4\times 10^{6}\textup{ m} m_{p}=6.2\times10^{23}\textup{ kg}

\phi _{s}=\frac{-GM}{r}=-\frac{(6.67\times10^{11})(6.2\times10^{23})}{(3.4\times10^{6})}=-1.216\times10^{7}=-1.22\times10^{7}\textup{J kg}^{-1}(to 3 s.f.)

(bii)

By COE, \textup{KE}_{\textup{s}}+\textup{PE}_{\textup{s}}=\textup{KE}_{\infty}+\textup{PE}_{\infty}

\frac{1}{2}\textup{m}_{\textup{r}}\textup{v}_{e}^{2}+\textup{m}_{\textup{r}}\phi _{\textup{s}}\geqslant 0

\frac{1}{2}\textup{m}_{\textup{r}}\textup{v}_{e}^{2}\geqslant -\textup{m}_{\textup{r}}\phi _{\textup{s}}=-(-1.216\times10^{7})

\textup{v}_{\textup{e}}\geqslant \sqrt{2(1.216\times10^{7})}

\textup{v}_{\textup{e min}}=4932.12\textup{m s}^{-1}

But v_{r}=3800\textup{ m s}^{-1}<\textup{v}_{\textup{e min}}. Hence, rock returns to surface.

[Maximum Marks: 10]

(a)

The internal energy of an ideal gas is only the microscopic molecular kinetic energies due to the random motion of its particles. The internal energy of an ideal gas depends only on its state, i.e., its pressure, temperature, volume and amount.

(bi)

From PV=nRT, \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

T_{2}=\frac{P_{2}V_{2}}{P_{1}V_{1}}\cdot T_{1}=\frac{(3.6\times10^{3})}{(3.2\times10^{-3})}\cdot (285.15\textup{ K})=320.793\textup{ K}

\therefore in ^{\circ}C = 47.643 = 47.6 ^{\circ}C

(bii)

\left | \textup{WD} \right |=p\Delta V = p(V_{f}-V_{i})=(1.0\times 10^{5})(3.6-3.2)(10^{-3})=40.0\textup{ J}

(ci)

From First Law of Thermodynamics,

\Delta U=Q_{\textup{to}}+W_{\textup{on}}=101+(-40)=61.0\textup{ J}

(cii)

From ideal gas equation, n=\frac{PV}{RT}=\frac{(1\times 10^{5})(3.2\times 10^{-3})}{(8.31)(285.15)}=0.135044\textup{ mol}

N=nN_{A}=(0.135044)(6.023\times 10^{23})=8.133705 \times 10^{22}\textup{ molecules}

Since ideal gas, \Delta U=\Delta KE = 61\textup{ J}

\therefore<KE> of 1 molecule = \frac{61}{8.133705\times 10^{22}} = 7.4996 \times 10^{-22} = 7.50 \times 10^{-22} J (to 3 s.f.)

[Maximum Marks: 8]

(ai)

x_{0}=0.500 (to 3 s.f.)

(aii)

w=\frac{2\pi}{T}=\frac{2\pi}{0.8}=7.8539=7.86 \text{ rad s}^{-1}

(aiii)

v_{0}=wx_{0}=(7.8539)(0.5)=3.9269=3.93 \text{ cm s}^{-1}

(bi)

Amplitude decreases

Period increases

[Maximum Marks: 8]

(a)

Diffraction refers to the spreading of wavefronts through an aperture (slit) or around an obstacle. It is observable when the wavelength is approximately of the same order of magnitude as the slit or obstacle size.

(bi)

a\sin \theta=n\lambda\sin\theta=\frac{n\lambda}{a}=\frac{(1)(590\times 10^{-9})}{(0.1\times 10^{-3})}=5.9\times 10^{-3}

Since \sin\theta is small, by small angle approximation, \therefore \sin\theta\simeq \theta=5.9\times 10^{-3}\text{ rad} and \tan\theta\simeq \theta

\tan\theta=5.9\times10^{-3}=\frac{\frac{w}{2}}{2.6}

w = 0.03068 =0.0307 m (3 s.f.)

(bii)

For Young’s Double Slit Experiment (YDSE),

w=\frac{\lambda D}{d}=\frac{(590\times 10^{-9})(2.6)}{(1.4\times 10^{-3})}=1.095714\times 10^{-3} m

N = \frac{W_{\text{SS}}}{W_{\text{DS}}}=\frac{0.03068 \text{ m}}{1.095714\text{ mm}} = 28

Checking 14th Bright Fringe for YDSE,

d\sin\theta=n\lambda

\sin\theta=\frac{n\lambda}{d}=\frac{(14)(590\times10^{-9})}{(1.4\times10^{-3})}=5.9\times 10^{-3}\text{ rad} (This means that the 14th order bright fringe coincides with the 1st order minima of the single slit diffraction)

Central bright fringe of single slit diffraction only contains 0 – 13 bright fringes of YDSE.

No. of YDSE bright fringes = 13 + 1 + 13 = 27

[Maximum Marks: 9]

(a)

(bi)

From Ohm’s Law, R=\frac{\text{V}}{\text{I}}=\frac{500}{100}=5.00\Omega (to 3 s.f.)

(bii)

P=I^{2}R=(100\times 10^{-3})^{2}(5)=0.0500 \text {W} (3 s.f.)

(biii)

When I = 0, V = 550 mV = Emf of solar cell

\varepsilon =V_{T}+V_{r}=I(R+r)

\therefore r=\frac{\varepsilon }{I}-R=\frac{550 \times 10^{-3}}{100\times 10^{-3}}-5=0.500\Omega (to 3 s.f.)

[Maximum Marks: 9]

(a)

Similarity: Electric and gravitational potential are both inversely proportional to the distance from the point change and point mass respectively.

Difference: Electric potential can be either negative or positive depending on the signs of the charges, but gravitational potential is always negative.

(bi)

Opposite signs. At any point along AB, the electric potential is the sum of the individual electric potentials due to charges A and B. When x=9cm, the net potential is 0V, suggesting that the 2 individual potentials due to A and B sum up to 0V, implying that the charges must be opposite in signs.

(bii)

When x=9cm, \text{V}_{\text{net}}=\text{V}_{\text{A}}+\text{V}_{\text{B}}=\frac{Q_{\text{A}}}{4\pi\varepsilon _{0}(0.09)}+\frac{Q_{\text{B}}}{4\pi\varepsilon _{0}(0.03)}=0

\frac{Q_{\text{A}}}{4\pi\varepsilon _{0}(0.09)}=-\frac{Q_{\text{B}}}{4\pi\varepsilon _{0}(0.03)}

\therefore \frac{Q_{\text{A}}}{Q_{\text{B}}}=\left | -\frac{0.09}{0.03} \right |=3.00 (to 3s.f.)

(biii)

E=-\frac{dV}{dr}=-\frac{\Delta V}{\Delta r}=-\frac{(y_{2}-y_{1})}{(x_{2}-x_{1})}=-\frac{(-70-180)}{(0.12-0.01)}=2272.727=2270 \text{ V m}^{-1} (to 3s.f.)

[Maximum Marks: 20]

(a)

The magnetic force acting per unit electric constant current per unit length on a long conductor wire placed at right angles to a uniform magnetic field. The SI Unit for magnetic flux density is the Tesla, T.

(bi)

When the proton travels in the field, the magnetic force, F_{\text{B}}, is always normal to the path of the proton, according to Fleming’s Left Hand Rule. This F_{\text{B}} always points to a fixed point, which is the centre of the circular motion and provides for the centripetal force required for circular motion.

(bii)

F_{\text{B}}=F_{\text{C}}Bqv=\frac{mv^{2}}{r}

B=\frac{mv}{qr}=\frac{(1.67\times 10^{-27})(6.2\times 10^{5})}{(1.6\times 10^{-19})(0.076)}=0.085148=0.0851\text{ T} (to 3 s.f.)

(ci)

(cii)

E=vB=(6.2\times10^{5})(0.085148)=52791.76=52.8 \text{k V m}^{-1} (to 3s.f.)

(di)

\text{RMS}=\frac{B_{0}}{\sqrt{2}}=\frac{6.4\text{ mT}}{\sqrt{2}}=4.5254\text{ mT}=4.53\text{ mT} (to 3s.f.)

(dii)

1 ms and 4 ms

(e)

\left | E_{0} \right |=\left | \frac{\Delta\Phi }{\Delta t} \right |=\left | \frac{\Delta BAN\cos\theta}{\Delta t} \right |=\left | AN\cos\theta\left ( \frac{\Delta B}{\Delta t} \right ) \right |=\left | \pi r_{\text{turn}}^{2}\cdot N\cdot \cos\theta\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right ) \right |

\left | E_{0} \right |=\left | \pi\left ( \frac{1.2}{100} \right )^{2}(270)(\cos0)\left ( \frac{-8-8}{3.6-1.4} \right )\left ( \frac{10^{-3}}{10^{-3}} \right )\right |=0.88832=0.888\textup{ V} (to 3s.f.)

(f)

[Maximum Marks: 20]

(ai)

1. Only Electromagnetic (EM) radiation of frequency larger than or equal to the threshold frequency of the metal is able to cause emission of photoelectrons from metal, regardless of radiation intensity.

2. The almost instantaneousness of photoelectron emission, when energy of EM radiation is larger than or equal to the work function of the metal .

(aii)

Emission line spectra provides evidence for electron transitions within an atom when electrons emit photons of specific energies (energy difference between any 2 energy levels) and transit to a more stable, quantised energy level. The well-defined separation of emission lines is experimental proof for the quantised energy levels of an atom, thereby suggesting that electrons cannot have energies between 2 energy levels, thus proving the discrete nature of electron energy levels.

(bi)

E=\frac{hc}{\lambda}=\frac{(1.989\times 10^{-25})}{340\times 10^{-9}}=5.85\times 10^{-19}\text{ J}

in eV, E_{\text{purple}}=\frac{5.85\times 10^{-19}}{1.6\times10{-19}}=3.6562=3.66{\text{eV}} (to 3s.f.)

(bii)

If electrons de-excite from -3.4eV energy level to -13.6eV energy level, it must release 10.2eV of photon energy, which is much larger than the energy of violet light (3.66eV). So, if electrons de-excite from higher energy levels to -13.6eV energy level, from Planck’s Law, since the energy is inversely proportional to the wavelength, photons having smaller wavelength than purple light would be emitted, thereby showing that visible line spectrum does not result.

(ci)

\lambda =\frac{\ln2}{t_{\frac{1}{2}}}=\frac{\ln2}{53}=0.013078=0.0131 (to 3s.f.)

(cii)

M=M_{0}e^{-\lambda t}=(5.7\times10^{-12})e^{0.0131\text{ day}^{-1}(120\text{ days})}=1.18348\text{ pkg} (Mass of undecayed nuclei)

From n=\frac{M}{A_{\text{r}}} and n=\frac{N}{N_{\text{A}}},

N=\frac{M\cdot N_{\text{A}}}{A_{\text{r}}}=\frac{(1.18348\times10^{-12})(1000)(6.023\times 10^{23})}{7}=1.0183\times 10^{14}=1.02\times 10^{14} (to 3s.f.)

(ciii)

Beryllium-7 decays by emitting \gamma-photons, which are massless and chargeless. This does not alter Beryllium-7.

(d)

By COM, p_{\text{i}}=p_{\text{f}} and E=pc

0=\frac{E}{c}-\text{m}_{\text{BE}}\text{v}_{\text{BE}}, \text{m}_{\text{BE}}=7u

v_{\text{BE}}=\frac{E}{c}\cdot \frac{1}{\text{m}_{\text{BE}}}=\frac{(0.48\times 10^{6}\times1.6\times 10^{-19})}{3\times 10^{8}}\times \frac{1}{7(1.66\times 10^{-27})}=22030.9=2.20\times 10^{4}\text{ m s}^{-1} (to 3s.f.)

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