## 2020 A-level Physics (9749) Paper 2 Suggested Solutions

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

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# Answer Key to 2020 A Level Physics Paper 2:

[Maximum Marks: 9]

(a)

The cyclist forward driving force equals to the opposing frictional forces. Hence, the net force is 0. By Newton’s First Law, an object will continue in its state of uniform motion in a straight line unless an external resultant force acts on it.

(b)

F=\frac{1}{2}c_{\text{D}}\rho Av^{2}\therefore v^{2}=\frac{2F}{c_{\text{D}}\rho A}

v_{\text{true}}=\sqrt{\frac{2(22)}{(0.88)(1.2)(0.32)}}=11.4108

2\frac{\Delta v}{v}=\frac{\Delta F}{F}+\frac{\Delta c_{\text{D}}}{c_{\text{D}}}+\frac{\Delta \rho}{\rho}+\frac{\Delta A}{A}

v=11.4108\pm 1.4155 = 11\pm 1

(ci)

Work done is defined as the product of the force and the distance moved in the direction of the force, i.e. WD = F•d

\text{P}=\frac{\Delta\text{Energy}}{t}=\frac{\text{WD}}{t}=\frac{F\cdot d}{t}=Fv(cii)

\text{P}=Fv=(22)(11.4108)=251.0376=251 \text{ W}[Maximum Marks: 8]

(a)

Gravitational potential at a point in a gravitational field is defined as the work done by an external agent in bringing unit mass from infinity to that point, without acceleration.

(b)

\Delta U=m\Delta\phi=m(\phi_{\text{f}}-\phi_{\text{i}})=m(\phi_{\text{0}}-\phi_{\text{E}})=m(-\frac{GM}{r_{0}}+\frac{GM}{r_{\text{E}}})\Delta U=GMm\left [\frac{1}{r_{\text{E}}}-\frac{1}{r_{0}} \right ]=(6.67\times 10^{-4})(6\times 10^{24})(1600)\left [ \frac{1}{6.4\times10^{6}}-\frac{1}{2.7\times10^{7}} \right ]=7.633\times 10^{10}=7.63\times 10^{10}\text{ J}

(ci)

Since satellite orbits planet, F_{\text{G}}=F_{\text{c}}

\frac{GMm}{r^{2}}=\frac{mv^{2}}{r}\therefore v^{2}=\frac{GM}{r} — (1)

KE = \frac{1}{2}mv^{2} — (2)

Substitute 1 into 2,

KE = \frac{1}{2}m(\frac{GM}{r})=\frac{GMm}{2r} (shown)

(cii)

Kinetic Energy = \frac{GMm}{2r}=\frac{(6.67\times 10^{-11})(6\times 10^{24})(1600)}{2(2.7\times 10^{7})}=1.1857\times 10^{10}\text{ J}=1.19\times 10^{10}\text{ J}

(ciii)

Incorrect. The expression in (ci) is only used when the object is moving in orbital or circular motion. If a satellite is launched from Earth surface to the orbit, there is a change in height and thus potential energy. Therefore, conservation of energy needs to be employed.

[Maximum Marks: 5]

By Gaussian Theorem, inside the conductor, E = 0 and V = constant

(a)

E=-\frac{dV}{dr}

(b)

[Maximum Marks: 8]

(a)

Diffraction is the spreading of wavefronts when a wave passes through an aperture (slits) or around an obstacle.

(b)

1. The waves from the 2 sources must be emitted from coherent sources to ensure constant phase difference.

2. The waves from the 2 sources must have roughly the same intensities.

(ci)

gradient = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{\lambda}{d}

\frac{(4.4-8.8)10^{-3}}{1-2}=\frac{\lambda}{0.12\times10^{-3}}\lambda=528 \text{ nm} (to 3 s.f.)

(cii)

A (central initial) = 2A

A (central final) = 1.5 A

A (1st dark initial) = 0

A (1st dark final) = 0.5 A (due to incomplete cancellations)

Ratio = (\frac{1.5\text{A}}{0.5\text{A}})^{2} = 9.00 (to 3 s.f.)

[Maximum Marks: 9]

(a)

I=\frac{Q}{t}=\eta AvqQ=\eta Avqt=(5.9\times10^{28})(\pi)(0.19\times10^{-3})^{2}(7.2\times 10^{-5})(1.6\times 10^{-19})(30\times 60)=138.75 \text{ C}=139 \text{ C} (to 3 s.f.)

(bi)

1. f=\frac{w}{2\pi}=\frac{120\pi}{2\pi}=60.0 \text{ Hz}

2. V_{\text{rms}}=\frac{V_{0}}{\sqrt{2}}=\frac{9}{\sqrt{2}}=6.3639=6.36 \text{ V} (to 3 s.f.)

(bii)

1. I_{0}=\frac{V_{0}}{R_{\text{eff}}}=\frac{9}{(\frac{1}{12}+\frac{1}{6})^{-1}+12}=0.5625=0.563\text{ A} (to 3 s.f.)

2. By current divider law, I_{6\Omega}=\frac{12}{12+6}\times 0.5625=0.375\text{ A}

P_{\text{rms}}=I_{\text{rms}}^{2}\cdot R=(\frac{I_{0}}{\sqrt{2}})^{2}R=(\frac{0.375}{\sqrt{2}})^{2}(6)=0.4218=0.422\text{ W} (to 3 s.f.)

[Maximum Marks: 8]

(a)

The magnetic flux density on unit length of a long conductor carrying unit charge current, placed perpendicular to the uniform magnetic field.

(b)

R=\frac{\rho_{\text{Cu}}L_{\text{wire}}}{A_{\text{wire}}}=\frac{\rho_{\text{Cu}}(2\pi r_{\text{sol}})(N)}{\pi r_{\text{wire}}^{2}}=\frac{(1.7\times 10^{-8})(2)(2\times 10^{-2})(520)}{(0.23\times 10^{-3})^{2}}=6.6843=6.68\Omega(c)

B_{\text{sol}}=\frac{\mu_{0}IN}{L}=\frac{\mu_{0}VN}{R\cdot D_{\text{wire}}N}=\frac{4\pi \times 10^{-7}(24)}{(6.6843)(0.46\times 10^{-3})}=9.8085\times 10^{-3}\text{ T}=9.81 \text{ mT}(to 3s.f.)

(d)

Force is 0. The current in the wire is parallel to the field of the solenoid. According to Fleming’s Left Hand Rule, the force is 0.

[Maximum Marks: 11]

(a)

Discrete packet or quantum of Electromagnetic (EM) radiation with each packet having energy, E=hf, where h is the Planck constant and f is the frequency of the incoming EM radiation.

(b)

This is an absorption spectrum. After white light passes through the cool gas, the remaining outgoing light passes through a spectroscope, revealing dark lines on a coloured background. The incoming photons from the incoming white light, whose energies are exactly equal to the difference between the discrete energy levels in the atoms of the cool gas are absorbed by the cool gas. The electrons from the lower energy levels are excited to the higher energy levels. Being in the unstable states, those electrons subsequently de-excite back to the lower, more stable energy levels, thereby releasing photons in random directions that do not pass through the spectroscopy. Since teh atom energy levels are discrete, only photons of certain frequencies are absorbed. As these frequencies of light are now missing, they account for the dark lines in a continuous coloured spectrum.

(ci)

E_{2}=\frac{-13.6}{2^{2}}=-3.40 eV

(cii)

\left | \Delta E \right |=\frac{hc}{\lambda}\therefore\lambda=\frac{hc}{\left | \Delta E \right |}

\lambda=\frac{1.989\times 10^{-25}}{\left | (3.40-1.511)\times 1.6\times10^{-19} \right |}=6.58125\times10^{-7}=658 \text{ nm} (to 3 s.f.)

(ciii)

The electrical energy level (EPE) of electrons in each energy level is defined as the work done by the external agent to bring the electron from infinity to that point without acceleration. Since the electric force is attractive in nature, the work done by the external agent is negative. Moreover, by convention, EPE at infinity is 0.

[Maximum Marks: 22]

(a)

v^{2}=u^{2}+2asa=\frac{v^{2}-u^{2}}{2s}=\frac{0-(\frac{185\times1000}{3600})^{2}}{2(80)}=-16.505=-16.5 \text{ m s}^{-2}

(bi)

a_{\text{c}}=4g=\frac{v^{2}}{r}v_{\text{max circular}}=\sqrt{a_{\text{c}}r}=\sqrt{4(9.81)(30)} = 34.3103 = 34.3 m/s (to 3s.f.)

(bii)

When the tyres are heated, the tyre compound becomes softer, thereby increasing the contact area by filling up the air gaps between the tyres and the ground, thus improving the grip between the tyres and the ground. Moreover, with a larger temperature, the coefficient of friction increases. thereby increasing the friction between the tyres and the ground. This helps to achieve greater cornering speeds and larger accelerations.

(c)

At the lower side of the wing, the air moves faster, suggesting lower pressure and thus smaller upward force, which means the rate of change of momentum of the air on the lower side of the wings must be smaller.

On the upper side of the wing, the air moves slower, suggesting higher pressure, and larger downward force, which means that the rate of change of momentum of the air on the upper side of the wings must be longer. By Bernoulli Principle, this causes a larger downward net force to add traction to the car, thus appearing to increase the weight of the car.

(di)

Total Upward Force = Total Downward Force

N_{\text{R}}+N_{\text{F}}=W— (1)

Total Clockwise Moments = Total Anti-clockwise Moments

N_{\text{R}}x_{\text{R}}=N_{\text{F}}x_{\text{F}}+Dh — (2)

From 1, N_{\text{F}}=W-N_{\text{R}} — (3)

Substitute 3 into 2,

N_{\text{R}}x_{\text{R}}=(W-N_{\text{R}})x_{\text{F}}+DhN_{\text{R}}x_{\text{R}}=Wx_{\text{F}}-N_{\text{R}}x_{\text{F}}+DhN_{\text{R}}(x_{\text{R}}+x_{\text{F}})=Wx_{\text{F}}+Dh\therefore N_{\text{R}}=\frac{Wx_{\text{F}}+Dh}{x_{\text{R}}+x_{\text{F}}} (shown)

(dii)

N_{\text{R}}+N_{\text{F}}=W — (1)

N_{\text{R}}x_{\text{R}}=N_{\text{F}}x_{\text{F}}+Dh — (2)

From 1, N_{\text{R}}=W-N_{\text{F}} — (3)

Substitute 3 into 2,

Wx_{\text{R}}-N_{\text{F}}x_{\text{R}}=N_{\text{F}}x_{\text{F}}+Dh

\therefore N_{\text{F}}=\frac{Wx_{\text{R}}-Dh}{x_{\text{R}}+x_{\text{F}}}

(diii)

N_{\text{R}}=\frac{Wx_{\text{F}}+Dh}{x_{\text{R}}+x_{\text{F}}} and N_{\text{F}}=\frac{Wx_{\text{R}}-Dh}{x_{\text{R}}+x_{\text{F}}}. When the car accelerates, D\neq 0, and N_{p} increases by \frac{Dh}{x_{\text{R}}+x_{\text{F}}} whereas N_{\text{F}} decreases by \frac{Dh}{x_{\text{R}}+x_{\text{F}}}, therby causing weight trasnfer to the rear wheels.

(ei)

By COE, \Delta \text{KE}=Q=mc\Delta \theta

\frac{1}{2}m_{\text{car}}(v_{\text{f}}-v_{\text{c}})^{2}=m_{\text{pad}}Nc\Delta \thetaHence, \Delta \theta=\frac{m_{\text{car}}(v_{\text{f}}-v_{\text{i}})^{2}}{2m_{\text{pad}}Nc}

\therefore \Delta \theta=\frac{(750)(0-\frac{185(1000)}{3600})^{2}}{2(1.2)(4)(1130)}=182.57=183 \text{ K}=183^{\circ}\text{C}

(eii)

The entire kinetic energy loss of the car goes entirely into raising the temperature of the 4 brake pads and there are no energy losses.

(eiii)

1. To allow for a more effective heat dissipation

2. To allow water to be ejected out during rainy weather.

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