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2021 A Level Physics 9749 Paper 2

2021 A-level Physics (9749) Paper 2 Suggested Solutions

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Answer Key to 2021 A Level Physics Paper 2:

[Maximum Marks: 5]

(a)

The block slides down the slope with constant acceleration.

 

(bi)

\displaystyle v=\frac{{\Delta s}}{{\Delta t}}=\frac{{{{y}_{1}}-{{y}_{2}}}}{{{{x}_{1}}-{{x}_{2}}}}=\frac{{0.52-0}}{{0.5-0.1}}=1.30\,\text{m }{{\text{s}}^{{\text{-1}}}}\,(\text{to 2 s}\text{.f}\text{.})

 

(bii)

\begin{array}{l}v=u+at\\a=\frac{{v-u}}{t}=\frac{{1.30-u}}{{0.2}}=6.50\,\text{m}\,{{\text{s}}^{{\text{-2}}}}\end{array}

[Maximum Marks: 10]

(a)

E=\frac{{{{p}^{2}}}}{{2m}}=\frac{{{{{(3.2)}}^{2}}}}{{2(0.62)}}=8.25806=8.26\,\text{J}\,\text{(3 s}\text{.f}\text{.)}

 

(b)

By N2L, {{F}_{{\text{net}}}}=\frac{{\Delta p}}{{\Delta t}}=\frac{{{{p}_{{\text{f}}}}-{{p}_{{\text{i}}}}}}{{\Delta t}}=\frac{{-1.8-3.2}}{{0.68-0.53}}=-33.333\,\text{N}

\begin{array}{l}{{F}_{{\text{net}}}}=mg-\text{N}\\\text{N}=mg-{{F}_{{\text{net}}}}\\=(0.62)(9.81)-(-33.333)\\=39.415\\=39.4\,\text{N}\,(\text{3 s}\text{.f}\text{.})\end{array}

 

(c)

Kinetic energy before hitting ground = 8.25806 J

Kinetic energy after hitting ground = \frac{{{{p}^{2}}}}{{2m}}=\frac{{{{{(-1.8)}}^{2}}}}{{2(0.62)}}=2.612\,\text{J}

% KE loss = \frac{{8.25806-2.612}}{{8.25806}}\times 100\%=68.359\%

% KE remaining after each bounce = 100 % – 68.359 % = 31.6406 %

Hence, 8.25806{{\left( {\frac{{31.6406}}{{100}}} \right)}^{n}}<(0.05)(8.25806)

Take lg on both sides

\begin{array}{l}n\lg (0.316406)<\lg (0.05)\\n>2.603,n\in {{\mathbb{Z}}^{+}}\\\therefore n=3\end{array}

[Maximum Marks: 5]

(a)

Let L be the length of PQ.

Total CW moments = Total ACW moments

F\left( {\frac{2}{3}L\sin 43{}^\circ } \right)=W\left( {\frac{1}{2}L\cos 58{}^\circ } \right)F=\frac{{(2.3)(9.81)(\tfrac{1}{2})(\cos 58{}^\circ )(\tfrac{3}{2})}}{{\sin 43{}^\circ }}=13.148=13.1\,\text{N}\,(\text{3 s}\text{.f}\text{.})

 

(b)

For the sign to be in equilibrium, the net forces and the net torque about any point must be zero. These forces form a closed loop polygon and must pass through a single point.

For force F, its weight and the force acting on the sign at Q to pass through a single point, the force acting on the sign at Q cannot be vertical.

[Maximum Marks: 8]

(a)

By definition, g is defined as the gravitational force acting on a unit mass at a point.

g=\frac{{GM(1)}}{{{{r}^{2}}}}=\frac{{GM}}{{{{r}^{2}}}}\,(\text{derived})


(b)

F_{G}=F_{c}

\frac{GMm}{r^{2}}=mr\left (\frac{2\pi}{T} \right)^{2}

r^{3}=\frac{GMT^{2}}{4\pi^{2}}

r=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}}}=\sqrt[3]{\frac{(6.67\times 10^{11})(6\times 10^{24})(6600)^{2}}{4\pi^{2}}}=7.6149\times 10^{6}\text{ m}

g=\frac{GM}{r^{2}}=\frac{(6.67\times 10^{-11})(6\times 10^{24})}{(7.6149\times 10^{6})^{2}}=6.901=6.90 \textup{ N} \textup{ kg}^{-1}


(c)

An object of mass, m, is free falling when acted only by the gravitational attractive force, F_{{\text{G}}}. By Newton’s Second Law, {{g}_{{\text{free}\,\text{fall}}}}=\frac{{{{F}_{\text{G}}}}}{m}. By definition, gravitational field strengthF_{{\text{G}}}. By Newton’s Second Law, g=\frac{{{{F}_{\text{G}}}}}{m}. Hence, the acceleration of free fall equals the gravitational field strength near the strength of the Earth.

[Maximum Marks: 8]

(a)

v=f\lambda


\lambda =\frac{v}{f}=\frac{{340}}{{1700}}=0.2

I\propto {{A}^{2}}

 

(b)

\displaystyle \text{T}=\frac{1}{\text{f}}=\frac{1}{{250}}=4\,\text{ms}

 

(c)

{{B}_{{\text{coil}}}}=\frac{{{{\mu }_{0}}IN}}{{2r}}=\frac{k}{r}


\therefore {{B}_{{\text{coil}}}}\propto \frac{1}{r}

[Maximum Marks: 11]

(ai)

{{V}_{\text{L}}}={{I}_{\text{L}}}{{R}_{\text{L}}}=(0.3)(5)=1.5\,\text{V}

By Potential Divider Principle, {{\text{V}}_{{3\,\Omega }}}=6-1.5=4.5\,\text{V}

\displaystyle {{I}_{{3\,\Omega }}}=\frac{{{{V}_{{3\,\Omega }}}}}{{{{R}_{{3\,\Omega }}}}}=\frac{{4.5}}{3}=1.5\,\text{A}

I_{R}=1.5-0.3=1.2 \text{ A}

V_{R}=I_{R}R_{R}

\therefore {{R}_{\text{R}}}=\frac{{{{V}_{\text{R}}}}}{{{{I}_{\text{R}}}}}=\frac{{{{V}_{\text{L}}}}}{{{{I}_{\text{R}}}}}=\frac{{1.5}}{{1.2}}=1.25\,\Omega \,\,(\text{3 s}\text{.f}\text{.})


(aii)

{{E}_{\text{T}}}={{V}_{\text{T}}}{{I}_{\text{T}}}t=120\,\text{J}

\displaystyle \therefore t=\frac{{120}}{{{{V}_{\text{T}}}{{I}_{\text{T}}}}}=\frac{{120}}{{(6)(1.5)}}=13.3333\,\text{s}

{{E}_{\text{L}}}={{V}_{\text{L}}}{{I}_{\text{L}}}t=(1.5)(0.3)(13.3333)=6.00\,\text{J}(\text{3 s}\text{.f}\text{.})


(aiii)

When lamp is first switched on, its temperature is low and hence its resistance is low. Hence, by Ohm’s law, current is largest when lamp is first switched on. As current passes through the lamp, the lattice ions vibrate vigorously, thereby increasing the impedance between the charge flow, hence decreasing current flow.


(b)

\frac{r_{X}}{r_{Y}}=\frac{1}{2},\frac{v_{X}}{v_{Y}}=3

From I=nAvq and since I flowing through X and Y is the same,

\frac{{{{I}_{Y}}}}{{{{I}_{X}}}}=\frac{{{{n}_{Y}}{{A}_{Y}}{{v}_{Y}}q}}{{{{n}_{X}}{{A}_{X}}{{v}_{X}}q}}=\frac{{{{n}_{Y}}\pi {{r}_{Y}}^{2}{{v}_{Y}}q}}{{{{n}_{X}}\pi {{r}_{X}}^{2}{{v}_{X}}q}}

\frac{{{{n}_{Y}}}}{{{{n}_{X}}}}=\frac{3}{4}=0.750\,(\text{3 s}\text{.f}\text{.})

[Maximum Marks: 11]

(a)

Charge Distribution: When \alpha-particles (positively charged Helium nucleus) were directed towards the thin gold foil, a small number of \alpha-particles were deflected at a scattering angle of about 5{}^\circ -90{}^\circ. Very few \alpha-particles (\sim 1 in 8000) were deflected at a scattering angle of approximately 180^\circ, suggesting a near head on collision with the gold nuclei. This means that the nucleus of the atom is small (\sim5% of atom), positively charged and located at approximately the centre of atom.

mass distribution: Very few \alpha-particles (\sim 1 in 8000) were deflected at a scattering angle of about 180^\circ, suggesting a near head on collision with the gold nucleus. This means that the nucleus of the atom must be very small (\sim 5% of atom), positive, containing almost entire mass of the atom (protons and neutrons) to even cause the \alpha-particles to recoil almost backwards.

 

(b)

By Conservation of Energy, KE loss = EPE gain

5.59\,\text{M}eV=\frac{{Qq}}{{4\pi {{\varepsilon }_{o}}r}}

\therefore r=\frac{1}{{4\pi {{\varepsilon }_{o}}}}\cdot \frac{{Qq}}{1}\cdot \frac{1}{{5.59\text{M}eV}}=\frac{{(8.99\times {{{10}}^{9}})(79)(1.6\times {{{10}}^{{-19}}})(2)(1.6\times {{{10}}^{{-19}}})}}{{5.59\times {{{10}}^{6}}\times 1.6\times {{{10}}^{{-19}}}}}=4.065\times {{10}^{{-14}}}\text{ m}=4.07\times {{10}^{{-14}}}\text{ m }(\text{3 s}\text{.f}\text{.})

 

(c)

{}_{{86}}^{{222}}Rn\to {}_{2}^{4}He+{}_{{84}}^{{218}}Po

E_{LHS} = (7.69)(222) = 1707.18 MeV

E_{RHS} = (4)(7.08)+ (218)(\text{B}{{\text{E}}_{{\text{Po}}}}) = 28.32 + 218 \text{B}{{\text{E}}_{{\text{Po}}}}

Since energy released, E_{RHS}>E_{LHS} by 6.62 MeV

\therefore1707.18 MeV + 6.62 MeV = 28.32 MeV + 218 \text{B}{{\text{E}}_{{\text{Po}}}}

\therefore \text{B}{{\text{E}}_{{\text{Po}}}} = 7.7315 = 7.73 MeV/nucleon (3 s.f.)

[Maximum Marks: 22]

(a)

It is to prevent the electrons from being scattered by air particles and not being able to reach the target metal.


(b)


(ci)

Thermal Energy = 99% of Electron Energy = (0.99)VIt = (0.99)(65000)(0.12)(1.1) = 8494.2 = 8490 J (3 s.f.)


(cii)

Q=mc\theta

\theta =\frac{Q}{{mc}}=\frac{{8494.2}}{{\left( {\frac{{12}}{{1000}}} \right)(130)}} = 5445 = 5450 ^{\circ}C (3 s.f.)


(ciii)

To ensure that the electron beam hits all parts of the target anode evenly, rather than at one specific portion.


(di)


(dii)

Wavelength increases. When the photon is incident on the electron, it loses some of its energy. Since energy is inversely proportional to wavelength, from Planck’s Law, the wavelength increases, resulting in a redshift.


(diii)

Due to Compton Scattering, the photons’ intensity is reduced and they travel in random directions. Hence, the X-ray image have poor contrast.


(e)

Attenuation \propto {{Z}^{3}}

{{Z}_{{\text{ST}}}}=7,{{Z}_{\text{B}}}=14

Ratio = {{\left( {\frac{{14}}{7}} \right)}^{3}}={{2}^{3}}=8.00 (3 s.f.)


(f)

From I=I_{o}e^{-\mu x},

\frac{I_{o}}{2}=I_{o}e^{-\mu x_{\frac{1}{2}}} \frac{1}{e^{\mu x_{\frac{1}{2}}}}=\frac{1}{2}

So, e^{\mu x_{\frac{1}{2}}}=2

Take ln on both sides,

\mu x_{\frac{1}{2}}=\ln 2

\therefore x_{\frac{1}{2}}=\frac{\ln2}{\mu}


(g)

It allows very small medical issues, that would have not been detected, to be noticed.


(h)

Barium which has a Z value of 56, would produce a larger attenuation of incident X-rays (56^{3}). Hence, there would now be significant differences between the X-ray intensities in the different regions, thereby improving the contrast.


(i)

In X-ray imaging, only a single exposure of X-rays is used to produce the 2D image. In CT scans, X-ray beams spirals down the object taking many X-ray images to produce the 3D image.

The patient has more exposure to the X-rays in a CT scan. The X-rays, a form of ionizing radiation, can damage living cells and cause cancer.

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