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Paper 1 Answers

Time2 hours

Marks60

SetSet 4

PaperLevel 1 - Physics Paper 1

Information

Section A: Mechanics
Section B: Waves and Optics
Section C: Oscillations

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Constants

Gravitational acceleration	\( g=9.81\,\mathrm{m\,s^{-2}} \)
Speed of light	\( c=3.00\times10^8\,\mathrm{m\,s^{-1}} \)
Elementary charge	\( e=1.60\times10^{-19}\,\mathrm{C} \)
Electron mass	\( m_e=9.11\times10^{-31}\,\mathrm{kg} \)
Planck constant	\( h=6.63\times10^{-34}\,\mathrm{J\,s} \)
Permittivity of free space	\( \epsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \)
Magnetic constant	\( \mu_0=1.26\times10^{-6}\,\mathrm{N\,A^{-2}} \)
Boltzmann constant	\( k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}} \)

Section A: Mechanics

Answer 1. Projectile launch, stopping work, and impulse

[15 marks]

A robotic arm releases a sealed sample cartridge of mass \(0.180\,\mathrm{kg}\) from the end of a horizontal conveyor. At release the cartridge is \(3.20\,\mathrm{m}\) horizontally from the centre of a padded catch tray, and the tray surface is \(0.080\,\mathrm{m}\) below the release point. The launch speed is \(5.80\,\mathrm{m\,s^{-1}}\) at \(32.0^\circ\) above the horizontal. Air resistance is negligible until the cartridge enters the tray padding, which brings it to rest over \(0.060\,\mathrm{m}\) measured along the direction of its motion.

a) Show that the projectile path carries the cartridge to the catch tray.

[4 marks]

Resolve the launch velocity: \[ v_{0x}=5.80\cos32.0^\circ=4.92\,\mathrm{m\,s^{-1}}, \qquad v_{0y}=5.80\sin32.0^\circ=3.07\,\mathrm{m\,s^{-1}}. \] The time to reach the tray centre horizontally is \[ t=\frac{x}{v_{0x}}=\frac{3.20}{4.92}=0.651\,\mathrm{s}. \] The vertical displacement at this time is \[ y=v_{0y}t-\frac12gt^2 =(3.07)(0.651)-\frac12(9.81)(0.651)^2 =-8.2\times10^{-2}\,\mathrm{m}. \] This is \(0.082\,\mathrm{m}\) below the release point, so the cartridge reaches the tray surface to within the stated data precision.

Answer 1. Projectile launch, stopping work, and impulse

[15 marks]

b) Calculate the velocity of the cartridge just before it touches the tray padding.

[3 marks]

The horizontal velocity remains \[ v_x=4.92\,\mathrm{m\,s^{-1}}. \] The vertical velocity at the tray is \[ v_y=v_{0y}-gt=3.07-9.81(0.651)=-3.31\,\mathrm{m\,s^{-1}}. \] Therefore \[ v=\sqrt{4.92^2+3.31^2}=5.93\,\mathrm{m\,s^{-1}}. \] The direction is \[ \tan^{-1}\left(\frac{3.31}{4.92}\right)=33.9^\circ \] below the horizontal.

Answer 1. Projectile launch, stopping work, and impulse

[15 marks]

c) Estimate the average stopping force exerted by the tray padding. State the main approximation.

[3 marks]

Ignoring the small work done by weight during the short stopping distance, the padding removes the cartridge kinetic energy: \[ F_{\mathrm{avg}}s=\frac12mv^2. \] Thus \[ F_{\mathrm{avg}}=\frac{\frac12(0.180)(5.93)^2}{0.060} =52.7\,\mathrm{N}. \] The force exerted by the padding on the cartridge is directed approximately opposite to the incoming velocity.

Answer 1. Projectile launch, stopping work, and impulse

[15 marks]

d) Find the impulse vector exerted by the padding on the cartridge, using \(\hat{\imath}\) horizontally toward the tray and \(\hat{\jmath}\) upward.

[3 marks]

Use \(\vec J=\Delta\vec p=m\vec v_f-m\vec v_i\). With \(\vec v_f=0\), \[ \vec J=-m\vec v_i =-0.180(4.92\hat{\imath}-3.31\hat{\jmath}) =(-0.886\hat{\imath}+0.596\hat{\jmath})\,\mathrm{N\,s}. \] The magnitude is \[ |\vec J|=\sqrt{0.886^2+0.596^2}=1.07\,\mathrm{N\,s}. \]

Answer 1. Projectile launch, stopping work, and impulse

[15 marks]

e) Interpret the signs of the impulse components physically.

[2 marks]

The impulse has a negative horizontal component because the padding removes the forward momentum. Its positive vertical component means the padding also pushes upward while stopping the downward motion. The impulse is therefore directed backward and upward, opposite to the incoming velocity vector.

Answer 2. Banked circular motion and changing speed

[15 marks]

A cyclist rides on a banked indoor velodrome bend of radius \(28.0\,\mathrm{m}\). The track is banked at \(12.0^\circ\) to the horizontal. For the tyres on the timber surface, the coefficient of static friction is \(0.350\). During one half-turn, the cyclist speeds up uniformly from \(9.00\,\mathrm{m\,s^{-1}}\) to \(15.0\,\mathrm{m\,s^{-1}}\). Treat the cyclist and bicycle as a point mass.

a) Find the speed for which no cross-slope friction is needed.

[3 marks]

With no cross-slope friction required, \[ \tan\theta=\frac{v_0^2}{rg}. \] Hence \[ v_0=\sqrt{rg\tan\theta} =\sqrt{(28.0)(9.81)\tan12.0^\circ} =7.64\,\mathrm{m\,s^{-1}}. \]

Answer 2. Banked circular motion and changing speed

[15 marks]

b) Calculate the tangential acceleration while the cyclist speeds up.

[3 marks]

The half-turn arc length is \[ s=\pi r=\pi(28.0)=88.0\,\mathrm{m}. \] Using \(v^2=u^2+2a_ts\), \[ a_t=\frac{15.0^2-9.00^2}{2(88.0)} =0.819\,\mathrm{m\,s^{-2}}. \] The tangential acceleration is forward, in the direction of motion.

Answer 2. Banked circular motion and changing speed

[15 marks]

c) Find the radial acceleration and the magnitude of the total acceleration when the speed is \(15.0\,\mathrm{m\,s^{-1}}\).

[3 marks]

At \(15.0\,\mathrm{m\,s^{-1}}\), the inward radial acceleration is \[ a_r=\frac{v^2}{r}=\frac{15.0^2}{28.0}=8.04\,\mathrm{m\,s^{-2}}. \] The radial and tangential components are perpendicular, so \[ a=\sqrt{a_r^2+a_t^2} =\sqrt{8.04^2+0.819^2} =8.08\,\mathrm{m\,s^{-2}}. \]

Answer 2. Banked circular motion and changing speed

[15 marks]

d) Determine the direction of the cross-slope friction at high speed and test whether \(15.0\,\mathrm{m\,s^{-1}}\) is possible without slipping. Let \(k=v^2/(rg)\) and \(q=f/N\). For friction down the slope, the required friction ratio is \[ q=\frac{k\cos\theta-\sin\theta}{\cos\theta+k\sin\theta}. \]

[4 marks]

Since both speeds are above \(7.64\,\mathrm{m\,s^{-1}}\), the cyclist would tend to move up the bank if there were no friction. The required cross-slope static friction therefore acts down the bank, giving an additional inward component. Let \(q=f/N\) and \(k=v^2/(rg)\). For friction down the slope, \[ k=\frac{\sin\theta+q\cos\theta}{\cos\theta-q\sin\theta}, \] so \[ q=\frac{k\cos\theta-\sin\theta}{\cos\theta+k\sin\theta}. \] At \(15.0\,\mathrm{m\,s^{-1}}\), \[ k=\frac{15.0^2}{(28.0)(9.81)}=0.819, \] and \[ q=\frac{(0.819)\cos12.0^\circ-\sin12.0^\circ}{\cos12.0^\circ+(0.819)\sin12.0^\circ} =0.516. \] Because \(0.516>0.350\), static friction is not sufficient at this speed.

Answer 2. Banked circular motion and changing speed

[15 marks]

e) Find the safe-speed range for the banked bend. At the high-speed limit, the equation for \(k_{\max}\) is \[ k_{\max}=\frac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}, \] where \(k=v^2/(rg)\).

[2 marks]

At the maximum safe speed, friction is down the slope with \(f=\mu_sN\): \[ k_{\max}=\frac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}. \] Thus \[ k_{\max}=\frac{\sin12.0^\circ+0.350\cos12.0^\circ}{\cos12.0^\circ-0.350\sin12.0^\circ}=0.608, \] and \[ v_{\max}=\sqrt{k_{\max}rg} =\sqrt{(0.608)(28.0)(9.81)} =12.9\,\mathrm{m\,s^{-1}}. \] Since \(\mu_s>\tan12.0^\circ\), the lower safe-speed limit includes rest; the safe range is approximately \(0\) to \(12.9\,\mathrm{m\,s^{-1}}\).

Section B: Waves and Optics

Answer 3. Grating spectrometer resolution and uncertainty

[15 marks]

A diffraction grating spectrometer is used at normal incidence to compare two close green spectral lines of wavelengths \(540.0\,\mathrm{nm}\) and \(541.8\,\mathrm{nm}\). The grating has \(750\,\mathrm{lines\,mm^{-1}}\), and a screen is placed \(1.50\,\mathrm{m}\) from the grating. The illuminated width of the grating is \(4.00\,\mathrm{mm}\).

a) Find the grating spacing and the highest possible order for these wavelengths.

[3 marks]

The line density is \[ n_L=750\times10^3=7.50\times10^5\,\mathrm{m^{-1}}. \] Therefore \[ d=\frac{1}{n_L}=1.33\times10^{-6}\,\mathrm{m}. \] Using the longer wavelength for the most restrictive order condition, \[ m_{\max}=\left\lfloor\frac{d}{\lambda}\right\rfloor =\left\lfloor\frac{1.33\times10^{-6}}{541.8\times10^{-9}}\right\rfloor =2. \]

Answer 3. Grating spectrometer resolution and uncertainty

[15 marks]

b) Calculate the first-order and second-order angles for the \(540.0\,\mathrm{nm}\) line.

[3 marks]

For \(\lambda=540.0\,\mathrm{nm}\), the grating equation is \[ d\sin\theta_m=m\lambda. \] For first order, \[ \sin\theta_1=\frac{540.0\times10^{-9}}{1.33\times10^{-6}}=0.405, \] so \[ \theta_1=23.9^\circ. \] For second order, \[ \sin\theta_2=\frac{2(540.0\times10^{-9})}{1.33\times10^{-6}}=0.810, \] so \[ \theta_2=54.1^\circ. \]

Answer 3. Grating spectrometer resolution and uncertainty

[15 marks]

c) Estimate the first-order separation of the two spectral lines on the screen. For a small wavelength difference at fixed order, the angular separation is given by \[ d\cos\theta\,\Delta\theta=m\Delta\lambda. \]

[3 marks]

For small wavelength separation at fixed order, \[ d\cos\theta\,\Delta\theta=m\Delta\lambda. \] In first order, using \(\theta=23.9^\circ\), \[ \Delta\theta=\frac{1(1.8\times10^{-9})}{(1.33\times10^{-6})\cos23.9^\circ} =1.48\times10^{-3}\,\mathrm{rad}. \] On the screen, \(y=L\tan\theta\), so \[ \Delta y\approx L\sec^2\theta\,\Delta\theta =(1.50)\sec^2(23.9^\circ)(1.48\times10^{-3}) =2.66\times10^{-3}\,\mathrm{m}. \] The first-order lines are separated by about \(2.7\,\mathrm{mm}\).

Answer 3. Grating spectrometer resolution and uncertainty

[15 marks]

d) Use resolving power to decide whether the two wavelengths can be resolved in first and second order. The resolving power of a grating is \[ R=mN \] and also \[ R=\frac{\lambda}{\Delta\lambda_{\min}}, \] where \(N\) is the number of illuminated grating lines.

[3 marks]

The number of illuminated grating lines is \[ N=(750\,\mathrm{lines\,mm^{-1}})(4.00\,\mathrm{mm})=3000. \] The resolving power is \(R=mN\). In first order, \[ R_1=3000, \qquad \Delta\lambda_{\min,1}=\frac{540.9\,\mathrm{nm}}{3000}=0.180\,\mathrm{nm}. \] In second order, \[ R_2=6000, \qquad \Delta\lambda_{\min,2}=\frac{540.9\,\mathrm{nm}}{6000}=0.090\,\mathrm{nm}. \] The actual separation \(1.8\,\mathrm{nm}\) is larger than both limits, so the two lines should be resolvable in either first or second order if the optical alignment is good.

Answer 3. Grating spectrometer resolution and uncertainty

[15 marks]

e) If the grating line density is \(750\pm3\,\mathrm{lines\,mm^{-1}}\) and the first-order angle reading has uncertainty \(\pm0.05^\circ\), estimate the uncertainty in a wavelength determined from the first-order angle. The fractional uncertainty is \[ \left(\frac{\Delta\lambda}{\lambda}\right)^2=\left(\frac{\Delta d}{d}\right)^2+(\cot\theta\,\Delta\theta)^2. \]

[3 marks]

For a wavelength found from \(\lambda=d\sin\theta/m\), the fractional uncertainty from grating spacing and angle is approximately \[ \left(\frac{\Delta\lambda}{\lambda}\right)^2 =\left(\frac{\Delta d}{d}\right)^2+\left(\cot\theta\,\Delta\theta\right)^2. \] Since \(d=1/n_L\), \[ \frac{\Delta d}{d}=\frac{\Delta n_L}{n_L}=\frac{3}{750}=4.00\times10^{-3}. \] With \(\Delta\theta=0.05^\circ=8.73\times10^{-4}\,\mathrm{rad}\) and \(\theta=23.9^\circ\), \[ \cot\theta\,\Delta\theta=(\cot23.9^\circ)(8.73\times10^{-4})=1.97\times10^{-3}. \] Therefore \[ \frac{\Delta\lambda}{\lambda}=\sqrt{(4.00\times10^{-3})^2+(1.97\times10^{-3})^2} =4.46\times10^{-3}, \] and \[ \Delta\lambda=(540.0\,\mathrm{nm})(4.46\times10^{-3})=2.4\,\mathrm{nm}. \]

Section C: Oscillations

Answer 4. Driven damped oscillator in a seismometer

[15 marks]

A laboratory seismometer contains a proof mass of \(0.800\,\mathrm{kg}\) attached to a spring of stiffness \(320\,\mathrm{N\,m^{-1}}\). A magnetic damper provides approximately linear damping, so the relative displacement \(x\) of the mass can be modelled as a driven damped oscillator. In one damping setting, a free oscillation has its amplitude reduced by a factor of 2 after 18 complete cycles. Later, a steady sinusoidal ground motion drives the instrument near resonance and the relative displacement amplitude is \(12.0\,\mathrm{mm}\).

a) Write the driven damped oscillator equation for the seismometer and calculate its undamped natural frequency.

[4 marks]

The driven damped oscillator model is \[ m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=F_0\cos\omega t, \] where the driving term represents the periodic forcing associated with the ground motion. The undamped natural angular frequency is \[ \omega_0=\sqrt{\frac{k}{m}} =\sqrt{\frac{320}{0.800}} =20.0\,\mathrm{rad\,s^{-1}}. \] The corresponding frequency is \[ f_0=\frac{\omega_0}{2\pi}=\frac{20.0}{2\pi}=3.18\,\mathrm{Hz}. \]

Answer 4. Driven damped oscillator in a seismometer

[15 marks]

b) Estimate the damping rate and the resonant angular frequency for this lightly damped setting. The free-oscillation amplitude envelope is \[ A=A_0e^{-\beta t}. \] For light damping, the resonant angular frequency is approximately \[ \omega_{\mathrm{res}}\approx\sqrt{\omega_0^2-2\beta^2}. \]

[3 marks]

The period is \[ T=\frac{2\pi}{\omega_0}=\frac{2\pi}{20.0}=0.314\,\mathrm{s}. \] For weak damping, the amplitude envelope is \(A=A_0e^{-\beta t}\). After 18 cycles, \[ \frac12=e^{-\beta(18T)}. \] Thus \[ \beta=\frac{\ln2}{18T} =\frac{0.693}{18(0.314)} =0.123\,\mathrm{s^{-1}}. \] The resonance angular frequency for light damping is approximately \(\omega_{\mathrm{res}}\approx\omega_0\), more exactly \(\sqrt{\omega_0^2-2\beta^2}\), so here \(\omega_{\mathrm{res}}\approx20.0\,\mathrm{rad\,s^{-1}}\).

Answer 4. Driven damped oscillator in a seismometer

[15 marks]

c) Estimate the percentage of mechanical energy lost per cycle and the quality factor \(Q\). The logarithmic decrement per cycle is \[ \delta=\beta T. \] For weak damping, the quality factor is approximately \[ Q\approx\frac{\pi}{\delta}. \] The oscillator energy is proportional to amplitude squared.

[3 marks]

The logarithmic decrement per cycle is \[ \delta=\frac{\ln2}{18}=0.0385. \] Because oscillator energy is proportional to amplitude squared, the energy ratio after one cycle is \[ \frac{E_{n+1}}{E_n}=e^{-2\delta}=e^{-0.0770}=0.926. \] The fractional energy loss per cycle is therefore \[ 1-0.926=0.074, \] or \(7.4\%\). A corresponding quality-factor estimate is \[ Q\approx\frac{2\pi}{\text{fractional energy loss per radian cycle denominator}}=\frac{2\pi}{0.0770}=81.6, \] equivalently \(Q\approx\pi/\delta\).

Answer 4. Driven damped oscillator in a seismometer

[15 marks]

d) Find the mechanical energy stored in the oscillator at the stated steady-state amplitude.

[2 marks]

For amplitude \(A=12.0\,\mathrm{mm}=0.0120\,\mathrm{m}\), the mechanical energy associated with the relative oscillator motion is \[ E=\frac12kA^2 =\frac12(320)(0.0120)^2 =2.30\times10^{-2}\,\mathrm{J}. \]

Answer 4. Driven damped oscillator in a seismometer

[15 marks]

e) Compare this setting with one in which the damping coefficient is doubled. Discuss the effect on \(Q\), energy loss per cycle, and resonant amplitude.

[3 marks]

Doubling the damping coefficient doubles \(\beta\) and doubles the logarithmic decrement per cycle: \[ \delta'=2\delta=0.0770. \] The quality factor is approximately halved: \[ Q'\approx\frac{\pi}{\delta'}=40.8. \] The energy ratio per cycle becomes \[ \frac{E_{n+1}}{E_n}=e^{-2\delta'}=e^{-0.154}=0.857, \] so the energy loss per cycle rises to about \(14.3\%\). The resonance curve would be lower and broader; for the same driving force near resonance the amplitude would be roughly halved, so the stored energy at the peak would be roughly one quarter as large.