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Paper 1 Answers

Time2 hours

Marks60

SetSet 2

PaperLevel 1 - Physics Paper 1

Information

Section A: Mechanics
Section B: Waves and Optics
Section C: Oscillations and Collisions, Conservation and Fields

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. Rescue sled on a curved valley track

[15 marks]

A rescue sled of total mass \(180\,\mathrm{kg}\) crosses a mountain valley on a smooth curved track. It is released from rest at a station \(14.0\,\mathrm{m}\) above the lowest point of the track. At the lowest point the track is locally circular with radius \(22.0\,\mathrm{m}\). Treat the sled as a particle and neglect rolling resistance.

a) Use energy conservation to find the sled's speed at the lowest point.

[4 marks]

The sled starts from rest and mechanical energy is conserved. Taking the lowest point as zero gravitational potential energy, \[ mgh=\frac12mv^2. \] The mass cancels, so \[ v=\sqrt{2gh}. \] With \(h=14.0\,\mathrm{m}\), \[ v=\sqrt{2(9.81)(14.0)} =16.6\,\mathrm{m\,s^{-1}}. \]

Answer 1. Rescue sled on a curved valley track

[15 marks]

b) Calculate the magnitude and direction of the sled's centripetal acceleration at the lowest point.

[3 marks]

At the lowest point the acceleration toward the centre of curvature is centripetal: \[ a_c=\frac{v^2}{R}. \] Using \(v^2=2gh=274.68\,\mathrm{m^2\,s^{-2}}\), \[ a_c=\frac{274.68}{22.0} =12.5\,\mathrm{m\,s^{-2}}. \] The direction is upward, toward the centre of the local circular arc.

Answer 1. Rescue sled on a curved valley track

[15 marks]

c) Find the normal force exerted by the track on the sled at the lowest point.

[4 marks]

At the lowest point the radial direction is upward. The forces in this direction are the normal force upward and weight downward, so \[ N-mg=m\frac{v^2}{R}. \] Therefore \[ N=m\left(g+\frac{v^2}{R}\right). \] Numerically, \[ N=180(9.81+12.5) =4.01\times10^3\,\mathrm{N}. \]

Answer 1. Rescue sled on a curved valley track

[15 marks]

d) The track is rated for a maximum normal force of \(4.50\times10^3\,\mathrm{N}\) on this loaded sled at the lowest point. Find the maximum safe speed there and the corresponding maximum release height above the lowest point.

[4 marks]

The largest allowed normal force gives the largest allowed speed at the bottom: \[ N_{\max}=m\left(g+\frac{v_{\max}^2}{R}\right). \] Thus \[ v_{\max}=\sqrt{R\left(\frac{N_{\max}}{m}-g\right)}. \] With \(N_{\max}=4.50\times10^3\,\mathrm{N}\), \[ v_{\max}=\sqrt{22.0\left(\frac{4500}{180}-9.81\right)} =18.3\,\mathrm{m\,s^{-1}}. \] The corresponding release height follows from \(mgh_{\max}=\frac12mv_{\max}^2\): \[ h_{\max}=\frac{v_{\max}^2}{2g} =\frac{(18.3)^2}{2(9.81)} =17.0\,\mathrm{m}. \] The planned \(14.0\,\mathrm{m}\) release height is therefore within the limit.

Answer 2. Winch drum with changing torque

[15 marks]

A motor drives a winch drum of radius \(0.120\,\mathrm{m}\) and moment of inertia \(0.360\,\mathrm{kg\,m^2}\). A light cord wrapped around the drum lifts a \(25.0\,\mathrm{kg}\) crate vertically without slipping. From rest, the motor torque is \(\tau(t)=36.0+4.00t\,\mathrm{N\,m}\), where \(t\) is in seconds. Friction in the bearings is negligible.

a) Derive an expression for the angular acceleration of the drum as a function of time.

[4 marks]

Let upward motion of the crate and unwinding rotation of the drum be positive. For the crate, \[ T-mg=ma=mR\alpha. \] For the drum, \[ \tau-TR=I\alpha. \] Eliminate \(T\) using \(T=m(g+R\alpha)\): \[ \tau-mgR-mR^2\alpha=I\alpha. \] Therefore \[ \alpha(t)=\frac{\tau(t)-mgR}{I+mR^2}. \] Here \(mgR=(25.0)(9.81)(0.120)=29.4\,\mathrm{N\,m}\) and \(I+mR^2=0.360+25.0(0.120)^2=0.720\,\mathrm{kg\,m^2}\), so \[ \alpha(t)=\frac{6.57+4.00t}{0.720} =9.13+5.56t\,\mathrm{rad\,s^{-2}}. \]

Answer 2. Winch drum with changing torque

[15 marks]

b) Find the speed of the cord and crate after \(3.00\,\mathrm{s}\).

[3 marks]

Starting from rest, \[ \omega(t)=\int_0^t\alpha(t')\,dt' =9.13t+2.78t^2. \] At \(t=3.00\,\mathrm{s}\), \[ \omega=9.13(3.00)+2.78(3.00)^2 =52.4\,\mathrm{rad\,s^{-1}}. \] The cord speed is \(v=R\omega\), so \[ v=(0.120)(52.4) =6.29\,\mathrm{m\,s^{-1}}. \]

Answer 2. Winch drum with changing torque

[15 marks]

c) Find the work done by the motor during the first \(3.00\,\mathrm{s}\).

[4 marks]

The angular displacement is \[ \theta(t)=\int_0^t\omega(t')\,dt' =\frac{9.13t^2}{2}+\frac{2.78t^3}{3}. \] At \(3.00\,\mathrm{s}\), \[ \theta=66.1\,\mathrm{rad},\qquad h=R\theta=(0.120)(66.1)=7.93\,\mathrm{m}. \] The work done by the motor becomes rotational kinetic energy, translational kinetic energy, and gravitational potential energy: \[ W=\frac12I\omega^2+\frac12mv^2+mgh. \] Thus \[ W=\frac12(0.360)(52.4)^2+\frac12(25.0)(6.29)^2+(25.0)(9.81)(7.93) =2.93\times10^3\,\mathrm{J}. \]

Answer 2. Winch drum with changing torque

[15 marks]

d) Calculate the instantaneous power delivered by the motor at \(t=3.00\,\mathrm{s}\).

[3 marks]

Instantaneous motor power is torque times angular speed: \[ P=\tau\omega. \] At \(t=3.00\,\mathrm{s}\), \[ \tau=36.0+4.00(3.00)=48.0\,\mathrm{N\,m}. \] Therefore \[ P=(48.0)(52.4) =2.51\times10^3\,\mathrm{W}. \]

Answer 2. Winch drum with changing torque

[15 marks]

e) State what would happen if the applied torque were less than \(mgR\).

[1 mark]

The motor torque must exceed \(mgR\) to give a positive upward angular acceleration from rest. If \(\tau<mgR\), the crate would not start lifting upward; if it were already moving upward, it would slow down.

Section B: Waves and Optics

Answer 3. Camera calibration and diffraction limit

[15 marks]

A compact camera is calibrated by photographing a flat test chart \(2.40\,\mathrm{m}\) in front of a thin converging lens of focal length \(12.0\,\mathrm{mm}\). The entrance pupil is a circular aperture of diameter \(3.00\,\mathrm{mm}\), and the calibration light has wavelength \(550\,\mathrm{nm}\). For a circular aperture, the Airy/Rayleigh angular radius is \[ \theta_R=1.22\frac{\lambda}{D}, \] where \(D\) is the aperture diameter.

a) Find the image distance from the lens to the sensor plane.

[4 marks]

Use the thin-lens equation \[ \frac1f=\frac1s+\frac1{s'}. \] With \(f=0.0120\,\mathrm{m}\) and \(s=2.40\,\mathrm{m}\), \[ \frac1{s'}=\frac1{0.0120}-\frac1{2.40} =82.9\,\mathrm{m^{-1}}. \] Therefore \[ s'=1.206\times10^{-2}\,\mathrm{m} =12.1\,\mathrm{mm}. \]

Answer 3. Camera calibration and diffraction limit

[15 marks]

b) Find the magnification and the image height of a \(1.20\,\mathrm{m}\) high calibration chart.

[3 marks]

The lateral magnification is \[ m=-\frac{s'}{s}. \] Thus \[ m=-\frac{0.01206}{2.40} =-5.03\times10^{-3}. \] For a \(1.20\,\mathrm{m}\) high chart, the image height is \[ h'=mh=(-5.03\times10^{-3})(1.20) =-6.03\times10^{-3}\,\mathrm{m}. \] The image is inverted and has height \(6.03\,\mathrm{mm}\).

Answer 3. Camera calibration and diffraction limit

[15 marks]

c) Estimate the diffraction-limited angular radius of the Airy disk and its radius on the sensor.

[3 marks]

For a circular aperture, the angular radius of the Airy disk to the first minimum is approximately \[ \theta=1.22\frac{\lambda}{D}. \] With \(\lambda=550\times10^{-9}\,\mathrm{m}\) and \(D=3.00\times10^{-3}\,\mathrm{m}\), \[ \theta=1.22\frac{550\times10^{-9}}{3.00\times10^{-3}} =2.24\times10^{-4}\,\mathrm{rad}. \] The radius of the central bright spot on the sensor is approximately \[ r=s'\theta=(0.01206)(2.24\times10^{-4}) =2.70\times10^{-6}\,\mathrm{m}. \] So the Airy radius on the sensor is \(2.70\,\mu\mathrm{m}\).

Answer 3. Camera calibration and diffraction limit

[15 marks]

d) Using the Rayleigh criterion, decide whether two fine marks \(0.800\,\mathrm{mm}\) apart on the chart can be resolved by the aperture.

[3 marks]

The Rayleigh criterion requires the angular separation to be at least about \[ \theta_R=1.22\frac{\lambda}{D}=2.24\times10^{-4}\,\mathrm{rad}. \] At an object distance of \(2.40\,\mathrm{m}\), the corresponding separation on the chart is \[ x_{\min}=s\theta_R=(2.40)(2.24\times10^{-4}) =5.37\times10^{-4}\,\mathrm{m} =0.537\,\mathrm{mm}. \] Two marks separated by \(0.800\,\mathrm{mm}\) have a larger angular separation than this, so they should be resolvable by the aperture alone.

Answer 3. Camera calibration and diffraction limit

[15 marks]

e) State how the diffraction-limited resolution changes if the aperture diameter is reduced to \(1.50\,\mathrm{mm}\).

[2 marks]

Since \(\theta_R=1.22\lambda/D\), reducing the aperture diameter increases the diffraction angle. If the diameter is halved to \(1.50\,\mathrm{mm}\), the diffraction-limited angular radius doubles and the smallest resolvable separation on the chart also doubles.

Section C: Oscillations and Collisions, Conservation and Fields

Answer 4. Floating platform oscillations

[15 marks]

A small floating service platform has vertical sides, horizontal waterline area \(3.20\,\mathrm{m^2}\), and mass \(240\,\mathrm{kg}\). It floats in fresh water of density \(1000\,\mathrm{kg\,m^{-3}}\). For small vertical displacements, neglect damping and assume the waterline area remains constant.

a) Find the platform's equilibrium draft.

[3 marks]

At equilibrium the buoyant force equals the weight: \[ \rho gAx_0=Mg. \] Therefore the equilibrium draft is \[ x_0=\frac{M}{\rho A} =\frac{240}{(1000)(3.20)} =7.50\times10^{-2}\,\mathrm{m}. \] So the platform sinks \(0.0750\,\mathrm{m}\) below the water surface at equilibrium.

Answer 4. Floating platform oscillations

[15 marks]

b) Show that a small vertical displacement gives simple harmonic motion and find the effective spring constant.

[3 marks]

If the platform is displaced downward by \(y\) from equilibrium, the extra displaced volume is \(Ay\). The extra buoyant force is upward and has magnitude \[ \rho gAy. \] Thus the restoring force is \[ F=-\rho gAy, \] which has the SHM form \(F=-ky\) with \[ k_{\mathrm{eff}}=\rho gA=(1000)(9.81)(3.20)=3.14\times10^4\,\mathrm{N\,m^{-1}}. \]

Answer 4. Floating platform oscillations

[15 marks]

c) Calculate the platform's angular frequency and frequency of vertical oscillation.

[3 marks]

The angular frequency is \[ \omega=\sqrt{\frac{k_{\mathrm{eff}}}{M}} =\sqrt{\frac{3.1392\times10^4}{240}} =11.4\,\mathrm{rad\,s^{-1}}. \] Therefore \[ f=\frac{\omega}{2\pi} =\frac{11.4}{2\pi} =1.82\,\mathrm{Hz}. \]

Answer 4. Floating platform oscillations

[15 marks]

d) A \(160\,\mathrm{kg}\) payload is placed on the platform. Find the new equilibrium draft and the new oscillation frequency.

[4 marks]

With the payload, the total mass is \[ M'=240+160=400\,\mathrm{kg}. \] The new equilibrium draft is \[ x_0'=\frac{400}{(1000)(3.20)} =0.125\,\mathrm{m}. \] The effective spring constant is unchanged because the waterline area is unchanged. The new frequency is \[ f'=\frac{1}{2\pi}\sqrt{\frac{3.1392\times10^4}{400}} =1.41\,\mathrm{Hz}. \] The added payload increases the inertia, so the oscillation frequency decreases.

Answer 4. Floating platform oscillations

[15 marks]

e) With the payload on board, the platform is pushed \(0.0400\,\mathrm{m}\) below its new equilibrium position and released from rest. Find the oscillation energy and maximum speed.

[2 marks]

For the loaded platform, \(M'=400\,\mathrm{kg}\) and \(k_{\mathrm{eff}}=3.1392\times10^4\,\mathrm{N\,m^{-1}}\). If it is released from rest at amplitude \(A_0=0.0400\,\mathrm{m}\), the mechanical energy is \[ E=\frac12k_{\mathrm{eff}}A_0^2 =\frac12(3.1392\times10^4)(0.0400)^2 =25.1\,\mathrm{J}. \] The maximum speed occurs at equilibrium: \[ v_{\max}=\omega' A_0=\sqrt{\frac{k_{\mathrm{eff}}}{M'}}A_0. \] Thus \[ v_{\max}=\sqrt{\frac{3.1392\times10^4}{400}}(0.0400) =0.354\,\mathrm{m\,s^{-1}}. \]