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Paper 1 Answers

Time2 hours

Marks60

SetSet 1

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

\( g=9.81\,\mathrm{m\,s^{-2}} \)

Section A: Mechanics

Answer 1. Package drop, stopping work, and impulse

[15 marks]

A delivery drone travels horizontally at \(11.0\,\mathrm{m\,s^{-1}}\) when it releases a \(0.750\,\mathrm{kg}\) instrument package. The release point is \(28.0\,\mathrm{m}\) above level ground. Air resistance is negligible until the package reaches a foam capture pad.

a) Find the time taken to reach the capture pad and the horizontal distance from the release point to the landing point.

[4 marks]

Take upward as positive. The package has no initial vertical velocity, so its vertical position relative to the ground is \[ y=28.0-\frac12gt^2. \] At the ground, \(y=0\), giving \[ t=\sqrt{\frac{2(28.0)}{9.81}}=2.39\,\mathrm{s}. \] The horizontal velocity is constant while the package is in flight, so \[ x=v_xt=(11.0)(2.39)=26.3\,\mathrm{m}. \]

Answer 1. Package drop, stopping work, and impulse

[15 marks]

b) Find the package velocity components immediately before it reaches the pad. Hence find its speed and direction.

[4 marks]

The horizontal component remains \[ v_x=11.0\,\mathrm{m\,s^{-1}}. \] The vertical component just before the pad is \[ v_y=-gt=-(9.81)(2.39)=-23.4\,\mathrm{m\,s^{-1}}. \] Hence the speed is \[ v=\sqrt{11.0^2+23.4^2}=25.8\,\mathrm{m\,s^{-1}}. \] The angle below the horizontal is \[ \theta=\tan^{-1}\left(\frac{23.4}{11.0}\right)=64.8^\circ. \]

Answer 1. Package drop, stopping work, and impulse

[15 marks]

c) The foam brings the package to rest over a distance of \(0.320\,\mathrm{m}\), measured approximately along the incoming path. Estimate the average resisting force exerted by the foam, ignoring the weight impulse during this short stopping interval.

[4 marks]

The kinetic energy just before capture is \[ K=\frac12mv^2=\frac12(0.750)(25.8)^2=2.50\times10^2\,\mathrm{J}. \] During the short stopping distance, ignore the small change in gravitational potential energy compared with the work done by the pad. The average resisting force magnitude \(F\) then satisfies \[ -Fs=0-K. \] Thus \[ F=\frac{K}{s}=\frac{2.50\times10^2}{0.320}=7.80\times10^2\,\mathrm{N}. \] The force acts approximately opposite the incoming velocity.

Answer 1. Package drop, stopping work, and impulse

[15 marks]

d) Estimate the impulse exerted by the capture pad on the package. Give the vector impulse using \(+\hat{\imath}\) horizontally forward and \(+\hat{\jmath}\) upward.

[3 marks]

Ignoring the weight impulse during the short capture, the impulse from the pad is the change in momentum: \[ \vec J=m(\vec v_f-\vec v_i). \] With \(\vec v_f=0\) and \(\vec v_i=11.0\hat{\imath}-23.4\hat{\jmath}\), \[ \vec J=0.750(-11.0\hat{\imath}+23.4\hat{\jmath}) =(-8.25\hat{\imath}+17.6\hat{\jmath})\,\mathrm{N\,s}. \] Its magnitude is \[ |\vec J|=0.750(25.8)=19.4\,\mathrm{N\,s}. \] The impulse points opposite the package's incoming momentum.

Answer 2. Torque, rotational work, and stopping

[15 marks]

A calibration flywheel has moment of inertia \(0.620\,\mathrm{kg\,m^2}\) about its axle. A light cord is wrapped around a drum of radius \(0.180\,\mathrm{m}\) fixed to the flywheel. A technician pulls the free end of the cord with a constant force of \(24.0\,\mathrm{N}\). A bearing friction torque of magnitude \(1.10\,\mathrm{N\,m}\) opposes the rotation.

a) Find the angular acceleration of the flywheel while the cord is being pulled, and find the linear acceleration of the cord.

[5 marks]

The pull gives a driving torque \[ \tau_{\mathrm{pull}}=RF=(0.180)(24.0)=4.32\,\mathrm{N\,m}. \] The net torque while the cord is being pulled is \[ \tau_{\mathrm{net}}=4.32-1.10=3.22\,\mathrm{N\,m}. \] Using \(\sum\tau=I\alpha\), \[ \alpha=\frac{3.22}{0.620}=5.19\,\mathrm{rad\,s^{-2}}. \] Since the cord unwinds without slipping, its linear acceleration is \[ a=R\alpha=(0.180)(5.19)=0.934\,\mathrm{m\,s^{-2}}. \]

Answer 2. Torque, rotational work, and stopping

[15 marks]

b) Starting from rest, the flywheel is pulled until \(2.40\,\mathrm{m}\) of cord has unwound. Use rotational work and energy to find the angular speed at that instant.

[4 marks]

The cord length unwound is related to angular displacement by \[ \theta=\frac{s}{R}=\frac{2.40}{0.180}=13.3\,\mathrm{rad}. \] The net rotational work is \[ W_{\mathrm{net}}=\tau_{\mathrm{net}}\theta=(3.22)(13.3)=42.9\,\mathrm{J}. \] This becomes rotational kinetic energy: \[ W_{\mathrm{net}}=\frac12I\omega^2. \] Therefore \[ \omega=\sqrt{\frac{2W_{\mathrm{net}}}{I}} =\sqrt{\frac{2(42.9)}{0.620}} =11.8\,\mathrm{rad\,s^{-1}}. \]

Answer 2. Torque, rotational work, and stopping

[15 marks]

c) After the \(2.40\,\mathrm{m}\) of cord has unwound, the cord leaves the drum and the flywheel slows under the same bearing friction torque. Find the time taken to stop and the additional angle turned.

[4 marks]

After the cord leaves, the only torque is the resisting bearing torque. The angular deceleration magnitude is \[ \alpha_f=\frac{1.10}{0.620}=1.77\,\mathrm{rad\,s^{-2}}. \] The stopping time is \[ t=\frac{\omega_0}{\alpha_f}=\frac{11.8}{1.77}=6.65\,\mathrm{s}. \] The additional angle turned before stopping is \[ \Delta\theta=\frac{\omega_0^2}{2\alpha_f} =\frac{(11.8)^2}{2(1.77)} =39.0\,\mathrm{rad}. \]

Answer 2. Torque, rotational work, and stopping

[15 marks]

d) Explain why increasing the drum radius would increase the driving torque but would not increase the work done by the pull for the same pulled cord length.

[2 marks]

For a fixed pull force and fixed cord length, the work done by the technician is \(W=Fs\), so the driving work does not depend on drum radius. A larger radius gives a larger torque \(RF\), but the same cord length corresponds to a smaller angular displacement \(s/R\). These two effects cancel for the work done by the pull. The resisting bearing work would not cancel in the same way, because it depends on the angular displacement.

Section B: Waves and Optics

Answer 3. Standing waves and grating calibration

[15 marks]

A laboratory rig uses a stretched string and a diffraction grating in the same alignment test. The string has length \(0.640\,\mathrm{m}\), is fixed at both ends, and has linear density \(2.40\times10^{-3}\,\mathrm{kg\,m^{-1}}\). Its fundamental resonance is \(220\,\mathrm{Hz}\). A laser of wavelength \(532\,\mathrm{nm}\) is incident normally on a grating with \(500\,\mathrm{lines\,mm^{-1}}\). A screen is \(1.80\,\mathrm{m}\) from the grating.

a) Derive the relation between the fundamental frequency and the wave speed for this string, then calculate the wave speed.

[3 marks]

For the fundamental standing wave on a string fixed at both ends, \[ L=\frac{\lambda_1}{2}, \] so \(\lambda_1=2L\). The wave speed is therefore \[ v_w=f_1\lambda_1=2Lf_1. \] With the given values, \[ v_w=2(0.640)(220)=2.82\times10^2\,\mathrm{m\,s^{-1}}. \]

Answer 3. Standing waves and grating calibration

[15 marks]

b) Calculate the tension in the string.

[3 marks]

For a transverse wave on a stretched string, \[ v_w=\sqrt{\frac{T}{\mu}}. \] Hence \[ T=\mu v_w^2. \] Substituting, \[ T=(2.40\times10^{-3})(281.6)^2 =1.90\times10^2\,\mathrm{N}. \]

Answer 3. Standing waves and grating calibration

[15 marks]

c) Find the third-harmonic frequency and the node positions along the string.

[3 marks]

For a string fixed at both ends, the harmonic frequencies are \[ f_n=nf_1. \] For \(n=3\), \[ f_3=3(220)=660\,\mathrm{Hz}. \] The third harmonic has three half-wavelengths along the string, so adjacent nodes are separated by \(L/3\). The node positions are \[ x=0,\quad \frac{L}{3},\quad \frac{2L}{3},\quad L, \] which gives \[ x=0,\ 0.213\,\ 0.427\,\ 0.640\,\mathrm{m}. \]

Answer 3. Standing waves and grating calibration

[15 marks]

d) Find the grating spacing, the first-order diffraction angle, and the distance on the screen from the central maximum to the first-order maximum.

[4 marks]

The grating spacing is the reciprocal of the line density: \[ d=\frac{1}{500\times10^3}=2.00\times10^{-6}\,\mathrm{m}. \] For the first order, the grating equation gives \[ d\sin\theta_1=\lambda. \] Thus \[ \sin\theta_1=\frac{532\times10^{-9}}{2.00\times10^{-6}}=0.266, \] so \[ \theta_1=15.4^\circ. \] The screen displacement from the central maximum is \[ y_1=(1.80)\tan15.4^\circ=0.496\,\mathrm{m}. \]

Answer 3. Standing waves and grating calibration

[15 marks]

e) Find the highest possible diffraction order and explain why the next order is impossible.

[2 marks]

Allowed orders must satisfy \[ |m|\lambda\le d. \] Therefore \[ m_{\max}=\left\lfloor\frac{d}{\lambda}\right\rfloor =\left\lfloor\frac{2.00\times10^{-6}}{532\times10^{-9}}\right\rfloor=3. \] The \(m=4\) order is impossible because it would require \(\sin\theta>1\).

Section C: Oscillations and Collisions, Conservation and Fields

Answer 4. Damped sensor motion and floating equilibrium

[15 marks]

A floating environmental buoy contains a vertical spring-mass accelerometer. The oscillating sensor mass is \(0.450\,\mathrm{kg}\), the spring constant is \(18.0\,\mathrm{N\,m^{-1}}\), and the linear damping coefficient is \(0.180\,\mathrm{kg\,s^{-1}}\). After a wave passes, the sensor is displaced by \(0.0600\,\mathrm{m}\) from equilibrium and released from rest. The complete buoy, including the sensor, has mass \(3.65\,\mathrm{kg}\) and a vertical cylindrical hull of cross-sectional area \(3.00\times10^{-2}\,\mathrm{m^2}\). The damping rate is \[ \beta=\frac{b}{2m} \] and the damped angular frequency is \[ \omega_d=\sqrt{\omega_0^2-\beta^2}. \] For underdamped motion, the amplitude envelope is \[ A(t)=A_0e^{-\beta t}. \]

a) Calculate the undamped angular frequency and the damping rate. Decide whether the motion is underdamped.

[4 marks]

The undamped angular frequency is \[ \omega_0=\sqrt{\frac{k}{m}} =\sqrt{\frac{18.0}{0.450}} =6.32\,\mathrm{rad\,s^{-1}}. \] The damping rate is \[ \beta=\frac{b}{2m}=\frac{0.180}{2(0.450)}=0.200\,\mathrm{s^{-1}}. \] Since \(\beta<\omega_0\), the motion is underdamped.

Answer 4. Damped sensor motion and floating equilibrium

[15 marks]

b) Find the damped period and write the amplitude envelope for the motion.

[3 marks]

For underdamped motion, \[ \omega_d=\sqrt{\omega_0^2-\beta^2}. \] Thus \[ \omega_d=\sqrt{(6.32)^2-(0.200)^2} =6.32\,\mathrm{rad\,s^{-1}} \] to three significant figures. The damped period is \[ T_d=\frac{2\pi}{\omega_d}=\frac{2\pi}{6.32}=0.994\,\mathrm{s}. \] The amplitude envelope is \[ A(t)=A_0e^{-\beta t}. \]

Answer 4. Damped sensor motion and floating equilibrium

[15 marks]

c) Estimate the amplitude after three damped periods and the mechanical energy lost over those three periods.

[4 marks]

After three damped periods, \[ t=3T_d=3(0.994)=2.98\,\mathrm{s}. \] The amplitude is \[ A=0.0600e^{-(0.200)(2.98)}=0.0331\,\mathrm{m}. \] The mechanical energy scale at amplitude \(A\) is \[ E=\frac12kA^2. \] Initially, \[ E_0=\frac12(18.0)(0.0600)^2=3.24\times10^{-2}\,\mathrm{J}. \] After three cycles, \[ E_3=\frac12(18.0)(0.0331)^2=9.86\times10^{-3}\,\mathrm{J}. \] So the energy lost is \[ E_0-E_3=2.25\times10^{-2}\,\mathrm{J}. \]

Answer 4. Damped sensor motion and floating equilibrium

[15 marks]

d) In fresh water of density \(1000\,\mathrm{kg\,m^{-3}}\), find the depth of the cylindrical hull below the waterline.

[3 marks]

For a floating body, the buoyant force balances the weight: \[ \rho gAh=mg. \] Canceling \(g\), \[ h=\frac{m}{\rho A}. \] In fresh water, \[ h=\frac{3.65}{(1000)(3.00\times10^{-2})} =0.122\,\mathrm{m}. \]

Answer 4. Damped sensor motion and floating equilibrium

[15 marks]

e) State how the immersion depth changes if the buoy is placed in seawater of density \(1025\,\mathrm{kg\,m^{-3}}\).

[1 mark]

In denser water, less volume must be displaced to support the same weight. In seawater of density \(1025\,\mathrm{kg\,m^{-3}}\), the depth would be \[ h=\frac{3.65}{(1025)(3.00\times10^{-2})}=0.119\,\mathrm{m}, \] so the buoy would sit slightly higher.