Physics Higher level Paper 2 Monday 3 May 2021 (afternoon) 2 hours 15 minutes Yue: Five explanation
1. Two players are playing table tennis. Player A hits the ball at a height of 0.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. The initial speed of the ball is 12.0 m s-1 horizontally. Assume that air resistance is negligible. (a) Show that the time taken for the ball to reach the surface of the table is about 0.2 s. (b) Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity vv of the ball until it reaches the table surface. Take g to be +10ms-2. [2] 0 Consider vertical motion between point 0 & P s= UH ) 1- Kath 024=0 ( t) % (9.8×1-2) p t2= ' 2¥82 0.2sec =t § Vg=ug + Agt Vg -104 ) vy=o gt Vy= gt ,tank 10 Vg = 1011-7
ball
speed of
a peak height of
m
mass of the ball
(Question 1 continued) (c) The net is stretched across the middle and the net has a height of 15.0 cm. Show that the ball will go over the net of the table. The table has a length of 2.74 m. (d) The
bounces and then reaches
0.18 m above the table with a horizontal
10.5
s-1. The
is 2.7 g. (i) Determine the kinetic energy of the ball immediately after the bounce. (ii) Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic. Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures. [3] M = en , } @ ' 24-015 ) 7- o.ie mt 315cm ¥4 V L sa = Unt + % art ← 2- 74 Mt → ¥ 4 = (2) (t) t= 2 24 ← → Sg = oxt b. ✗ 9-8×1-2 ↳ Net Middle of Sg = ¥ ✗ 9.0×229*-3-474=6-38 Mt = v06 Mt the table ↳ much over tore height of the net Abbey conservation of energy between point M & N mgh + t.mg?-o-tmvi2;7-opo✗ 10 ✗ • 24 + { Fg 6123 = { MVP • 00640+1.944 = 1- moi 1- 955 = Iz MII ANS F. At = M (DD = MC Vf Vi ) d (c) f = } ˢoog4°÷ ◦ ,8"=%ooE ¥ % = 5 67N I 5- 7 N
2. A planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given: Mass of planet = 8.0 × 1024 kg Mass of star = 3.2 × 1030 kg Distance from the star to the planet R = 4.4 × 1010 m. (a) Explain why a centripetal force is needed for the planet to be in a circular orbit. 7 24 30 ☒ Aaaaa qq.gg * %☆@•t I % ANS 2 (a) ^ Fe for circular motion LfgGMs Mp mob ( bttne STAR 82 = -5 Centre ] → circular motion involves charge in velocity → centripetal acceleration is perpendicular to the torrential velocity
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(b) Calculate the value of the centripetal force
f- = 9 Mi Mz f= 667 × 15 " " ✗ 8.0×10>4×3-2×1030 (4.4×1010) * = £67 ✗ 8.0×3.2 ✗ 10-111-241-30 4.4×4.4 ✗ 1020 = (6.67×8-0×3.2) ✗ 10-11-124 -130-20 4-4 × 4-4 = 8.81 ✗ 1033 Newton
(c) A spacecraft is to be launched from the surface of the planet to escape from the star system.The radius of the planet is (i) Show that the gravitational potential due to the planet and the star at the surface of the planet is about J kg-1. (ii) Estimate the escape speed of the spacecraft from the planet–star system. . 9 'll ✗ 103km ' 5×109 tlotential U = Ggb= 6.67×1534×8.0×1024 ° 91×103 ×103 = 5.8×107 %g Us = G- = 6-67 ✗ 10-34×3.2×1030 4^4×1610 = -4-85×109%9 Total potential V= Up -1 Us = -5 × 109 Jlkg F- TOTAL > 0 , F- K > / Ep / = mug m÷ > Hug V2 > zvg > 2×4-9×109 U > 1-2×4.9×109 = 9.91×104 m 5 '
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3. A mass of 1.0 kg of water is brought to its boiling point of 100 °C using an electric heater of power 1.6kW. (a) (i) The molar mass of water is 18 g mol-1. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas
KT mob a' 3¥ = fÑa) mass __ ¥a = [3×1.38×-10%3] 3) ✗ 6.02×10231 18×10-3 = J 3×1.38×373×6.02 18 ✗ ( 4- 720m51
(ii) State one assumption of the kinetic model of an ideal gas. (b) A mass of 0.86 kg of water remains after it has boiled for 200 s. (i) Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer. Explain why the temperature of water remains at 100 °C during this time particle can be considered points ( ullth out dimension) NO Intermolecular forces volume of the particle is negligible to the volume of the gas water vapor is ed = 1- 0.86 P = ¥ w =P Lt > M L = 1.6×103×200 = 014kg L = 1^6×103×200 = 2-3×106 %g014 ( Ii ) All the em ogg added to use to break the bond
The heater is removed and a mass of 0.30 kg of pasta at -10 °C is added to the boiling water. Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible. Specific heat capacity of pasta = 1.8 kJ kg-1 K-1 Specific heat capacity of water = 4.2 kJ kg-1 K-1 C. Heat lost by the water = Heat gained by the pasta M water swale ✓ ( 100 e.) = Mpasta Spas ta ( T C- 107 ? 086 × 42 × (100-7) = 0-30 ✗ 1.8 ✗ (7-110) 3 6 12 ✗ 100 3.612T = O ' 54T + 054 ✗ 10 361.2 5-4 = (3-6 121-054) T T= 355.8 4. 152 ʰ 86%
(ii) Calculate the power transferred by the heater when both switches are closed. P=V¥= I -611103W 1.6 × 103 ¥2 11 R= 220×220 1^6×103 = ¥¥ ! = 12¥ I 30 ohhh R Tmr nÉp' = ¥ ,-52 ( %) zaovoltLmf ↳ a = 2× 1.6× 103 w ↳ ← 220 Volt = 3.2×103 W' =3 -2kW = 3200 Watt
4. A planet orbits at a distance d from a star. The power emitted by the star is P. The total surface area of the planet is A. (a) (i) Explain why the power incident on the planet is P ×A. → PLANET ORBat Ffr surface Area of thanet A- = 47T R& fSTAR / d → )4- = TTR ' ✓ * A P= PowerPower radiated by the emitted by the = SURFACE star on unit area at Star AREA distanced is = -P of 49 DI PLANET Power incident on the planet __ ¥d ) (F)
(ii) The albedo of the planet is αp . The equilibrium surface temperature of the planet is T. Derive the expression where e is the emissivity of the planet. Permitted = e 6 AT 4 albedo = Total scattered power total incident power = Lpfrom previous tart P scattered = @ p ) ( Pine ident ? P incident =Pscattered + Permitted ¥ ) (F) µ P incident = Xp Pine ident + e • At 4 P incident ( a- Xp ) = e. 6 AT 4 74 = P incident ( a- Xp ) e. 6 A P A- ( e xp ) _ Proved 1- = 4 Fadi ETA
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(b) On average, the Moon is the same distance from the Sun as the Earth. The Moon can be assumed to have an emissivity e = 1 and an albedo αM = 0.13. The solar constant is 1.36 × 103 W m-2. Calculate the surface temperature of the Moon.
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5. A device sends an impulse of electrical energy to maintain a regular heartbeat in a person. The device is powered by an alternating current (ac) supply connected to a step-up transformer that charges a capacitor of capacitance 30μF. (a) Explain the role of the diode in the circuit when the switch is at position A. [2] a "° " ↑ ✗ OUTPUT 11↑ N N DIODE WILL ACT AS A HALF WAVE RECTIFIER
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(b) The voltage across the primary coil of the transformer is 220 V. The number of turns on the secondary coil is 15 times greater than the number of turns on the primary coil. (i) Show that the maximum energy stored by the capacitor is about 160 J.
Up = 220 Volt Ha = N Hz = 15N N=%15¥ = ¥-0 <= 30µF V2 = (5) ( 220) Volt U= Iz c. v2 = ¥ (30×10-6) (15 × 220) ≥ = 163 35 I I 160J
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The switch is moved to position B.
Describe what happens to the energy stored in the capacitor when the switch is moved to position B.
Il)Calculate the maximum charge Q stored in the capacitor. III) Identify, using the label + on the diagram, the polarity of the capacitor. (c)
0 Go = CV = 30×10-6×3300 = • 099 & Lower plate is positive ↳ the energy stored in tire capacitor transferred to the load ( heart]
Show that the charge remaining in the capacitor after a time equal to one time constant τ of the circuit will be 0.37 Q0. The graph shows the variation with time of the charge in the capacitor as it is being discharged through the heart. Determine the electrical resistance of the closed circuit with the switch in position B. *☒ & D= Qo eHe t=Z ② = Go e% = % = 0368 Qo = 037 Qo iii) 0-37 ← 1^6×1.0-3 See = RE R = { = 1.6×10-3 30×10--6 = 53 -0hm
Explain the effect of the electrode capacitance on the discharge time.
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(d) In practice, two electrodes connect the heart to the circuit. These electrodes introduce an additional capacitance.
I!= electrode makes a capacitor and this 1- Capacitor is connected parallel to the previous capacitor hence total capacitor increases there f- we E- RE Increases hence discharging time increases
6. A painting is protected behind a transparent glass sheet of refractive index nglass. A coating of thickness w is added to the glass sheet to reduce reflection. The refractive index of the coating ncoating is such that nglass > ncoating > 1. The diagram illustrates rays normally incident on the coating. Incident angles on the diagram are drawn away from the normal for clarity. State the phase change when a ray is reflected at B. B) Explain the condition for w that eliminates reflection for a particular light wavelength in air λair. a.) Phase change at the denser medium = 180° C. boundary ) M= tour ( Mtk ) ✗ = 2W ( path difference ? ✗ coating + coating ✗ coating -_×¥ w=(m+±)¥=(m+±)×aˢ 2M coating
State the Rayleigh criterion for resolution The painting contains a pattern of red dots with a spacing of 3 mm. Assume the wavelength of red light is 700 nm. The average diameter of the pupil of a human eye is 4 mm. Calculate the maximum possible distance at which these red dots are distinguished. ② (C) ↳ central maxima of one diffraction pattern lies over the central/first minima of other diffraction pattern c Iii) D= 112b¥ & 01 for circular aperture • * D= 1.22×700×159 4 × 10-3 = 214×10-4 radian , @ = D= g- = 3×10-3 2-14×10--4 = 14Mt
qz V23 8- -1h > 34 + , He 't 90 2 06 20 PbTl → • z 1a81 to measure Radioactive decay for either
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High temp = more KE , more speed ↳ energy required to overcome electrostatic repulsion
energy Is released when binding enemy bet nucleon increases 144 q , V > 35 Kyle + Ba 1- Q I 1↓ 89× 8 6. trev 144 × 8 2. Her 235×76 MeV Q = ( (23530-6) ( 8 (8-6) (144 × 8-2) ] = 160 Mev
8. On a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger. (a) Outline how a standing wave is produced on the string. [2] b) The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm. (i) Show that the speed of the wave on the string is about 240 m s-1. Travelling wave moves along the length of the string and reflects at fired end And because of the superposition of Incident hid reflected wave , the standing wave is produced N H / A / Amplitude = 046M , f , = 195 Hz , L = 062Mt ← L = % L = ✗ = 2 ( L) = 2 ( 0627 = 1.24 Mt D= f- ✗ = (1-95) (1-24)=242 Mlseo
in m s-1, the maximum velocity of vibration of point P when it is vibrating with a frequency of 195Hz.
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Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position. Calculate,
Iii ) For SHM A ✗ x dii ) V= hw cos wt Umar = hw = h ( 29 f) = • 004×2×3.14 ✗ 195 = 4.9 M 5-
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(iv) Calculate, in terms of g, the maximum acceleration of P.
Max Acceleration = WZR = ( 29 f) > ( %) = (27 × 1955 × 1.004) = 6004 ME = 6%-008 = 600g F- = 1- m what f- ✗ set , X ✗ JET ÷ ñ=-= FX ( o -4cm ) = 06cm
The string is made to vibrate in its third harmonic. State the distance between consecutive nodes.
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② a * ← % → ← ✗a → ← % 1L = ¥ ¥ = § = 6÷m = 21 cm
9. Conservation of energy and conservation of momentum are two examples of conservation laws. (a) Outline the significance of conservation laws for physics. (b) When a pi meson π- (du ) and a proton (uud) collide, a possible outcome is a sigma baryon Σ0 (uds) and a kaon meson Κ0 (ds ). Apply three conservation laws to show that this interaction is possible. → Both express to in cities of nature → feat = d÷ , If fexternae = 0 , b= contrail → ( TE ) = ( KE) + ( P E) = Constant , can aptly to two variable position and fend out unknown quantity •• qq0 DT + UUD → Ud Stds Using idea that particle and antiparticle has opposite number 2d tñtw + U →2d -1k -1s -15 u=U Conservation : charge ( Yes ) 0=0 baryon number Yes 1-3=1-3 strangeness → Yes
photoelectric effect
a phenomenon
the surface of a metal when light is incident on it. These ejected
ejected
are called photoelectrons.
the surface of
is important to
on the frequency of the light
that the emission of photoelectrons
the kinetic energy of the ejected photoelectrons
metal’s
The process through
photoelectrons
10. In an electric circuit used to investigate the photoelectric effect, the voltage is varied until the reading in the ammeter is zero. The stopping voltage that produces this reading is 1.40 V. Describe the photoelectric effect. Show that the maximum velocity of the photoelectrons is 700kms-1. The
is
in which electrons are ejected from
electrons
It
note
and
is dependent
that is incident on the
surface.
which
are
from
the metal due to the action of light is commonly referred to as photoemission. a.) b) e. Vo = 1- Monnaie " 2×1.6 ✗ 10-15×1^47Phrase = 2 e V0 a= 9. 1 ✗ 10-3 I ≥ Too kmtseo
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·
and
F- = Wo + e. V0 1- = Wo + evo h÷ = 2- 3eV -11^40 EV = 3- Toei ✗ = 3- 70×1.6×10-19 6- 6 ✗ 10-34×3×108 = 3-4×10-7 Mt
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