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2023 · H1 Physics · Paper 1

2023 H1 Physics Paper 1, worked

30 multiple-choice questions · 30 marks

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Q1Answer C
Power is energy per unit time. - J s−1^{-1} and kJ min−1^{-1} are energy divided by time, so they are units of power. - kW is a unit of power. - kW h is a power multiplied by a time, so it is a unit of energy, not power.
Q2Answer A
The volume of the stone is a difference, V=Vw+s−VwV = V_{w + s} - V_{w}. For a sum or a difference, the absolute uncertainties add (they never subtract):ΔV=ΔVw+s+ΔVw\Delta V = \Delta V_{w + s} + \Delta V_{w}Percentage uncertainties are added only for products and quotients.
Q3Answer C
Speed is a scalar, so it has magnitude only. Acceleration, displacement and velocity are vectors, which have direction.
Q4Answer D
For constant acceleration,v2=u2+2asv^{2} = u^{2} + 2asCompare with y=c+mxy = c + mx, where y=v2y = v^{2} and x=sx = s: the intercept is u2u^{2} and the gradient is 2a2a.
Q5Answer D
The aeroplane starts from rest:s=12at2⇒a=2st2=2(560)352=0.91  m s-2s = \frac{1}{2}at^{2}\quad \Rightarrow \quad a = \frac{2s}{t^{2}} = \frac{2(560)}{35^{2}} = 0.91\text{\: m }\text{s}^{\text{-2}}v=at=(0.914)(35)=32  m s-1v = at = (0.914)(35) = 32\text{\: m }\text{s}^{\text{-1}}
Q6Answer C
For isolated objects, momentum is conserved in every collision, and total energy is always conserved. In an inelastic collision, some kinetic energy is converted into other forms (heat, sound, deformation), so kinetic energy is not conserved.
Q7Answer D
Speed just before impact (falling 3.0 m from rest):v1=2gh1=2(9.81)(3.0)=7.67  m s-1(downwards)v_{1} = \sqrt{2gh_{1}} = \sqrt{2(9.81)(3.0)} = 7.67\text{\: m }\text{s}^{\text{-1}} \text{(downwards)}Speed just after the bounce (rising 2.0 m):v2=2gh2=2(9.81)(2.0)=6.26  m s-1(upwards)v_{2} = \sqrt{2gh_{2}} = \sqrt{2(9.81)(2.0)} = 6.26\text{\: m }\text{s}^{\text{-1}} \text{(upwards)}Taking upwards as positive, the change in momentum isΔp=(0.050)(6.26)−(0.050)(−7.67)=0.697  kg m s-1\Delta p = (0.050)(6.26) - (0.050)( - 7.67) = 0.697\text{\: kg m }\text{s}^{\text{-1}}F=ΔpΔt=0.69710×10−3=70  NF = \frac{ \Delta p }{ \Delta t } = \frac{0.697}{10 \times 10^{- 3}} = 70\text{\: N}
Q8Answer B
The object is in equilibrium, so WW, FF and TT form a closed triangle with arrows head to tail. - WW acts vertically downwards and FF horizontally to the right. - TT acts along the string, upwards and to the left, at 30∘^{\circ} to the vertical. Because TT is at 30∘^{\circ} to the vertical, F=Wtan⁡30∘=0.58WF = W\tan 30{^\circ} = 0.58W, so the horizontal side of the triangle is shorter than the vertical side. In B, WW points down, FF continues to the right from its head, and TT closes the triangle back up and to the left at 30∘^{\circ} to the vertical. In D the arrows are head to tail, but TT is at about 60∘^{\circ} to the vertical. In A and C, TT points downwards.
Q9Answer C
From the graph, the length increases from 5 cm to 30 cm when 400 g is added:x=30−5=25  cm=0.25  m,F=(0.400)(9.81)=3.92  Nx = 30 - 5 = 25\text{\: cm} = 0.25\text{\: m},\quad\quad F = (0.400)(9.81) = 3.92\text{\: N}k=Fx=3.920.25=16  N m-1k = \frac{F}{x} = \frac{3.92}{0.25} = 16\text{\: N }\text{m}^{\text{-1}}
Q10Answer A
Take moments about the hinge at the wall. The bar meets the shelf 20 cm from the wall, at 40∘^{\circ} to the horizontal. Only the vertical component of its push has a moment about the hinge:Fv=50sin⁡40∘=32.1  NF_{\text{v}} = 50\sin 40{^\circ} = 32.1\text{\: N}The weight of the shelf (10 N) acts at its centre, 15 cm from the wall. The load LL is at X, 25 cm from the wall.(32.1)(20)=(10)(15)+L(25)(32.1)(20) = (10)(15) + L(25)L=643−15025=20  NL = \frac{643 - 150}{25} = 20\text{\: N}
Q11Answer D
The object falls from rest with constant acceleration, so the distance fallen is proportional to t2t^{2}. At time t/2t/2 it has fallen 14\frac{1}{4} of the height hh. Its height above the ground is then 34h\frac{3}{4}h:P.E.=mg(3h4)=3E4,K.E.=E−3E4=E4\text{P.E.} = mg\left( \frac{3h}{4} \right) = \frac{3E}{4},\quad\quad\text{K.E.} = E - \frac{3E}{4} = \frac{E}{4}
Q12Answer A
An angle in radians is arc length divided by radius:θ=sr\theta = \frac{s}{r}
Q13Answer B
The centripetal force is provided by gravity:F=GMmr2F = \frac{GMm}{r^{2}}F′=G(3M)(2m)(4r)2=616GMmr2=0.375FF\prime = \frac{G(3M)(2m)}{(4r)^{2}} = \frac{6}{16} \frac{GMm}{r^{2}} = 0.375F
Q14Answer D
The rock stays above the same point, so its period equals the length of the Martian day, T=8.86×104T = 8.86 \times 10^{4} s. Its orbital radius isr=3.40×106+1.72×107=2.06×107  mr = 3.40 \times 10^{6} + 1.72 \times 10^{7} = 2.06 \times 10^{7}\text{\: m}Gravity provides the centripetal force:GMr2=ω2r⇒GM=ω2r3=(2π8.86×104)2(2.06×107)3=4.40×1013  N m2 kg-1\frac{GM}{r^{2}} = \omega^{2}r\quad \Rightarrow \quad GM = \omega^{2}r^{3} = \left( \frac{2\pi}{8.86 \times 10^{4}} \right)^{2}(2.06 \times 10^{7})^{3} = 4.40 \times 10^{13}\text{\: N }\text{m}^{\text{2}}\text{ k}\text{g}^{\text{-1}}At the surface (R=3.40×106R = 3.40 \times 10^{6} m):g=GMR2=4.40×1013(3.40×106)2=3.80  m s-2g = \frac{GM}{R^{2}} = \frac{4.40 \times 10^{13}}{(3.40 \times 10^{6})^{2}} = 3.80\text{\: m }\text{s}^{\text{-2}}
Q15Answer D
The satellite's direction of motion changes continuously, so its velocity is changing and it is accelerating. There is a resultant force on it (the gravitational force, towards the Earth), so it is not in equilibrium. The gravitational force is the centripetal force. It is not balanced by a separate centripetal force, so B is wrong.
Q16Answer B
F=mv2r⇒v=Frm=(2.3×10−8)(1.1×10−10)9.11×10−31=1.7×106  m s-1F = \frac{mv^{2}}{r}\quad \Rightarrow \quad v = \sqrt{\frac{Fr}{m}} = \sqrt{\frac{(2.3 \times 10^{- 8})(1.1 \times 10^{- 10})}{9.11 \times 10^{- 31}}} = 1.7 \times 10^{6}\text{\: m }\text{s}^{\text{-1}}
Q17Answer D
R=ρlπr2⇒R∝lr2R = \frac{ \rho l }{\pi r^{2}}\quad \Rightarrow \quad R \propto \frac{l}{r^{2}}- A: 1/(1/2)2=41/(1/2)^{2} = 4, so 4R4R - B: 1/22=1/41/2^{2} = 1/4, so R/4R/4 - C: 2/(1/2)2=82/(1/2)^{2} = 8, so 8R8R - D: 2/22=1/22/2^{2} = 1/2, so R/2R/2
Q18Answer B
The charge passing per second is 43 C. Each electron carries 1.60×10−191.60 \times 10^{- 19} C:N=431.60×10−19=2.7×1020N = \frac{43}{1.60 \times 10^{- 19}} = 2.7 \times 10^{20}The 12 V is not needed.
Q19Answer A
As RR decreases, the total resistance decreases, so the current increases: the ammeter reading increases. The voltmeter measures the p.d. across RR, which is the terminal p.d., E−IrE - Ir. A larger current means larger lost volts IrIr, so the voltmeter reading decreases.
Q20Answer C
To vary the p.d. across the lamp all the way from 0 to EE, the circuit must be a potential divider: a potentiometer connected across the whole battery, with the lamp connected between one end and the sliding contact. In C, with the slider at one end the lamp has the full EE across it, and with the slider at the other end it has zero. In A and B a fixed resistor is always in series with part of the circuit, so the lamp can never reach EE (or zero). D is a variable resistor in series with the lamp, which can reduce the p.d. but never to zero.
Q21Answer B
The terminal p.d. is 12 V, which is half the e.m.f., so the lost volts across the internal resistance are also 12 V. The external resistance must equal the internal resistance, 6.0 Ω\Omega. - Two 4.0 Ω\Omega resistors give either 8.0 Ω\Omega (series) or 2.0 Ω\Omega (parallel). - Three resistors: one in series with two in parallel gives 4.0+2.0=6.04.0 + 2.0 = 6.0 Ω\Omega. The least number is 3.
Q22Answer A
From the graph, at 50 ∘^{\circ}C the thermistor's resistance is 1.0 kΩ\Omega. The output VV is the p.d. across the thermistor. The 20.0 V is shared in the ratio of the resistances:V=1.01.0+5.00(20.0)=3.3  VV = \frac{1.0}{1.0 + 5.00}(20.0) = 3.3\text{\: V}
Q23Answer D
An α-particle has charge 2e=3.20×10−192e = 3.20 \times 10^{- 19} C.E=Fq=353.20×10−19=1.1×1020  N C-1E = \frac{F}{q} = \frac{35}{3.20 \times 10^{- 19}} = 1.1 \times 10^{20}\text{\: N }\text{C}^{\text{-1}}The gold nucleus is positive, so its field points away from it, to the right at X.
Q24Answer D
A magnetic field exerts a force only on a moving charge (F=Bqvsin⁡θF = Bqv\sin \theta). The charge is stationary, so the force is zero.
Q25Answer B
Parallel wires with currents in opposite directions repel. Y is pushed away from X, so the force on Y is to the right (direction B).
Q26Answer B
The force on the electron is perpendicular to its velocity. - A magnetic force is perpendicular to both the velocity and the field. A field perpendicular to the velocity and to FF gives exactly this (B). A field along FF (A) would give a force perpendicular to FF, not along it. - An electric force on an electron is along the field line (opposite to the field). For C, the force would be perpendicular to FF. For D, the force would be along the velocity, not at right angles to it.
Q27Answer B
In every valid equation, both the nucleon numbers and the proton numbers must balance. - A: 210=206+4210 = 206 + 4, 83=81+283 = 81 + 2 ✓ - B: 208=208+0208 = 208 + 0 ✓, but 82≠81+(−1)=8082 \neq 81 + ( - 1) = 80 ✗ - C: 212=208+4212 = 208 + 4, 84=82+284 = 82 + 2 ✓ - D: 208=208+0208 = 208 + 0, 81=82+(−1)81 = 82 + ( - 1) ✓ (β−^{-}-decay increases the proton number by 1, so lead cannot become thallium this way.)
Q28Answer B
An α-particle has 2 protons and 2 neutrons.Δm=2(1.6726×10−27)+2(1.6749×10−27)−6.6442×10−27\Delta m = 2(1.6726 \times 10^{- 27}) + 2(1.6749 \times 10^{- 27}) - 6.6442 \times 10^{- 27}Δm=6.6950×10−27−6.6442×10−27=5.080×10−29  kg\Delta m = 6.6950 \times 10^{- 27} - 6.6442 \times 10^{- 27} = 5.080 \times 10^{- 29}\text{\: kg}
Q29Answer B
Only a very massive, concentrated positive charge can turn an α-particle back through more than 90∘^{\circ}, and the fact that only a small number did so shows that the nucleus is very small compared with the atom. A shows that most of the atom is empty space, but not that there is a small massive nucleus. C and D are features of how the experiment is carried out.
Q30Answer A
Subtract the background count of 50 min−1^{-1} to get the count rate due to the sample. - At t=0t = 0: 250−50=200250 - 50 = 200 min−1^{-1}. - Half of this is 100 min−1^{-1}, which is a measured count of 100+50=150100 + 50 = 150 min−1^{-1}. From the graph this occurs at t=50t = 50 s. (Check: a quarter, 50 min−1^{-1}, is a measured 100 min−1^{-1} at t=100t = 100 s, one more half-life.)

No question text is reproduced anywhere in this section. The A-Level H1 Physics papers are © UCLES/SEAB — refer to the official papers for the wording. The worked solutions, diagrams and interactive models are original work by The Learning Cave.