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2024 · H1 Physics · Paper 1

2024 H1 Physics Paper 1, worked

30 multiple-choice questions · 30 marks

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Q1Answer B
The prefix deci (d) means 10−110^{- 1}:1  dm=0.1  m=100  mm=1×102  mm1\text{\: dm} = 0.1\text{\: m} = 100\text{\: mm} = 1 \times 10^{2}\text{\: mm}
Q2Answer C
Both velocities make 30∘^{\circ} with the wall, so their components perpendicular to the wall are vsin⁡30∘v\sin 30{^\circ}. Take the direction away from the wall as positive:u⊥=−4.0sin⁡30∘=−2.0  m s-1,v⊥=+3.0sin⁡30∘=+1.5  m s-1u_{\bot} = - 4.0\sin 30{^\circ} = - 2.0\text{\: m }\text{s}^{\text{-1}},\quad\quad v_{\bot} = + 3.0\sin 30{^\circ} = + 1.5\text{\: m }\text{s}^{\text{-1}}Δp⊥=m(v⊥−u⊥)=(1.0)(1.5−(−2.0))=3.5  kg m s-1\Delta p_{\bot} = m(v_{\bot} - u_{\bot}) = (1.0)(1.5 - ( - 2.0)) = 3.5\text{\: kg m }\text{s}^{\text{-1}}
Q3Answer B
Air resistance depends on the speed of the ball, and the speed changes throughout the flight. So the air resistance is not constant in either direction (A and C are wrong). Vertically, the resultant force is the weight plus the changing vertical component of air resistance, so the vertical acceleration is not uniform. Horizontally, the only force is the changing horizontal component of air resistance, so the horizontal acceleration is not uniform either (D is wrong).
Q4Answer B
For constant acceleration from rest: - aa is constant: a horizontal line. - v=atv = at: a straight line through the origin. - s=12at2s = \frac{1}{2}at^{2}: a parabola through the origin, curving upwards. B shows constant aa with a parabolic ss. A pairs constant velocity with a curved ss, C pairs a curved vv with constant aa, and D pairs an increasing aa with a straight-line vv. These pairs contradict each other.
Q5Answer B
Speed is the gradient of the distance–time graph. - A: between 15 s and 19 s the graph rises by about 180 m, a gradient of about 45 m s−1^{-1}, so the maximum speed is well above 10 m s−1^{-1}. - B: the car starts from rest (speed 0) and later reaches about 45 m s−1^{-1}. Speed changes continuously, so at some instant it is exactly 20 m s−1^{-1}. Correct. - C: the first 200 m takes about 14.5 s, an average speed of about 14 m s−1^{-1}. - D: from 20 s to 25 s the distance increases by about 60 m (12 m s−1^{-1}); from 15 s to 20 s it increases by about 220 m (44 m s−1^{-1}). D is wrong.
Q6Answer A
- While PP is less than the maximum friction, friction equals PP, so the resultant force is zero and a=0a = 0. The graph runs along the PP-axis. - Once PP exceeds the maximum friction fmax⁡f_{\max }, the resultant force is P−fmax⁡P - f_{\max }, soa=P−fmax⁡ma = \frac{P - f_{\max }}{m}This is a straight line with gradient 1/m1/m, starting from the PP-axis at P=fmax⁡P = f_{\max }. Graph A shows this.
Q7Answer D
The two forces act on the same object and cancel, so there is no resultant force. By Newton's first law, the object continues at constant velocity. The first law is used. Newton's third law describes forces that two bodies exert on each other, which act on different bodies. These two forces act on the same object, so they are not a third-law pair. The third law is not used.
Q8Answer B
Take right as positive. For an elastic collision, the relative speed of approach equals the relative speed of separation:5−(−2)=v2−v1⇒v2−v1=7  m s-15 - ( - 2) = v_{2} - v_{1}\quad \Rightarrow \quad v_{2} - v_{1} = 7\text{\: m }\text{s}^{\text{-1}}Only B and D satisfy this. Now apply conservation of momentum, with large mass MM and small mass mm:5M−2m=Mv1+mv25M - 2m = Mv_{1} + mv_{2}- B: 5M−2m=3M+10m5M - 2m = 3M + 10m, so M=6mM = 6m. The large mass is larger, so this is possible. - D: 5M−2m=5M+12m5M - 2m = 5M + 12m, so 14m=014m = 0. This is impossible.
Q9Answer B
The work done in stretching the spring (the increase in elastic potential energy) is ∫Fdl\int F \text{d}l, the area between the graph and the length axis between l1l_{1} and l2l_{2}. Here length is on the vertical axis, so this area is the horizontal strip between l1l_{1} and l2l_{2}, bounded on the left by the length axis and on the right by the line: P+QP + Q.
Q10Answer C
For equilibrium, F3F_{3} must be equal and opposite to the resultant of F1F_{1} (upwards) and F2F_{2} (to the right). The resultant of F1F_{1} and F2F_{2} points up and to the right, so F3F_{3} points down and to the left. F2F_{2} is longer than F1F_{1}, so F3F_{3} is closer to the horizontal than to the vertical. Diagram C shows F3F_{3} pointing down and to the left at this angle.
Q11Answer C
- Friction between two solid surfaces does not depend on the speed, so the frictional force stays the same. - Viscous drag in a fluid increases as the speed increases.
Q12Answer C
Speed of X at the bottom (energy conservation):v=2ghv = \sqrt{2gh}X and Y stick together. By conservation of momentum (each of mass mm):mv=2mv′⇒v′=v2mv = 2mv\prime\quad \Rightarrow \quad v\prime = \frac{v}{2}The combined masses rise until their kinetic energy becomes potential energy:t=v′22g=(v/2)22g=14(v22g)=h4t = \frac{v\prime^{2}}{2g} = \frac{(v/2)^{2}}{2g} = \frac{1}{4}\left( \frac{v^{2}}{2g} \right) = \frac{h}{4}
Q13Answer D
FF is the centripetal force:F=mv2r⇒v=FrmF = \frac{mv^{2}}{r}\quad \Rightarrow \quad v = \sqrt{\frac{Fr}{m}}
Q14Answer D
Gravity provides the centripetal force:GMmr2=mv2r⇒v2=GMr⇒r∝1v2\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}\quad \Rightarrow \quad v^{2} = \frac{GM}{r}\quad \Rightarrow \quad r \propto \frac{1}{v^{2}}If the speed becomes v/4v/4, the radius becomes 42=164^{2} = 16 times larger: 16r16r.
Q15Answer A
The force on 1 kg is GM/d2GM/d^{2}, where dd is the distance from the centre of the planet. P is at d=Rd = R. Q is 2R2R from the surface, so d=3Rd = 3R.FQFP=R2(3R)2=19\frac{F_{\text{Q}}}{F_{\text{P}}} = \frac{R^{2}}{(3R)^{2}} = \frac{1}{9}
Q16Answer B
Arc length and angle (in radians) are related by s=rθs = r\theta. The graph of ss against θ\theta is a straight line through the origin with gradient rr, the radius of the circle.
Q17Answer C
Power supplied by the cell:Pcell=2402.00×60=2.00  WP_{\text{cell}} = \frac{240}{2.00 \times 60} = 2.00\text{\: W}Power delivered to the resistor:PR=VI=(1.50)(0.50)=0.75  WP_{R} = VI = (1.50)(0.50) = 0.75\text{\: W}The rest is converted into thermal energy in the internal resistance:Pr=2.00−0.75=1.25  WP_{r} = 2.00 - 0.75 = 1.25\text{\: W}
Q18Answer C
The volume stays constant, so when the length becomes 4 times larger the cross-sectional area becomes 4 times smaller.R=ρlA⇒R′=ρ(4l)A/4=16ρlA=16RR = \frac{ \rho l }{A}\quad \Rightarrow \quad R\prime = \frac{\rho(4l)}{A/4} = 16\frac{ \rho l }{A} = 16R
Q19Answer D
Use E=I(R+r)E = I(R + r) for both readings:E=0.50(4.0+r)=0.25(10+r)E = 0.50(4.0 + r) = 0.25(10 + r)2.0+0.50r=2.5+0.25r⇒r=2.0 Ω2.0 + 0.50r = 2.5 + 0.25r\quad \Rightarrow \quad r = 2.0 \:\OmegaE=0.50(4.0+2.0)=3.0  VE = 0.50(4.0 + 2.0) = 3.0\text{\: V}
Q20Answer A
For every component to have the full 12 V, each must be connected directly across the battery, in parallel. - The heater and the motor each need their own switch. - The four lamps must be in parallel with each other (each with 12 V), all controlled by a single switch. A satisfies all of these. In B each lamp has its own switch. In C the lamps are in series, so each gets only 3 V. In D the lamps are in two series pairs, so each gets only 6 V.
Q21Answer A
The voltmeter measures the p.d. across the 3.0 Ω\Omega resistor. - With the variable resistor at 0 Ω\Omega, the whole 6.0 V is across the 3.0 Ω\Omega resistor. - With it at 12 Ω\Omega, the 6.0 V is shared in the ratio 3.0 : 12:V=3.03.0+12(6.0)=1.2  VV = \frac{3.0}{3.0 + 12}(6.0) = 1.2\text{\: V}The reading decreases to 1.2 V.
Q22Answer A
As the light intensity increases, the resistance of the LDR decreases. The cell's p.d. is shared in the ratio of the resistances, so the LDR's share (its p.d.) decreases. The cell has negligible internal resistance, so the two p.d.s always add up to the e.m.f. The p.d. across the variable resistor therefore increases.
Q23Answer D
The field of a point charge is radial, and its strength depends only on the distance from the charge (E∝1/r2E \propto 1/r^{2}). So at equal distances it has equal magnitude. - A: the field of a positive charge points away from it. - B: the field strength decreases with distance. - C: the field lines are radial, not circular.
Q24Answer D
Parallel currents in the same direction attract, and currents in opposite directions repel. The force is weaker at a larger distance. Let the separation of neighbouring wires be dd. - Q (out of the page): P repels it to the right and R repels it to the left, with equal forces. The resultant force on Q is zero. - P: Q repels it to the left (distance dd); R attracts it to the right, but more weakly (distance 2d2d). The resultant force on P is to the left. - R: by symmetry, the resultant force on R has the same size as on P but is to the right. P and R have equal resultant forces in opposite directions.
Q25Answer B
The magnetic force is F=qv×BF = qv \times B in direction. Reversing the field reverses the force. Changing the charge from negative to positive reverses it again. Two reversals bring the force back to its original direction: down and to the right, as in the first diagram. This is direction B.
Q26Answer D
The field is into the page and the electron moves in the plane of the page, so its velocity is perpendicular to the field. The 60∘^{\circ} is the angle to the edge of the field region, not to the field.F=Bev=(3.0×10−5)(1.60×10−19)(1.5×107)=7.2×10−17  NF = Bev = (3.0 \times 10^{- 5})(1.60 \times 10^{- 19})(1.5 \times 10^{7}) = 7.2 \times 10^{- 17}\text{\: N}
Q27Answer C
Most α-particles pass straight through, and only a very few are deflected through large angles. This is direct evidence that the positive charge and most of the mass are concentrated in a nucleus much smaller than the atom. The other statements are true, but they are not deduced from this experiment.
Q28Answer A
Isotopes have the same number of protons. Number of protons = nucleons - neutrons: - P: 22−11=1122 - 11 = 11 - Q: 23−12=1123 - 12 = 11 - R: 23−11=1223 - 11 = 12 - S: 22−12=1022 - 12 = 10 P and Q both have 11 protons, so they are isotopes.
Q29Answer A
mbefore=235.044+1.00863=236.05263  um_{\text{before}} = 235.044 + 1.00863 = 236.05263\text{\: u}mafter=140.914+91.926+3(1.00863)=235.86589  um_{\text{after}} = 140.914 + 91.926 + 3(1.00863) = 235.86589\text{\: u}Δm=0.18674  u=(0.18674)(1.66×10−27)=3.10×10−28  kg\Delta m = 0.18674\text{\: u} = (0.18674)(1.66 \times 10^{- 27}) = 3.10 \times 10^{- 28}\text{\: kg}E=Δmc2=(3.10×10−28)(3.00×108)2=2.79×10−11  JE = \Delta m c^{2} = (3.10 \times 10^{- 28})(3.00 \times 10^{8})^{2} = 2.79 \times 10^{- 11}\text{\: J}
Q30Answer B
- Random: the readings fluctuate about the trend, so the moment at which any particular nucleus decays cannot be predicted. - Spontaneous: the half-life stays the same even though the source gets warmer. The decay is not affected by external conditions such as temperature. (The activity decreasing is true of any decaying source, and does not show either property.)

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