2025 OCR AS Level Physics A - H156 Practice Paper with Answers

\(V_{out} = \frac{R_2}{R_1+R_2}V_{in} = \frac{2.0}{4.0+2.0}\times 12 = 4.0\,\text{V}\). The answer is B.

B1: B is correct. Common wrong answer A (\(\frac{1}{6}\times12\)) and D (voltage across the 4.0 kΩ resistor) score 0.

Higher light intensity lowers the LDR resistance. The total resistance falls, so the current rises. The p.d. across the fixed resistor, \(V = IR\), therefore increases (equivalently the LDR takes a smaller share of the battery p.d.). The answer is C.

B1: C is correct. Other options score 0.

Momentum is conserved: \(0.80\times1.5 = (0.80+0.40)v\), so \(v = 1.2/1.2 = 1.0\,\text{m s}^{-1}\). The answer is C.

B1: C is correct. Dividing by 0.80 kg only gives 1.5 (D); dividing by 1.6 gives 0.75 (B). Both score 0.

Resultant force \(= 3000 - 1200 = 1800\,\text{N}\). \(a = F/m = 1800/1200 = 1.5\,\text{m s}^{-2}\). The answer is B.

B1: B is correct. A uses only the resistive force, C ignores the resistive force and D adds the forces. All score 0.

Take moments about the left-hand end: \(R\times4.0 = 200\times2.0 + 100\times1.0 = 500\,\text{N m}\), so \(R = 125\,\text{N}\). (The left support supplies 175 N, so the forces sum to 300 N.) The answer is B.

B1: B is correct. D is the force at the left-hand support and scores 0.

Useful power \(= Fv = mgv = 250\times9.81\times0.40 = 981\,\text{W}\). Input power \(= 981/0.80 = 1226\,\text{W} \approx 1.2\,\text{kW}\). The answer is C.

B1: C is correct. A multiplies by 0.80 instead of dividing (0.78 kW); B is the useful output power. Both score 0.

\(\sin C = 1/n = 1/1.50 = 0.667\), so \(C = 41.8^\circ \approx 42^\circ\). The answer is B.

B1: B is correct. C (\(90^\circ - 42^\circ\)) scores 0.

\(v^2 = u^2 + 2as = 0 + 2\times9.81\times20 = 392\), so \(v = 19.8\,\text{m s}^{-1} \approx 20\,\text{m s}^{-1}\). The answer is B.

B1: B is correct. A (\(\sqrt{gs}\)), C (\(\sqrt{4gs}\)) and D (\(2\sqrt{2gs}\)) are wrong-method outcomes and score 0.

The p.d. divides in proportion to resistance: \(V_{out} = \frac{2.0}{4.0+2.0} \times 12 = 4.0\) V. Answer B.

B1 for B. Common errors: A (using 1/6 of supply), C (equal division), D (p.d. across the 4.0 kΩ resistor).

Increasing light intensity lowers the LDR resistance. In a series divider the component with the smaller share of the total resistance has the smaller share of the p.d., so the voltage across the LDR decreases. Answer C.

B1 for C. Rows A and B state the wrong resistance change; D has the correct resistance change but the wrong voltage change.

Momentum: \(2.0 \times 3.0 = 3.0\,v\), so \(v = 2.0\ \text{m s}^{-1}\). Initial \(E_k = \frac12 \times 2.0 \times 3.0^2 = 9.0\) J. Final \(E_k = \frac12 \times 3.0 \times 2.0^2 = 6.0\) J. Energy transferred = \(3.0\) J. Answer B.

B1 for B. C is the final kinetic energy; D is the initial kinetic energy.

Take moments about the left-hand support: \(R \times 4.0 = 200 \times 2.0 + 100 \times 1.0 = 500\) N m, so \(R = 125\) N. Answer C.

B1 for C. A (75 N) arises from placing the load 3.0 m from the left-hand end; B and D come from ignoring the load's distance from the supports.

At constant speed the driving force equals the resistive force. \(P = Fv\), so \(F = \frac{18\,000}{24} = 750\) N. Answer B.

B1 for B. A forgets to convert kW to W; C inverts the ratio (24/18); D multiplies power and speed.

\(\sin C = \frac{1}{n} = \frac{1}{1.60} = 0.625\), so \(C = 38.7^\circ \approx 39^\circ\). Answer B.

B1 for B. C (51°) is \(90^\circ - C\), the angle with the surface rather than the normal.

Taking upward as positive, the displacement is zero: \(0 = 14.0t - \frac12 \times 9.81\,t^2\), so \(t = \frac{2 \times 14.0}{9.81} = 2.85\) s. Answer C.

B1 for C. A is the time to reach maximum height; D uses 4u/g.

\(p = \rho g h = 1000 \times 9.81 \times 15 = 1.47 \times 10^{5}\ \text{Pa} \approx 1.5 \times 10^{5}\) Pa. Answer C.

B1 for C. D includes atmospheric pressure (about 1.0 × 10^5 Pa), which the question excludes. A and B are power-of-ten errors.

The voltmeter reads the p.d. across the lower resistor: \(V_{out} = \frac{R_2}{R_1+R_2} \times V_{in} = \frac{2.0}{4.0+2.0} \times 12 = 4.0\) V. Option D is the p.d. across the upper resistor; C assumes equal sharing.

B1: B selected. Guidance: \(V_{out} = 2.0/6.0 \times 12 = 4.0\) V. Other options score 0.

Momentum is conserved: \(2.0 \times 3.0 = (2.0 + 1.0)v\), so \(v = 6.0/3.0 = 2.0\) m s\(^{-1}\).

B1: C selected. Guidance: \(m_1u_1 = (m_1+m_2)v\) gives \(v = 2.0\) m s\(^{-1}\). Other options score 0.

The weight of the uniform beam acts at the pivot so has no moment. Taking moments about the pivot: \(60 \times 0.50 = 50 \times d\), so \(d = 30/50 = 0.60\) m.

B1: B selected. Guidance: clockwise moment = anticlockwise moment, \(50d = 30\) N m, \(d = 0.60\) m. Other options score 0.

(a) \(V_{out}=\frac{R_2}{R_1+R_2}V_{in}=\frac{4.0}{12.0+4.0}\times6.0=1.5\ \text{V}\).
(b) \(V_{out}=\frac{4.0}{2.0+4.0}\times6.0=4.0\ \text{V}\).
(c) The same current flows through both components. As temperature rises more charge carriers are released in the thermistor, so its resistance falls. Its share of the supply p.d. (\(V=IR\)) therefore falls and the fixed resistor takes a larger share, so \(V_{out}\) rises.
(d) \(3.0=\frac{4.0}{R_T+4.0}\times6.0\), so \(R_T+4.0=8.0\), \(R_T=4.0\ \text{k}\Omega\).
(e) Parallel resistance: \(1/R=1/4.0+1/8.0=3/8\), \(R=2.67\ \text{k}\Omega\). \(V=\frac{2.67}{12.0+2.67}\times6.0=1.09\ \text{V}\). The voltmeter draws current, lowering the resistance of the lower part of the divider, so a smaller share of the p.d. is across it.

(a) M1: substitution 4.0/(12.0+4.0) x 6.0 (or ratio V1/V2 = R1/R2). A1: 1.5 V. Allow 1.50 V.
(b) M1: substitution 4.0/(2.0+4.0) x 6.0. A1: 4.0 V.
(c) B1: resistance of thermistor decreases (as temperature rises). B1: p.d. is shared in proportion to resistance / thermistor has smaller share (so fixed resistor has larger share) AW. Not: 'current increases so V increases' without reference to share of p.d.
(d) M1: 3.0 = 4.0/(R+4.0) x 6.0 or recognising Vout = half of supply so R = 4.0 kOhm. A1: 4.0 kOhm. Allow 4000 Ohm.
(e) B1 (M1): combined resistance 2.67 kOhm (2.7 kOhm); A1 V = 1.09 V (allow 1.1 V) ecf. Explanation: B1 voltmeter in parallel lowers the resistance of that part of the divider / draws current (so smaller share of p.d.). Marks: M1 parallel resistance and substitution, A1 1.09 V; the explanation is credited together with the A1 (correct value without explanation max 1).

(a) For a closed (isolated) system the total momentum before an interaction equals the total momentum after it, provided no external resultant force acts.
(b) \(0.80\times1.5=(0.80+0.40)v\), so \(v=1.2/1.2=1.0\ \text{m s}^{-1}\).
(c) Initial \(E_k=\frac12\times0.80\times1.5^2=0.90\ \text{J}\). Final \(E_k=\frac12\times1.20\times1.0^2=0.60\ \text{J}\). Kinetic energy is not conserved (0.30 J lost), so the collision is inelastic, not perfectly elastic.
(d) \(F=\Delta p/\Delta t=0.40\times1.0/0.15=2.67\approx2.7\ \text{N}\).
(e) 2.7 N, opposite to the direction of motion of A, by Newton's third law.

(a) B1: total momentum before = total momentum after. B1: for a closed/isolated system (no external force) AW.
(b) M1: 0.80 x 1.5 = (0.80+0.40)v. A1: 1.0 m/s.
(c) C1: initial Ek = 0.90 J. C1: final Ek = 0.60 J. A1: conclusion not perfectly elastic as Ek is lost (allow ecf from (b)).
(d) M1: F = delta p / delta t with delta p = 0.40 x 1.0 (ecf from (b)). A1: 2.7 N (2.67 N unrounded).
(e) B1: 2.7 N opposite direction (to A's motion) AND Newton's third law. Both needed.

(a) For a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
(b)(i) Moments about A: \(F_B\times3.0=(20\times9.81\times2.0)+(25\times9.81\times3.5)=392.4+858.4=1250.8\), so \(F_B=417\ \text{N}\).
(ii) Total weight \(=45\times9.81=441.5\ \text{N}\), so \(F_A=441.5-416.9=24.5\ \text{N}\).
(c) At the point of tipping \(F_A=0\) and the plank pivots about B. Moments about B: child at distance \(x\) from A: \(25g(x-3.0)=20g(3.0-2.0)\), so \(x-3.0=0.80\), \(x=3.8\ \text{m}\). Since \(3.8\ \text{m}<4.0\ \text{m}\), the child reaches a position where the plank tips before reaching the end, so the plank does tip over.
(d) \(p=F/A=416.9/2.0\times10^{-3}=2.1\times10^5\ \text{Pa}\).

(a) B1: sum of clockwise moments = sum of anticlockwise moments. B1: about the same point, for a body in (rotational) equilibrium AW.
(b)(i) C1: weights 196 N (plank) and 245 N (child) or correct moment expressions. M1: moments taken about A (or equivalent about B/centre) with correct distances 2.0 and 3.5. A1: 417 N (416.9 N unrounded). Allow 420 N (2 s.f.).
(b)(ii) B1: 24.5 N (ecf from (i)). Allow 25 N.
(c) C1: recognises F_A = 0 at tipping, pivot about B. M1: 25g(x-3.0) = 20g(1.0). A1: x = 3.8 m and conclusion that the plank tips because 3.8 m < 4.0 m. A conclusion that the plank does not tip scores max 2.
(d) B1: 2.1 x 10^5 Pa (2.08 x 10^5 unrounded), ecf from (b)(i). Allow N m^-2.

(a) Light travels from a medium of higher refractive index to one of lower refractive index, and the angle of incidence at the boundary is greater than the critical angle.
(b) \(\sin C=n_2/n_1=1.45/1.50=0.9667\), so \(C=75.2^\circ\).
(c) At the limiting case the angle of incidence at the boundary is \(C\). The angle of refraction at the end face is \(r=90^\circ-75.2^\circ=14.8^\circ\). \(\sin i=1.50\sin14.8^\circ=1.50\times0.2560=0.384\), so \(i=22.6^\circ\).
(d) \(\sin C=1.00/1.50=0.667\), \(C=41.8^\circ\). At the end face \(\sin r=\sin i/1.50\le0.667\), so \(r\le41.8^\circ\). The angle of incidence at the side boundary is \(90^\circ-r\ge48.2^\circ\), which is greater than \(C\), so total internal reflection always occurs.

(a) B1: light travelling from higher to lower refractive index (denser to less dense) medium. B1: angle of incidence greater than (or equal to) the critical angle. Ignore references to 'angle of refraction is 90 degrees'.
(b) M1: sin C = 1.45/1.50 (or 1/n form with n ratio). A1: 75.2 degrees (75.16 unrounded). Allow 75 degrees.
(c) C1: angle at end face r = 90 - C = 14.8 degrees (ecf from (b)). M1: n sin i = n sin r applied: sin i = 1.50 sin 14.8. A1: sin i = 0.384. A1: i = 22.6 degrees (22.59 unrounded). Allow 23 degrees. Common error: using 75.2 degrees directly as the angle of refraction scores max 1 mark.
(d) B1: critical angle = 41.8 degrees (sin C = 1/1.5) or r has maximum value 41.8 degrees. B1: angle of incidence at side boundary = 90 - r is at least 48.2 degrees so always greater than C. AW.

(a) \(P=W/t=Fx/t\). Since \(x/t=v\) for constant speed, \(P=Fv\).
(b) Constant speed so tension \(=mg=450\times9.81=4414.5\ \text{N}\). \(P=Fv=4414.5\times0.80=3532\ \text{W}\approx3.5\ \text{kW}\).
(c) \(P_{in}=P_{out}/\text{efficiency}=3531.6/0.65=5433\ \text{W}\approx5.4\ \text{kW}\).
(d) Time \(t=12/0.80=15\ \text{s}\). \(E=Pt=5433\times15=8.15\times10^4\ \text{J}\approx8.1\times10^4\ \text{J}\) (81.5 kJ).
(e) Any two of: friction in the pulleys or between cable and guides (thermal energy); resistive heating in the motor windings; sound produced; energy used in accelerating the load at the start.

(a) B1: P = W/t and W = Fx so P = Fx/t. B1: x/t = v so P = Fv. Both steps needed.
(b) C1: F = mg = 4410 N (4414.5 N). A1: 3.5 kW (3532 W unrounded). Allow 3.53 kW. Common error: using mgh/t without time scores 0.
(c) M1: input = output / 0.65 (ecf from (b)). A1: 5.4 kW (5433 W unrounded). Allow 5.43 kW.
(d) C1: t = 12/0.80 = 15 s. A1: 8.1 x 10^4 J (8.15 x 10^4 J unrounded) ecf from (c). Allow 81 kJ or 82 kJ.
(e) B1 each, max 2: friction (in pulley/cable/bearings) / thermal energy in motor windings (resistive heating) / sound / energy to accelerate load initially. Not: 'energy is lost' or 'energy is used up' without a form or mechanism.

(a) An NTC thermistor is a component whose resistance decreases as its temperature increases.

(b) \(V_{out}=\frac{4.7}{8.2+4.7}\times 6.0=2.19\approx 2.2\,\text{V}\).

(c) p.d. across thermistor = \(6.0-4.0=2.0\,\text{V}\). \(\frac{R_T}{4.7}=\frac{2.0}{4.0}\) so \(R_T=2.35\approx 2.4\,\text{k}\Omega\).

(d) With a larger fixed resistor, it takes a larger share of the supply p.d. at any given thermistor resistance. For \(V_{out}=3.5\,\text{V}\) the ratio \(R_T/R\) must equal \(2.5/3.5\), a fixed value, so a larger \(R\) means a larger \(R_T\) is needed. A larger thermistor resistance occurs at a lower temperature (NTC), so the alarm switches on at a lower temperature.

(a) B1 resistance decreases; B1 as temperature increases (AW). Note: 'resistance changes with temperature' alone scores 0.
(b) M1 substitution \(\frac{4.7}{8.2+4.7}\times6.0\); A1 2.2 V (unrounded 2.186). Note: 3.8 V (wrong resistor) scores 1 mark.
(c) M1 thermistor p.d. = 2.0 V and use of \(V_1/V_2=R_1/R_2\) (or current = 4.0/4.7k); A1 2.4 kΩ (2.35). Allow 2350 Ω.
(d) B1 the ratio \(R_T/R\) (or \(V_T/V_{out}\)) needed for 3.5 V is fixed / larger R takes larger share of p.d.; B1 so a larger \(R_T\) is needed to trigger; B1 NTC resistance larger at lower temperature so lower temperature. ecf from wrong reasoning not allowed for final mark.

(a) For a closed (isolated) system, the total momentum before a collision equals the total momentum after, provided no external resultant force acts.

(b) \(0.80\times1.5=0.80\times0.30+0.60\,v\) so \(1.2=0.24+0.60v\), \(v=1.6\,\text{m s}^{-1}\).

(c) Before: \(\frac12(0.80)(1.5)^2=0.90\,\text{J}\). After: \(\frac12(0.80)(0.30)^2+\frac12(0.60)(1.6)^2=0.036+0.768=0.804\,\text{J}\). Kinetic energy is not conserved (0.096 J lost), so the collision is inelastic.

(d) \(F=\frac{\Delta p}{\Delta t}=\frac{0.60\times1.6}{0.040}=24\,\text{N}\).

(a) B1 total momentum before = total momentum after; B1 for a closed/isolated system, no external force (AW).
(b) M1 \(0.80\times1.5=0.80\times0.30+0.60v\); A1 1.6 m s^-1. Note: 1.2 m s^-1 or 0.4 scores M0.
(c) C1 initial KE 0.90 J; C1 final KE 0.80 J (0.804); A1 KE not equal/lost so not perfectly elastic. ecf from (b).
(d) M1 \(F=\Delta p/\Delta t\) with \(\Delta p=0.60\times1.6\); A1 24 N (0.96/0.040). ecf from (b). Allow use of A: \(0.80\times1.2/0.040\)=24 N.

(a) For a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.

(b) Moments about the hinge: \(T\sin35^\circ\times2.4=150\times1.2+90\times1.8=342\,\text{N m}\). \(T=\frac{342}{2.4\times\sin35^\circ}=248\,\text{N}\).

(c) Vertical component of tension \(=T\sin35^\circ=142.5\,\text{N}\). Vertical hinge force \(=240-142.5=97.5\,\text{N}\) upwards. Horizontal hinge force \(=T\cos35^\circ=203\,\text{N}\) away from the wall. Resultant \(=\sqrt{203^2+97.5^2}=226\,\text{N}\).

(a) B1 sum of clockwise moments = sum of anticlockwise moments; B1 about any point, for a body in equilibrium (AW).
(b) C1 weight of beam acts at 1.2 m (centre); M1 moments equation about hinge with \(T\sin35^\circ\); A1 248 N (unrounded 248.4). Note: use of cos35 gives 174 N and scores 2 marks.
(c) C1 vertical: 240 − 142.5 = 97.5 N; C1 horizontal = \(T\cos35^\circ\) = 203 N; M1 Pythagoras (or vector triangle); A1 226 N (225.7). ecf from (b). Allow 2 s.f. 230 N.

(a) \(v=c/n=\frac{3.00\times10^8}{1.52}=1.97\times10^8\,\text{m s}^{-1}\).

(b) \(\sin C=1/n=1/1.52\), \(C=41.1^\circ\).

(c) \(1.52\sin C=1.46\sin90^\circ\), \(\sin C=0.9605\), \(C=73.8^\circ\approx74^\circ\).

(d) \(68^\circ<73.8^\circ\), so the angle of incidence is less than the critical angle: total internal reflection does not occur. Part of the light refracts into the cladding and is lost from the core.

(e) The cladding protects the core surface from scratches/dirt (which cause light to escape) / prevents light leaking between touching fibres (cross-talk).

(a) M1 \(v=c/n\); A1 1.97×10^8 m s^-1.
(b) M1 \(\sin C=1/n\); A1 41° (41.1°).
(c) M1 \(\sin C=1.46/1.52\) or \(n_1\sin C=n_2\sin90^\circ\); A1 73.8° shown with at least 3 s.f. or unrounded value before rounding.
(d) B1 angle (68°) is less than the critical angle (so no total internal reflection) ; B1 light partly refracts into cladding / is lost / signal not carried. Reject 'light is absorbed' alone.
(e) B1 any one sensible reason: protects the core from scratches, prevents cross-talk between fibres, prevents loss when fibre touches other objects (AW).

(a) \(P=\frac{W}{t}=\frac{Fx}{t}\). Since \(v=x/t\), \(P=Fv\).

(b) \(F=P/v=\frac{18000}{25}=720\,\text{N}\). Speed is constant so there is no resultant force; the driving force equals the resistive force, \(720\,\text{N}\).

(c) Component of weight down the slope \(=mg\sin5.0^\circ=1100\times9.81\times\sin5.0^\circ=940\,\text{N}\). Driving force \(=720+940=1660\,\text{N}\). \(P=Fv=1660\times25=4.15\times10^4\,\text{W}\approx41\,\text{kW}\).

(d) Efficiency \(=\frac{18}{60}\times100\%=30\%\). The wasted energy is transferred to the surroundings as thermal energy (heat, and some sound).

(a) B1 \(P=Fx/t\) (substituting W); B1 \(x/t=v\) so \(P=Fv\).
(b) B1 \(F=P/v\)=720 N; B1 constant speed so resultant force zero / driving force = resistive force.
(c) C1 \(mg\sin5.0^\circ=940\,\text{N}\) (940.5); C1 total driving force = 720 + 940 = 1660 N (ecf from (b)); A1 41 kW or 41.5 kW (4.15×10^4 W). Note: using cos5.0° scores max 1 mark. Allow extra power only 23.5 kW scores 2 marks.
(d) B1 \(18/60\times100=30\%\); B1 wasted energy transferred as thermal energy to surroundings (allow heat/sound). Allow 0.30 without %.

(a) \(h=\frac12gt^2\) so \(g=\frac{2h}{t^2}=\frac{2\times1.20}{0.49^2}=10.0\,\text{m s}^{-2}\).

(b) \(\%\,\text{unc in }h=\frac{0.01}{1.20}\times100=0.83\%\). \(\%\,\text{unc in }t^2=2\times\frac{0.01}{0.49}\times100=4.1\%\). Total \(=4.9\%\). Absolute uncertainty \(=0.049\times10.0=0.49\approx0.5\,\text{m s}^{-2}\), so \(g=10.0\pm0.5\,\text{m s}^{-2}\).

(c) The range \(9.5\) to \(10.5\,\text{m s}^{-2}\) includes \(9.81\), so the result is consistent.

(d) Use an electromagnet release and a light gate/electronic timer (or data-logger) to reduce timing uncertainty. Measure \(t\) for several different heights, repeating each and taking a mean. Plot \(h\) against \(t^2\): the graph is a straight line through the origin with gradient \(g/2\), so \(g=2\times\) gradient.

(a) M1 \(g=2h/t^2\) with substitution; A1 10.0 m s^-2 (9.996). Allow 10.
(b) C1 percentage uncertainties 0.83% and 2×2.0% (4.1%); M1 added; A1 4.9% giving ±0.5 m s^-2 (0.49). Note: forgetting to double the time uncertainty gives 2.9% and scores 1 mark max (C0 M1 A0).
(c) B1 consistent because 9.81 lies within 10.0 ± 0.5 (ecf).
(d) B1 more precise timing, e.g. light gate and electronic timer / electromagnet release; B1 vary h, repeat and average t, plot h against t² (or t² against h); B1 gradient = g/2 (or g = 2 × gradient; if t² against h, g = 2/gradient).

As temperature rises the NTC thermistor resistance falls, so its share of the 6.0 V supply falls and the share across the fixed resistor rises, so \(V_{out}\) increases.

At 20 \(^\circ\)C: \(V_{out} = \frac{4.7}{4.7+8.0} \times 6.0 = 2.22\) V (< 3.6 V, fan off).

At 60 \(^\circ\)C: \(V_{out} = \frac{4.7}{4.7+1.5} \times 6.0 = 4.55\) V (> 3.6 V, fan on).

The switching threshold lies between the two values, so the circuit switches the fan off when cool and on when warm: it is suitable. Experiment: place the thermistor in a water bath with a thermometer, connect a voltmeter across the fixed resistor, heat slowly while stirring, record \(V_{out}\) at regular temperature intervals, and note the temperature at which \(V_{out}\) reaches 3.6 V (fan switches on). Repeat and average.

Levels of response (6 marks).

Level 3 (5-6 marks): Clear, well-structured line of reasoning covering all three bullet points; both calculations correct and a valid conclusion reached; the practical description is detailed and logical.
Level 2 (3-4 marks): Some of the bullet points addressed with a mostly clear line of reasoning; may have one calculation error or a limited practical description.
Level 1 (1-2 marks): Isolated relevant statements; limited or no linking; line of reasoning unclear.
Level 0 (0 marks): No response worthy of credit.

Indicative scientific points (list not exhaustive):
- NTC thermistor resistance decreases as temperature increases (B1)
- Larger fraction of supply p.d. across the fixed resistor / smaller across the thermistor, so \(V_{out}\) increases (B1)
- 20 \(^\circ\)C: \(V_{out} = 4.7/(4.7+8.0) \times 6.0 = 2.2\) V (2.22 V) (M1 A1)
- 60 \(^\circ\)C: \(V_{out} = 4.7/(4.7+1.5) \times 6.0 = 4.5\) V (4.55 V) (M1 A1). Allow 4.6 V.
- Compare with 3.6 V: off at 20 \(^\circ\)C, on at 60 \(^\circ\)C, so suitable; the threshold temperature lies between (B1)
- Practical: voltmeter in parallel with the fixed resistor; thermistor and thermometer in stirred water bath; heat gradually and record \(V_{out}\) and temperature; find temperature where \(V_{out}\) = 3.6 V; repeat for reliability (B1 each, max 2)

Guidance: Ignore comments on the thermistor self-heating. Not: voltmeter in series. ecf for a correct method with arithmetic slip. Candidates who use V1/V2 = R1/R2 correctly receive full credit for the calculations.

Momentum before: \(0.80 \times 1.5 = 1.20\) kg m s\(^{-1}\).
Momentum after: \(0.80 \times 0.60 + 0.40 \times 1.8 = 0.48 + 0.72 = 1.20\) kg m s\(^{-1}\). Momentum is conserved.

Kinetic energy before: \(\frac{1}{2} \times 0.80 \times 1.5^2 = 0.90\) J.
Kinetic energy after: \(\frac{1}{2} \times 0.80 \times 0.60^2 + \frac{1}{2} \times 0.40 \times 1.8^2 = 0.144 + 0.648 = 0.792\) J.
Kinetic energy decreases by about 0.11 J, so the collision is inelastic. Conservation of momentum holds for all collisions provided no external resultant force acts, so it does not show a collision is elastic. The track must be (nearly) frictionless/level so there is no external horizontal force.

By Newton's third law the forces on P and Q are equal in magnitude, opposite in direction and act for the same time, so the impulses (\(F\Delta t\)) are equal and opposite; changes in momentum are equal and opposite, so total momentum is unchanged.

Levels of response (6 marks).

Level 3 (5-6 marks): Clear line of reasoning covering all three bullets; both momentum and kinetic energy calculations correct and the conclusion (not elastic) clearly supported.
Level 2 (3-4 marks): Addresses most bullets with correct calculations for at least one of momentum or kinetic energy; reasoning mostly clear.
Level 1 (1-2 marks): Isolated correct statements or a single calculation; little linking.
Level 0 (0 marks): No response worthy of credit.

Indicative scientific points (list not exhaustive):
- Total momentum before = 1.2 kg m s\(^{-1}\); after = 0.48 + 0.72 = 1.2 kg m s\(^{-1}\), so conserved (M1 A1)
- Principle: total momentum of a system is constant provided no resultant external force acts (B1)
- Track must be level/friction negligible so no external horizontal force (B1)
- Kinetic energy before = 0.90 J (B1); after = 0.79 J (0.792 J) (B1)
- Kinetic energy is not conserved, so inelastic (energy transferred to thermal/sound/deformation); the conclusion is wrong because momentum is conserved in elastic and inelastic collisions (B1)
- Forces on P and Q equal and opposite (Newton's third law), act for the same time, so impulses equal and opposite; changes in momentum equal and opposite (B1)

Guidance: Allow 0.8 J for final kinetic energy. Not: 'momentum is lost'. ecf from a wrong kinetic energy value for the final judgement. Max 4 marks if no judgement about the conclusion is stated.