Show that the time taken for B to pass I is approximately 28 s.

Calculate the distance travelled by B in this time.

B slows down while I remains at a constant speed. The driver in each car wears a seat belt. Using Newton’s laws of motion, explain the difference in the tension in the seat belts of the two cars.

Calculate the speed of O immediately before the collision.

The duration of the collision is 0.45 s. Determine the average force acting on O.

An ammeter and a voltmeter are used to investigate the characteristics of a variable resistor of resistance $R$. State how the resistance of the ammeter and of the voltmeter compare to $R$ so that the readings of the instruments are reliable.

Show that the current in the circuit is approximately 0.70 A when $R = 0.80{\text{ }}\Omega$.

Outline what is meant by the internal resistance of a cell.

Determine the internal resistance of the cell.

Calculate the electromotive force (emf) of the cell.

distances itemized; (it must be clear through use of ${s_I}$ or distance I etc)

distances equated;

$t = \frac{{2v}}{a}$ / cancel and re-arrange;

substitution $\left( {\frac{{2 \times 45}}{{3.2}}} \right)$ shown / 28.1(s) seen;

or

clear written statement that the average speed of B must be the same as constant speed of I;

as B starts from rest the final speed must be ${\text{2}} \times {\text{45}}$;

so $t = \frac{{\Delta v}}{a} = \frac{{90}}{{3.2}}$;

28.1 (s) seen; (for this alternative the method must be clearly described)

or

attempts to compare distance travelled by I and B for 28 s;

I distance $= (45 \times 28 = ){\text{ }}1260{\text{ (m)}}$;

B distance $= (\frac{1}{2} \times 3.2 \times {28^2} = ){\text{ }}1255{\text{ (m)}}$;

deduces that overtake must occur about $\left( {\frac{5}{{45}} = } \right){\text{ }}0.1{\text{ s}}$ later;

use of appropriate equation of motion;

$(1.26 \approx )$ 1.3 (km);

Award [2] for a bald correct answer.

driver I moves at constant speed so no net (extra) force according to Newton 1;

driver B decelerating so (extra) force (to rear of car) (according to Newton 1) / momentum/inertia change so (extra) force must be present;

(hence) greater tension in belt B than belt I;

Award [0] for stating that tension is less in the decelerating car (B).

$930 \times v + 850 \times 45 = 1780 \times 52$ or statement that momentum is conserved;

$v = 58{\text{ }}({\text{m}}\,{{\text{s}}^{ - 1}})$;

Allow [2] for a bald correct answer.

use of force $\frac{{{\text{change of momentum}}}}{{{\text{time}}}}$ (or any variant, eg: $\frac{{930 \times 6.4}}{{0.45}}$);

$13.2 \times {10^3}{\text{ (N)}}$; } (must see matched units and value ie: 13 200 without unit gains MP2, 13.2 does not)

Award [2] for a bald correct answer.

Allow use of 58 m s^–1 from (c)(i) to give 12 400 (N).

ammeter must have very low resistance/much smaller than $R$;

voltmeter must have very large resistance/much larger than $R$;

Allow [1 max] for zero and infinite resistance for ammeter and voltmeter respectively.

Allow [1 max] if superlative (eg: very/much/OWTTE) is missing.

power (loss in resistor) $= 0.36{\text{ (W)}}$; } (accept answers in the range of 0.35 to 0.37 (W) – treat value outside this range as ECF (could still lead to 0.7))

${I^2} \times 0.80 = 0.36$;

$I = 0.67{\text{ (A)}}$ or $\sqrt {\left( {\frac{{0.36}}{{0.8}}} \right)}$; (allow answers in the range of 0.66 to 0.68 (A).

resistance of the components/chemicals/materials within the cell itself; } (not “resistance of cell”)

leading to energy/power loss in the cell;

power (in cell with 0.7 A) $= 0.58{\text{ W}}$; } (allow answers in the range of 0.57 W to 0.62 W)

${0.7^2} \times r = 0.58$;

$r = 1.2{\text{ (}}\Omega {\text{)}}$; (allow answers in the range of 1.18 to 1.27 ($\Omega$))

or

when powers are equal;

${I^2}R = {I^2}r$;

so $r = R$ which occurs at 1.2(5) ($\Omega$);

Award [1 max] for bald 1.2(5) ($\Omega$).

$\left( {E = I(R + r)} \right) = 0.7(0.8 + 1.2)$;

1.4 (V);

Allow ECF from (e) or (f)(ii).

or

when $R = 0$, power loss $= 1.55$;

$E = (\sqrt {1.55 \times 1.2}  = ){\text{ }}1.4{\text{ (V)}}$;