*Question*

This question is in two parts. **Part 1** is about simple harmonic motion (SHM). **Part 2** is about current electricity.

**Part 1** Simple harmonic motion (SHM)

An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension *F* in the spring is given by *F = k x* where *x* is the extension of the spring and* k* is a constant.

**Part 2** Current electricity

a.

Show that \({\omega ^2} = \frac{k}{m}\).[2]

One cycle of the variation of displacement with time is shown for two separate mass–spring systems, A and B.

(i) Calculate the frequency of the oscillation of A.

(ii) The springs used in A and B are identical. Show that the mass in A is equal to the mass in B.[3]

The graph shows the variation of the potential energy of A with displacement.

On the axes,

(i) draw a graph to show the variation of kinetic energy with displacement for the mass in A. Label this A.

(ii) sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.[5]

A 24 Ω resistor is made from a conducting wire.

(i) The diameter of the wire is 0.30 mm and the wire has a resistivity of 1.7\( \times \)10^{–8}Ωm. Calculate the length of the wire.

(ii) On the axes, draw a graph to show how the resistance of the wire in (d)(i) varies with the diameter of the wire when the length is constant. The data point for the diameter of 0.30 mm has already been plotted for you.

[4]

The 24 Ω resistor is covered in an insulating material. Explain the reasons for the differences between the electrical properties of the insulating material and the electrical properties of the wire.[3]

An electric circuit consists of a supply connected to a 24Ω resistor in parallel with a variable resistor of resistance *R*. The supply has an emf of 12V and an internal resistance of 11Ω.

Power supplies deliver maximum power to an external circuit when the resistance of the external circuit equals the internal resistance of the power supply.

(i) Determine the value of *R* for this circuit at which maximum power is delivered to the external circuit.

(ii) Calculate the reading on the voltmeter for the value of *R* you determined in (f)(i).

(iii) Calculate the total power dissipated in the circuit when the maximum power is being delivered to the external circuit.[8]

**Answer/Explanation**

## Markscheme

a.

*ma \(= – \)kx*;

\(a = – \frac{k}{m}x\); (*condone lack of negative sign*)

\(\left( {{\omega ^2} = \frac{k}{m}} \right)\)

**or**

implied use of defining equation for simple harmonic motion \(a = – {\omega ^2}x\);

\(\left( {{\rm{so }}{\omega ^2} = \frac{k}{m}} \right)\)

\(ma = – kx\) so \(a = – \left( {\frac{k}{m}} \right)x\);

(i) 0.833 (Hz);

(ii) frequency/period is the same so ω is the same;*k* is the same (as springs are identical);

(so *m* is the same)

(ii) end displacements correct \( \pm \) 0.01m;

maximum lower than 0.16J;

maximum equal to 0.04J \( \pm \) half square;

(i) \(l = \frac{{\pi {d^2}R}}{{4\rho }}\) seen / correct substitution

into equation: \(24 = \frac{{l \times 1.7 \times {{10}^{ – 8}}}}{{\pi \times {{\left( {0.15 \times {{10}^{ – 3}}} \right)}^2}}}\); } (*condone use of r for \(\frac{d}{2}\) in first **alternative)*

99.7 (m); *Award [2] for bald correct answer.*

*Award*

**[1 max]**if area is incorrectly calculated, answer is 399 m if conversion*to radius ignored, ie: allow ECF for second marking point if area is incorrect*

*provided working clear.*

(ii) any line showing resistance decreasing with increasing diameter **and** touching

point;

correct curved shape showing asymptotic behavior on at least one axis;

current/conduction is (related to) flow of charge;

conductors have many electrons free/unbound / electrons are the charge carriers / insulators have few free electrons;

pd/electric field accelerates/exerts force on electrons;

smaller current in insulators as fewer electrons available / larger current in conductors as more electrons available;

(i) use of total resistance = 11Ω; *(can be seen in second marking point)*\(\frac{1}{{11}} = \frac{1}{R} + \frac{1}{{24}}\);

20.3(Ω) ;

(ii) as current is same in resistor network and cell and resistance is same, half of emf must appear across resistor network;

6.0 (V);

**or**

\(I = \frac{{12}}{{\left( {11 + 11} \right)}} = 0.545\left( {\rm{A}} \right)\);*V*=(0.545×11=) 6.0 (V);

*Other calculations are acceptable.**Award [2] for a bald correct answer.*

(iii) use of 22 (ohm) * or *11+11 (ohm) seen;

use of \(\frac{{{V^2}}}{R}\) or equivalent;

6.54 (W);

*Award*

**[3]**for bald correct answer.*Award*

**[2 max]**if cell internal resistance ignored, yields 3.27 V.*Question*

This question is in two parts. **Part 1** is about momentum. **Part 2** is about electric point charges.

**Part 1** Momentum

**Part 2** Electric point charges

a.

State the law of conservation of linear momentum.[2]

A toy car crashes into a wall and rebounds at right angles to the wall, as shown in the plan view.

The graph shows the variation with time of the force acting on the car due to the wall during the collision.

The kinetic energy of the car is unchanged after the collision. The mass of the car is 0.80 kg.

(i) Determine the initial momentum of the car.

(ii) Estimate the average acceleration of the car before it rebounds.

(iii) On the axes, draw a graph to show how the momentum of the car varies during the impact. You are not required to give values on the y-axis.

#### [9]

Two identical toy cars, A and B are dropped from the same height onto a solid floor without rebounding. Car A is unprotected whilst car B is in a box with protective packaging around the toy. Explain why car B is less likely to be damaged when dropped.[4]

Define *electric field strength* at a point in an electric field.[2]

Six point charges of equal magnitude *Q* are held at the corners of a hexagon with the signs of the charges as shown. Each side of the hexagon has a length *a*.

P is at the centre of the hexagon.

(i) Show, using Coulomb’s law, that the magnitude of the electric field strength at point P due to **one** of the point charges is

\[\frac{{kQ}}{{{a^2}}}\]

(ii) On the diagram, draw arrows to represent the direction of the field at P due to point charge A (label this direction A) and point charge B (label this direction B).

(iii) The magnitude of *Q* is 3.2 μC and length *a* is 0.15 m. Determine the magnitude and the direction of the electric field strength at point P due to all six charges.[8]

**Answer/Explanation**

**Answer/Explanation**

## Markscheme

a.

total momentum does not change/is constant; } *(do not allow “momentum is conserved”)*

provided external force is zero / no external forces / isolated system;

(i) clear attempt to calculate area under graph;

initial momentum is half change in momentum;

\(\left( {\frac{1}{2} \times \frac{1}{2} \times 24 \times 0.16} \right) = 0.96\left( {{\rm{kgm}}{{\rm{s}}^{ – 1}}} \right)\)*Award [2 max] for calculation of total change (1.92kg ms^{–1})*

(ii) initial speed \( = \left( {\frac{{0.96}}{{0.8}} = } \right)1.2{\rm{m}}{{\rm{s}}^{ – 1}}\);

\(a = \frac{{1.2 – \left( { – 1.2} \right)}}{{0.16}}\) * or *\(a = \frac{{ – 1.2 – 1.2}}{{0.16}}\);

–15(ms

^{–2});

*(must see negative sign or a comment that this is a deceleration)*

**or**

average force =12 N;

uses *F*=0.8×a ;

–15(ms^{–2}); *(must see negative sign or a comment that this is a deceleration)**Award [3] for a bald correct answer.*

*Other solution methods involving different kinematic equations are possible.*

(iii) goes through *t*=0.08s **and** from negative momentum to positive / positive momentum to negative;

constant sign of gradient throughout;

curve as shown;*Award marks for diagram as shown.*

impulse is the same/similar in both cases / momentum change is same;

impulse is force × time / force is rate of change of momentum;

time to come to rest is longer for car B;

force experienced by car B is less (so less likely to be damaged);

electric force per unit charge;

acting on a small/point positive (test) charge;

(i) states Coulomb’s law as \(\frac{{kQq}}{{{r^2}}}\) * or *\(\frac{F}{q} = \frac{{kQ}}{{{r^2}}}\)

states explicitly q=1;

states r=a;

(ii)

arrow labelled A pointing to lower right charge;

arrow labelled B point to lower left charge;*Arrows can be anywhere on diagram.*

(iii) overall force is due to +Q top left and -Q bottom right / top right and bottom left and centre charges all cancel; } *(can be seen on diagram)*force is therefore \(\frac{{2kQ}}{{{a^2}}}\);

2.6×106 (N C^{-1}) ;

towards bottom right charge; *(allow clear arrow on diagram showing direction)*

*Question*

This question is about electric and magnetic fields.

A proton travelling to the right with horizontal speed 1.6×10^{4}ms^{–1} enters a uniform electric field of strength *E*. The electric field has magnitude 2.0×10^{3}NC^{–1} and is directed downwards.

a.

Calculate the magnitude of the electric force acting on the proton when it is in the electric field.[2]

A uniform magnetic field is applied in the same region as the electric field. A second proton enters the field region with the same velocity as the proton in (a). This second proton continues to move horizontally.

(i) Determine the magnitude and direction of the magnetic field.

(ii) An alpha particle enters the field region at the same point as the second proton, moving with the same velocity. Explain whether or not the alpha particle will move in a straight line.[5]

**Answer/Explanation**

## Markscheme

*F*=

*qE*1.6×10

**or**^{-19}×2.0×10

^{3};

=3.2×10

^{-16}(N);

(i) \(\left( {F = qvB \Rightarrow } \right)B = \frac{F}{{qv}}\) or \(\left( {Eq = qvB \Rightarrow } \right)B = \frac{E}{v}\);

\(\left( { = \frac{{3.2 \times {{10}^{ – 16}}}}{{1.6 \times {{10}^{ – 19}} \times 1.6 \times {{10}^4}}}} \right) = 0.13\) or 0.125(T);

directed into the page / OWTTE;

(ii) both electric and magnetic forces double / both forces increase by the same factor / both forces scale with q/charges and cancel;

so straight line followed; (only award if first mark awarded)

or

straight line followed if qE = qvB ⇒E v=B;

E, v and B constant (so straight line followed);