State the law of conservation of linear momentum.

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.

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.

Define electric field strength at a point in an electric field.

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.

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)