| Date | May 2014 | Marks available | 2 | Reference code | 14M.2.SL.TZ1.6 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate | Question number | 6 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about two children on a merry-go-round. Part 2 is about electric circuits.
Part 1 Two children on a merry-go-round
Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.
Aibhe is moving at speed 1.0ms–1 relative to the ground.
Part 2 Orbital motion
A spaceship of mass m is moving at speed v in a circular orbit of radius r around a planet of mass M.
Determine the magnitude of the velocity of Aibhe relative to
(i) Euan.
(ii) the centre of the merry-go-round.
(i) Outline why Aibhe is accelerating even though she is moving at constant speed.
(ii) Draw an arrow on the diagram on page 22 to show the direction in which Aibhe is accelerating.
(iii) Identify the force that is causing Aibhe to move in a circle.
(iv) The diagram below shows a side view of Aibhe and Euan on the merry-go-round.
Explain why Aibhe feels as if her upper body is being “thrown outwards”, away from the centre of the merry-go-round.
Euan is rotating on a merry-go-round and drags his foot along the ground to act as a brake. The merry-go-round comes to a stop after 4.0 rotations. The radius of the merry-go-round is 1.5 m. The average frictional force between his foot and the ground is 45 N. Calculate the work done.
Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below.
Euan pushes the merry-go-round so that he is again moving at 1.0 ms–1 relative to the ground.
(i) Determine Aibhe’s speed relative to the ground.
(ii) Calculate the magnitude of Aibhe’s acceleration.
Define the electric resistance of a wire.
Using the following data, calculate the length of constantan wire required to make a resistor with a resistance of 6.0Ω.
Resistivity of constantan = 5.0×10–7Ωm
Average radius of wire = 2.5×10–5m
Three resistors, each of resistance 6.0Ω, are arranged in the circuit shown below. The cell has an emf of 12V and negligible internal resistance.
Determine the total power supplied by the cell.
The same resistors and cell are now re-arranged into a different circuit, as shown below.
Explain why the total power supplied by the cell is greater than for the circuit in (g).
Markscheme
(ii) 1.0 or 0(ms-1);
hence her velocity is changing;
or
since her direction/velocity is changing;
a resultant/unbalanced/net force must be acting on her (hence she is accelerating);
Ignore length of arrow.
upper body (initially) continues to move in a straight line at constant speed/ velocity is tangential to circle;
\(W\left( { = {F_{av}}d = 45 \times 37.70} \right) = 1700\left( {\rm{J}} \right)\);
from \(v = \frac{{2\pi r}}{T}\), since r is halved, v is halved;
v=0.5(ms-1);
Award [3] for a bald correct answer.
a=0.33(ms-2);
Allow ECF from (d)(i).
Award [2] for a bald correct answer.
Accept equation with symbols defined. Accept p.d. Do not accept voltage.
\(l = \left( {\frac{{RA}}{\rho } = } \right)\frac{{6.0 \times 1.963 \times {{10}^{ - 9}}}}{{5.0 \times {{10}^{ - 7}}}}\);
=2.4×10-2(m);
Award [3] for a bald correct answer.
resistance of two resistors in parallel = 3.0(Ω), so total resistance = 6.0 + 3.0 = 9.0(Ω);
\(I = \left( {\frac{V}{R} = } \right)\frac{{12}}{{9.0}}\left( { = 1.333} \right)\left( {\rm{A}} \right)\);
\(P = \left( {VI = 12 \times 1.333 = } \right)16\left( {\rm{W}} \right)\);
or
resistance of two resistors in parallel = 3.0(Ω), so total resistance = 6.0 + 3.0 = 9.0(Ω);
\(P = \left( {\frac{{{V^{\rm{2}}}}}{R} = } \right)\frac{{144}}{{9.0}}\);
P =16 (W);
total resistance is smaller (=4.0Ω);
p.d./voltage is the same so current is greater (=3.0A);
since P=VI or P=I2R,power is greater (=36W);
or
total resistance is smaller (= 4.0Ω); p.d./voltage is the same;
since P=\(\frac{{{V^{\rm{2}}}}}{R}\), power is greater (=36W);
Award [1] for a bald calculation of 36 (W). The marks are for an explanation.
Examiners report
Most were able to identify the relative speeds. The markscheme was amended to also include answers in terms of velocity.
i) This was well-answered with most identifying a change in direction and a change in velocity.
ii) The majority were able to show the direction of the centripetal acceleration.
iii) Few identified a force that would act on Aibhe. They did not realize that the centripetal force is the resultant of the forces acting.
iv) Few realized from the diagram that it would be difficult provide an inward directed force on Aibhe’s upper torso. The consequence of this is that it would tend to continue to move in a direction which is tangential to the circle.
This was well done by many.
i) Many scored three marks here.
ii) Most candidates were able to gain both marks.
The definition of resistance was poorly attempted with many describing some difficulty that a current has in travelling down a wire.
This calculation was generally well done although it was disappointing to see a significant proportion of candidates who did not know the formula for the area of a circle.
Most were able to calculate the equivalent resistance of the combination of resistors and progress successfully to find the power supplied by the cell.
Many recalculated the power but didn’t provide an explanation and so consequently only scored one mark. Explanations were often detailed enough to score full marks.
