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Date May 2011 Marks available 5 Reference code 11M.2.SL.TZ2.8
Level Standard level Paper Paper 2 Time zone Time zone 2
Command term Draw Question number 8 Adapted from N/A

Question

This question is in two parts. Part 1 is about power and efficiency. Part 2 is about electrical resistance.

Part 1 Power and efficiency

A bus is travelling at a constant speed of 6.2 m s–1 along a section of road that is inclined at an angle of 6.0° to the horizontal.

(i) The bus is represented by the black dot shown below. Draw a labelled sketch to represent the forces acting on the bus.

(ii) State the value of the rate of change of momentum of the bus.

[5]
a.
The total output power of the engine of the bus is 70kW and the efficiency of the engine is 35%. Calculate the input power to the engine.
[2]
b.
The mass of the bus is 8.5×103 kg. Determine the rate of increase of gravitational potential energy of the bus.
[3]
c.
Using your answer to (c) and the data in (b), estimate the magnitude of the resistive forces acting on the bus.
[3]
d.

The engine of the bus suddenly stops working.

(i)  Determine the magnitude of the net force opposing the motion of the bus at the instant at which the engine stops.

(ii)  Discuss, with reference to the air resistance, the change in the net force as the bus slows down.
[4]
e.

Markscheme

(i)

   identification of normal reaction/N and weight/W;
   identification of friction and driving force;
   correct directions of all four forces;
   correct relative lengths; { (friction ≅ driving force and N ≅ W but N must not be longer than W) (judge by eye)

(ii)  zero;

a.

input power =\(\frac{{{\rm{output power}}}}{{{\rm{efficiency}}}} = \frac{{70}}{{0.35}}\);
= 200 kW;
Award [2] for a bald correct answer.

b.

height gained in 1s=(6.2 sin 6=) 0.648(m);
rate of change of PE=8.5×103×9.81×0.648;
=5.4×104W;

c.

power used to overcome friction=(7×104−5.4×104=)1.6×104(W); {(allow ECF from (c))
\(F = \left( {\frac{p}{v} = } \right)\frac{{1.6 \times {{10}^4}}}{{6.2}}\);
=2.6kN;

d.

(i) component of weight down slope = 8.5×103×9.81 sin 6;
net force=2.6×103+8.5×103×9.81 sin 6
=11kN;
Watch for ECF from (d). 

(ii) air resistance decreases as speed drops;
so net force decreases;

e.

Examiners report

(i) Diagrams were poorly presented and ill-thought. 4 marks were assigned to this and candidates should have given much more care to it. Marks were given for appropriate descriptions, directions and lengths of the vectors. In particular, candidates should recognize that the term “acceleration” will not do for a driving force, and that “normal” simply implies “at 90°”. The essential point about the upwards force from the surface is that it is a reaction force.
(ii) About half the candidates realized that the momentum change was zero as the velocity was constant.
a.
The efficiency calculation was well done by many.
b.
This question produced a mixed response varying from excellent fully-explained solutions to incoherent attempts with an incompetent inclusion of components or attempts that focussed on the change in the kinetic energy.
c.
Many recognized that the way to estimate the forces was to access the net rate of change of energy and divide this by the speed, but there were two hurdles here: a determination of the correct net power and the correct speed. Very many failed at one or both of these and thus failed to provide a correct answer.
d.
 
e.

Syllabus sections

Core » Topic 2: Mechanics » 2.2 – Forces
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