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