| Date | May 2012 | Marks available | 1 | Reference code | 12M.2.SL.TZ2.6 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Show that | Question number | 6 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about kinematics and mechanics. Part 2 is about electric potential difference and electric circuits.
Part 1 Kinematics and mechanics
Define linear momentum.
State, in terms of momentum, Newton’s second law of motion.
Show, using your answer to (b), how the impulse of a force F is related to the change in momentum Δp that it produces.
Show, using your answer to (b), how the impulse of a force F is related to the change in momentum Δp that it produces.
A railway truck on a level, straight track is initially at rest. The truck is given a quick, horizontal push by an engine so that it now rolls along the track.
The engine is in contact with the truck for a time T = 0.54 s and the initial speed of the truck after the push is 4.3 ms–1. The mass of the truck is 2.2×103 kg.
Due to the push, a force of magnitude F is exerted by the engine on the truck. The sketch shows how F varies with contact time t.
(i) Determine the magnitude of the maximum force Fmax exerted by the engine on the truck.
(ii) After contact with the engine (t = 0.54 s) the truck moves a distance 15 m along the track. After travelling this distance the speed of the truck is 2.8 ms–1. Assuming a uniform acceleration, calculate the time it takes the truck to travel 15 m.
(iii) Calculate the average rate at which the kinetic energy of the truck is dissipated as it moves along the track.
(iv) When the speed of the truck is 2.8ms–1 it collides with a stationary truck of mass 3.0×103kg. The two trucks move off together with a speed V. Show that the speed V=1.2ms–1.
(v) Outline the energy transformations that take place during the collision of the two trucks.
Markscheme
mass×velocity; (allow mv with symbols defined)
the rate of change of momentum of a body is equal to/directly proportional to the force acting on the body;
Accept \(F = \frac{{\Delta p}}{{\Delta t}}\) only if all symbols are defined.
\(\left( {F = \frac{{\Delta p}}{{\Delta t}}} \right)\)
therefore impulse FΔt =Δp; (accept t for Δt)
\(\left( {F = \frac{{\Delta p}}{{\Delta t}}} \right)\)
therefore impulse FΔt =Δp ; (accept t for Δt)
(i) (impulse=) change in momentum=2.2×103×4.3(=9.46×103Ns);
impulse=area under graph=½FmaxΔt;
½Fmax×0.54=9.46×103;
Fmax=35k(N) or 3.5×104 (N);
(ii) (magnitude of) acceleration=\(\left( {\frac{{{u^2} - {v^2}}}{{2s}} = \frac{{{{4.3}^2} - {{2.8}^2}}}{{30}} = } \right)0.355{\rm{m}}{{\rm{s}}^{ - 2}}\);
time=\(\left( {\frac{{u - v}}{a} = \frac{{1.5}}{{0.355}} = } \right)4.2{\rm{s}}\);
Award [1 max] if an additional 0.54 s is added to answer.
(iii) \(\Delta KE = \left( {\frac{1}{2} \times 2.2 \times {{10}^3}\left[ {{{4.3}^2} - {{2.8}^2}} \right] = } \right)1.17 \times {10^4}{\rm{J}}\);
rate of change of \(\Delta KE = \frac{{1.17 \times {{10}^4}}}{{4.2}} = 2.8{\rm{kW}}\)
(mark is for division by 4.2 and correct calculation)
(iv) statement of momentum conservation:
e.g. momentum of the truck before collision=momentum of both trucks after collision;
(allow clear symbolism instead of words)
2.2×103×2.8=5.2×103V or \(V = \frac{{2.2}}{{5.2}} \times 2.8\);
to give V=1.2ms-1
(v) the first truck loses kinetic energy that is transferred to internal energy in the links between the trucks (and as sound);
and to kinetic energy of the stationary truck;
Award [0] for “lost as heat, light and sound”, or “in air resistance”.
Examiners report
Linear momentum was defined accurately. Only a handful used speed rather than velocity in the definition.
Newton's second law of motion appeared to be well understood but some failed to quote it in the context of momentum and gave the simpler statement in terms of F=ma.
Many were able to show that impulse is equal to the change in momentum.
Many were able to show that impulse is equal to the change in momentum.
(i) This was another question where candidates let themselves down very badly with their quality of explanation and presentation. Although many obtained the correct answer it was often not clear that they had fully appreciated the assumptions they were making. Examiners (for full credit) were looking for explanations typically in terms of the area under the F-t graph – it was rare to see this – with a full consideration of the evaluation of the isosceles triangle. Many candidates will have scored only one mark for an evaluation of the momentum change (usually by inference rather than by direct candidate statement) and one mark for the answer.
(ii) This straightforward application of the kinematic equations was often well done, but some candidates failed to read the question carefully and could not identify the correct values for the speed of the truck.
(iii) A common error was to evaluate (4.8-2.3)2 rather than the correct (4.22-2.82) in the route towards the change in kinetic energy. However, whether evaluating the correct change in kinetic energy or not, most were able to divide a value for the change by the time calculated in (ii) to determine an average rate of energy dissipation.
(iv) Candidates were required to indicate the basis for the calculation (conservation of momentum) and to identify the algebraic or numerical method they were using. Although many gained full marks, once again candidates did not make it easy for examiners to establish the basis for the method. Clear statements of momentum conservation were rare and examiners were frequently expected to infer the method from an undefined set of symbols sometimes unrelated to this particular problem.
(v) Most candidates gained one mark from this question by outlining the transfer in kinetic energy from first to second truck. Few recognized the role of the trucks‟ coupling mechanism preferring to emphasize the relatively minor and generalized roles of heat and sound dissipation in the collision. This is a frequent response in energy transfer questions such as this; candidates should consider specific examples in the question rather than making recourse to more general issues of energy transfer.
