CAIE M1 2024 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind power at constant speed
DifficultyStandard +0.3 This is a straightforward power-force-velocity problem requiring standard M1 techniques: resolving forces on a slope, applying P=Fv for constant speed, then using F=ma with reduced power. The calculations are routine with clearly defined steps and no conceptual subtlety beyond standard mechanics formulas.
Spec6.02l Power and velocity: P = Fv
A car of mass 1600 kg travels up a slope inclined at an angle of \(\sin^{-1}\) 0.08 to the horizontal. There is a constant resistance of magnitude 240 N acting on the car.
  1. It is given that the car travels at a constant speed of 32 ms\(^{-1}\). Find the power of the engine of the car. [3]
  2. Find the acceleration of the car when its speed is 24 ms\(^{-1}\) and the engine is working at 95\% of the power found in (a). [3]
Question 3:
AnswerMarks
3(a)Resolving up slope.
If correct should see
AnswerMarks Guidance
DF= 240+1600g0.08 =240+1280=1520 M1 Must have correct number of relevant terms (weight component
and 240 N resistance). Allow sign errors. Allow cos 4.58 or cos
4.6. Do not allow g missing. Using sin−10.08 or sin 0.08 scores
M0B0A0.
Must have either 0.08 or sin4.58 or sin 4.6, not just sin.
AnswerMarks Guidance
Power =their(1520)32B1 Power
OE. E.g. =their1 520 .
32
Allow any driving force provided it has a resistance and a weight
component.
AnswerMarks Guidance
Power =48640 WA1 Allow 48 600 W or 48.64 kW or 48.6 kW. Must state units if given
in kW.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(b)0.95their 48640 46208 5776
DF= or = or 1925.3
AnswerMarks Guidance
24 24 3B1FT Power
DF = oe e.g. 0.95their 48640 = DF × 24
v
FT their power from part (a) Do not allow if not using power from
part (a)
Note: candidates who use sin 0.08 in part (a) should get a DF of
332.3 N, which can score B1FT and use of sin−10.08 should get a
DF of 93299 N, can score B1FT. If candidate uses 48600 DF =
46170
=1923.75
24
Candidates who omit the weight component in part (a) should get
a DF of 304 N, and can score B1FT
AnswerMarks Guidance
their DF−240−1600g0.08=1600aM1 N2L Must have correct number of relevant terms (weight
component and 240 N resistance). Allow sign errors. Must be
dimensionally correct. Allow without using 95% or with using
5%. Must have either 0.08 or sin4.58 or sin 4.6, not just sin
or sin−10.08 or sin 0.08.
AnswerMarks Guidance
a=0.253 ms−2A1 19
Allow Note: 0.25 scores A0.
75
If candidate uses 48600 they must get 0.252(34…) rather than
0.253.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | Resolving up slope.
If correct should see
DF= 240+1600g0.08 =240+1280=1520  | M1 | Must have correct number of relevant terms (weight component
and 240 N resistance). Allow sign errors. Allow cos 4.58 or cos
4.6. Do not allow g missing. Using sin−10.08 or sin 0.08 scores
M0B0A0.
Must have either 0.08 or sin4.58 or sin 4.6, not just sin.
Power =their(1520)32 | B1 | Power
OE. E.g. =their1 520 .
32
Allow any driving force provided it has a resistance and a weight
component.
Power =48640 W | A1 | Allow 48 600 W or 48.64 kW or 48.6 kW. Must state units if given
in kW.
3
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | 0.95their 48640 46208 5776
DF= or = or 1925.3
24 24 3 | B1FT | Power
DF = oe e.g. 0.95their 48640 = DF × 24
v
FT their power from part (a) Do not allow if not using power from
part (a)
Note: candidates who use sin 0.08 in part (a) should get a DF of
332.3 N, which can score B1FT and use of sin−10.08 should get a
DF of 93299 N, can score B1FT. If candidate uses 48600 DF =
46170
=1923.75
24
Candidates who omit the weight component in part (a) should get
a DF of 304 N, and can score B1FT
their DF−240−1600g0.08=1600a | M1 | N2L Must have correct number of relevant terms (weight
component and 240 N resistance). Allow sign errors. Must be
dimensionally correct. Allow without using 95% or with using
5%. Must have either 0.08 or sin4.58 or sin 4.6, not just sin
or sin−10.08 or sin 0.08.
a=0.253 ms−2 | A1 | 19
Allow Note: 0.25 scores A0.
75
If candidate uses 48600 they must get 0.252(34…) rather than
0.253.
3
Question | Answer | Marks | Guidance
A car of mass 1600 kg travels up a slope inclined at an angle of $\sin^{-1}$ 0.08 to the horizontal. There is a constant resistance of magnitude 240 N acting on the car.

\begin{enumerate}[label=(\alph*)]
\item It is given that the car travels at a constant speed of 32 ms$^{-1}$.

Find the power of the engine of the car. [3]

\item Find the acceleration of the car when its speed is 24 ms$^{-1}$ and the engine is working at 95\% of the power found in (a). [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q3 [6]}}
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