CAIE M1 2005 June — Question 7 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2005
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration given power
DifficultyStandard +0.3 This is a standard M1 power question requiring (i) P=Fv to find driving force then F=ma, and (ii) work-energy principle with constant power over time. Both parts follow routine procedures taught in mechanics courses, though part (ii) requires careful bookkeeping of energy terms (KE change, work against resistance, work by engine). Slightly above average due to the multi-step energy calculation, but no novel insight required.
Spec6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

A car of mass \(1200 \text{ kg}\) travels along a horizontal straight road. The power provided by the car's engine is constant and equal to \(20 \text{ kW}\). The resistance to the car's motion is constant and equal to \(500 \text{ N}\). The car passes through the points \(A\) and \(B\) with speeds \(10 \text{ m s}^{-1}\) and \(25 \text{ m s}^{-1}\) respectively. The car takes \(30.5 \text{ s}\) to travel from \(A\) to \(B\).
  1. Find the acceleration of the car at \(A\). [4]
  2. By considering work and energy, find the distance \(AB\). [8]

Question 7:
AnswerMarks
7(i)
(ii)Driving force = 20 000/10
DF – R = ma
2000 – 500 = 1200a
-1
Acceleration is 1.25ms
KE change =
2 2
½ 1200 (25 – 10 )
Difference in KE is 315 000 J
20 000 = WD by car’s
engine/30.5
Work done is 610 000 J
610 000 =315 000 +
WD against resistance
500(AB) = 295 000
AnswerMarks
Distance is 590 mB1
M1
A1 ft
A1 4
M1
A1
M1
A1
M1
M1
A1 ft
AnswerMarks
A1 8For using Newton’s second law
(3 terms needed)
For using KE change
2 2
= ½ m(v – u )
May be implied
For using
(constant)Power = WD/Time
May be implied
For using
WD by car’s engine = Increase
in KE + WD against resistance
For using WD against
resistance
×
= Resistance AB
Question 7:
7 | (i)
(ii) | Driving force = 20 000/10
DF – R = ma
2000 – 500 = 1200a
-1
Acceleration is 1.25ms
KE change =
2 2
½ 1200 (25 – 10 )
Difference in KE is 315 000 J
20 000 = WD by car’s
engine/30.5
Work done is 610 000 J
610 000 =315 000 +
WD against resistance
500(AB) = 295 000
Distance is 590 m | B1
M1
A1 ft
A1 4
M1
A1
M1
A1
M1
M1
A1 ft
A1 8 | For using Newton’s second law
(3 terms needed)
For using KE change
2 2
= ½ m(v – u )
May be implied
For using
(constant)Power = WD/Time
May be implied
For using
WD by car’s engine = Increase
in KE + WD against resistance
For using WD against
resistance
×
= Resistance AB
A car of mass $1200 \text{ kg}$ travels along a horizontal straight road. The power provided by the car's engine is constant and equal to $20 \text{ kW}$. The resistance to the car's motion is constant and equal to $500 \text{ N}$. The car passes through the points $A$ and $B$ with speeds $10 \text{ m s}^{-1}$ and $25 \text{ m s}^{-1}$ respectively. The car takes $30.5 \text{ s}$ to travel from $A$ to $B$.

\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the car at $A$.
[4]

\item By considering work and energy, find the distance $AB$.
[8]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2005 Q7 [12]}}