CAIE M1 2010 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2010
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind power and resistance simultaneously
DifficultyStandard +0.3 This is a standard power-force-acceleration problem requiring P=Fv to find driving force, then F=ma for acceleration, followed by solving for the new speed when acceleration halves. It involves multiple steps but uses routine M1 techniques with no novel insight required, making it slightly above average difficulty.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

2 A car of mass 600 kg travels along a horizontal straight road, with its engine working at a rate of 40 kW . The resistance to motion of the car is constant and equal to 800 N . The car passes through the point \(A\) on the road with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The car's acceleration at the point \(B\) on the road is half its acceleration at \(A\). Find the speed of the car at \(B\).

AnswerMarks Guidance
\([F - R = ma]\)M1 For using Newton's second law (3 terms)
\(F_A - 800 = 600a_A\)A1
\(F_A = \frac{40000}{25}(1600)\)B1
\(\frac{40000}{v_B} - 800 = 600(\frac{400}{600})\)A1
Speed is 33.3 ms⁻¹A1 [5]
$[F - R = ma]$ | M1 | For using Newton's second law (3 terms)
$F_A - 800 = 600a_A$ | A1 |
$F_A = \frac{40000}{25}(1600)$ | B1 |
$\frac{40000}{v_B} - 800 = 600(\frac{400}{600})$ | A1 |
Speed is 33.3 ms⁻¹ | A1 | [5]
2 A car of mass 600 kg travels along a horizontal straight road, with its engine working at a rate of 40 kW . The resistance to motion of the car is constant and equal to 800 N . The car passes through the point $A$ on the road with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car's acceleration at the point $B$ on the road is half its acceleration at $A$. Find the speed of the car at $B$.

\hfill \mbox{\textit{CAIE M1 2010 Q2 [5]}}