| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, inclined road |
| Difficulty | Standard +0.3 This is a standard two-part connected particles problem requiring routine application of energy methods (work-energy theorem) and Newton's second law. Part (a) involves straightforward calculation of work done against gravity, resistance, and change in kinetic energy. Part (b) requires separating the system to find tension, which is a textbook technique. The calculations are multi-step but follow standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | KE (final) = ½ × 1500 × 202 + ½ × 750 × 202 | |
| KE (initial) = ½ × 1500 × 302 + ½ × 750 × 302 | B1 | Use KE = ½mv2 for any two of the four elements |
| PE gain = 2250 × 10 × 800 × 0.08 | B1 | |
| WD against friction = 600 × 800 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| + ½ × 2250 × 202 + 2250 × 10 × 800 × 0.08 | M1 | Use energy equation. |
| DF = 1700 N | A1 | DF = 1696.875 N |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(b) | 2400 – 600 = 2250a |
| Answer | Marks | Guidance |
|---|---|---|
| T – 200 = 750a and 2400 – 400 – T = 1500a | M1 | Apply Newton’s second law to the system or to each of the car |
| Answer | Marks |
|---|---|
| A1 | Two correct equations |
| Attempting to solve for a or for T | M1 |
| T = 800 N and a = 0.8 ms–2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Mark |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PPMMTT
9709/41 Cambridge International AS & A Level – Mark Scheme October/November 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 13
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | KE (final) = ½ × 1500 × 202 + ½ × 750 × 202
KE (initial) = ½ × 1500 × 302 + ½ × 750 × 302 | B1 | Use KE = ½mv2 for any two of the four elements
PE gain = 2250 × 10 × 800 × 0.08 | B1
WD against friction = 600 × 800 | B1
½ × 2250 × 302 + DF × 800 = 600 × 800
+ ½ × 2250 × 202 + 2250 × 10 × 800 × 0.08 | M1 | Use energy equation.
DF = 1700 N | A1 | DF = 1696.875 N
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 2400 – 600 = 2250a
or
T – 200 = 750a and 2400 – 400 – T = 1500a | M1 | Apply Newton’s second law to the system or to each of the car
and trailer separately
A1 | Two correct equations
Attempting to solve for a or for T | M1
T = 800 N and a = 0.8 ms–2 | A1
4
Question | Answer | Mark | GuidanceA car of mass 1500 kg is pulling a trailer of mass 750 kg up a straight hill of length 800 m inclined at an angle of $\sin^{-1} 0.08$ to the horizontal. The resistances to the motion of the car and trailer are 400 N and 200 N respectively. The car and trailer are connected by a light rigid tow-bar. The car and trailer have speed $30 \text{ m s}^{-1}$ at the bottom of the hill and $20 \text{ m s}^{-1}$ at the top of the hill.
\begin{enumerate}[label=(\alph*)]
\item Use an energy method to find the constant driving force as the car and trailer travel up the hill. [5]
\item After reaching the top of the hill the system consisting of the car and trailer travels along a straight level road. The driving force of the car's engine is 2400 N and the resistances to motion are unchanged.
Find the acceleration of the system and the tension in the tow-bar. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q6 [9]}}