| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work-energy over time interval |
| Difficulty | Standard +0.3 This is a standard multi-part work-energy question requiring application of P=Fv, work-energy theorem, and equations of motion. While it involves several steps and careful bookkeeping across three sections, each part uses routine A-level mechanics techniques without requiring novel insight or complex problem-solving strategies. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | 36000 = 800v | M1 |
| v = 45ms–1 | A1 | Speed of the car |
| AB = 45 × 120 = 5400m | A1 | |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | −800 = 900a [a = –8/9] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | M1 | Using v2 =u2 +2as |
| v=35 ms–1 | A1 | Speed of the car at C |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.5 × 900 × (45 – v2) | M1 | Attempt change in KE |
| 0.5 × 900 × (45 – v2) = 800 × 450 | M1 | KE loss = WD against Friction |
| v=35 ms–1 | A1 | Speed of the car at C |
| Total: | 3 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iii) | CD = 6637.5 – 5400 – 450 = 787.5 | B1 |
| 0 = 352 – 2d × 787.5 | M1 | Using v2 =u2 +2as , a = –d |
| d = 7/9 = 0.778ms–2 | A1 | d = deceleration |
| P = 900 × (7/9) = 700 | A1 | Using F = ma |
| Total: | 4 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | 36000 = 800v | M1 | Using P = Fv
v = 45ms–1 | A1 | Speed of the car
AB = 45 × 120 = 5400m | A1
Total: | 3
--- 4(ii) ---
4(ii) | −800 = 900a [a = –8/9] | M1 | Using Newton’s 2nd law
16
v2 =452 − ×450
9 | M1 | Using v2 =u2 +2as
v=35 ms–1 | A1 | Speed of the car at C
Total: | 3
Alternative method for Question 4(ii)
0.5 × 900 × (45 – v2) | M1 | Attempt change in KE
0.5 × 900 × (45 – v2) = 800 × 450 | M1 | KE loss = WD against Friction
v=35 ms–1 | A1 | Speed of the car at C
Total: | 3
Question | Answer | Marks | Guidance
--- 4(iii) ---
4(iii) | CD = 6637.5 – 5400 – 450 = 787.5 | B1
0 = 352 – 2d × 787.5 | M1 | Using v2 =u2 +2as , a = –d
d = 7/9 = 0.778ms–2 | A1 | d = deceleration
P = 900 × (7/9) = 700 | A1 | Using F = ma
Total: | 4
Question | Answer | Marks | GuidanceA car of mass $900$ kg is moving on a straight horizontal road $ABCD$. There is a constant resistance of magnitude $800$ N in the sections $AB$ and $BC$, and a constant resistance of magnitude $R$ N in the section $CD$. The power of the car's engine is a constant $36$ kW.
\begin{enumerate}[label=(\roman*)]
\item The car moves from $A$ to $B$ at a constant speed in $120$ s. Find the speed of the car and the distance $AB$. [3]
\item The distance $BC$ is $450$ m. Find the speed of the car at $C$. [3]
\item The car comes to rest at $D$. The distance $AD$ is $6637.5$ m. Find the deceleration of the car and the value of $R$. [4]
\end{enumerate}
The car's engine is switched off at $B$.
\hfill \mbox{\textit{CAIE M1 2017 Q4 [10]}}