| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Single particle, Newton's second law – scalar (1D, horizontal or inclined) |
| Difficulty | Easy -1.2 This is a straightforward application of Newton's second law and the power formula P=Fv. Part (a) requires F=ma+resistance (one-step calculation), and part (b) requires recognizing that constant speed means driving force equals resistance, then P=Fv. Both parts are direct formula application with no problem-solving insight needed, making this easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F - 900 = 4000 \times 0.5\) | M1 | M1 for use of Newton's second law, 3 terms |
| \(F = 2900 \text{ N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(900 \times 25\) | M1 | M1 for use of \(P = Fv\) with \(F =\) resistance only |
| \(22\,500 \text{ W}\) or \(22.5 \text{ kW}\) | A1 |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F - 900 = 4000 \times 0.5$ | M1 | M1 for use of Newton's second law, 3 terms |
| $F = 2900 \text{ N}$ | A1 | |
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**Question 2(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $900 \times 25$ | M1 | M1 for use of $P = Fv$ with $F =$ resistance only |
| $22\,500 \text{ W}$ or $22.5 \text{ kW}$ | A1 | |2 A minibus of mass 4000 kg is travelling along a straight horizontal road. The resistance to motion is 900 N .
\begin{enumerate}[label=(\alph*)]
\item Find the driving force when the acceleration of the minibus is $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the power required for the minibus to maintain a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q2 [4]}}