| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power and resistance simultaneously |
| Difficulty | Standard +0.3 This is a standard M1 work-energy-power question requiring application of P=Fv and F=ma in two contexts. Part (a) involves setting up simultaneous equations from the acceleration condition (routine algebraic manipulation), while part (b) requires resolving forces on an incline at constant speed. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For use of \(P = Fv\) | B1 | \(P = 20F\) or \(P = 25F\) OE (e.g. \(F = \frac{P}{20}\) or \(F = \frac{P}{25}\)). But not with wrong \(F\) substituted (e.g. 6000). |
| Attempt to use Newton's second law in at least one case | M1 | Must have 3 terms. Allow sign errors. Allow \(F\). |
| \(\frac{P}{20} - 6000 = 15000a\) and \(\frac{P}{25} - 6000 = 15000\left(\frac{1}{2}a\right)\) | A1 | OE for both. Allow \(2a'\) and \(a'\). Must be the same \(P\) for both. |
| For solving simultaneously | M1 | Dependent on 2 equations of the correct form with the correct number of relevant terms. Must get to '\(P=\)' or '\(a=\)', but \(P = 200\text{kW}\) or \(200000\text{W}\) with no attempt at \(a\) gets M0. Must be the same \(P\) for both. |
| Power \([= 200\,000\text{W}] = 200\text{kW}\), \(a = \frac{4}{15}\ [\text{ms}^{-2}]\) | A1 | AG. OE awrt 0.267. Do not allow 200000 [W] as final answer. Must show some working when they find \(P\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For use of \(P = Fv\) | B1 | \(P = 20F\) or \(P = 25F\) OE. But not with wrong \(F\) substituted. |
| For one expression for \(P\) in terms of \(a\) only | M1 | Allow sign errors. Need 2 term expression. |
| \((15000a + 6000)\times 20 = (15000\times 0.5a + 6000)\times 25\) | A1 | Correct equation. |
| For solving for \(a\) | M1 | Must get to '\(a=\)'. |
| Power \([= 200\,000\text{W}] = 200\text{kW}\), \(a = \frac{4}{15}\ [\text{ms}^{-2}]\) | A1 | AG. OE awrt 0.267. Must show some working when they find \(P\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For use of \(P = Fv\) | B1 | e.g. \(200\,000 = 20F\) or \(200\,000 = 25F\) OE. e.g. \(F = \frac{200000}{20}[=10000]\) or \(F = \frac{200000}{25}[=8000]\). |
| Attempt to use Newton's second law in at least one case | M1 | Must have 3 terms. Allow sign errors. Allow with \(F\). Allow 200 in place of 200 000. |
| \(\frac{200000}{20} - 6000 = 15000a\) and \(\frac{200000}{25} - 6000 = 15000\left(\frac{1}{2}a\right)\) | A1 | For both. Allow \(2a'\) and \(a'\) here. |
| For solving for \(a\) in both cases | M1 | |
| For showing that both equations lead to \(a = \frac{4}{15}\ [\text{ms}^{-2}]\) | A1 | awrt 0.267. |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For attempt at resolving up hill | M1 | Or \(\frac{200000}{v} - 6000 - 2618 = 0\). May see \(\frac{200000}{8618}\). Must have correct number of terms. Allow sin/cos mix. Allow sign errors. Allow \(g\) missing, but not a different acceleration. Do not allow \(F\). |
| \(\frac{200000}{v} - 6000 - 15000g\sin 1 = 0\) | ||
| Steady speed \(= 23.2\ [\text{m s}^{-1}]\) | A1 | |
| 2 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| For use of $P = Fv$ | B1 | $P = 20F$ or $P = 25F$ OE (e.g. $F = \frac{P}{20}$ or $F = \frac{P}{25}$). But not with wrong $F$ substituted (e.g. 6000). |
| Attempt to use Newton's second law in at least one case | M1 | Must have 3 terms. Allow sign errors. Allow $F$. |
| $\frac{P}{20} - 6000 = 15000a$ and $\frac{P}{25} - 6000 = 15000\left(\frac{1}{2}a\right)$ | A1 | OE for both. Allow $2a'$ and $a'$. Must be the same $P$ for both. |
| For solving simultaneously | M1 | Dependent on 2 equations of the correct form with the correct number of relevant terms. Must get to '$P=$' or '$a=$', but $P = 200\text{kW}$ or $200000\text{W}$ with no attempt at $a$ gets M0. Must be the same $P$ for both. |
| Power $[= 200\,000\text{W}] = 200\text{kW}$, $a = \frac{4}{15}\ [\text{ms}^{-2}]$ | A1 | AG. OE awrt 0.267. Do not allow 200000 [W] as final answer. Must show some working when they find $P$. |
**Alternative method (using two expressions for $P$):**
| Answer | Mark | Guidance |
|--------|------|----------|
| For use of $P = Fv$ | B1 | $P = 20F$ or $P = 25F$ OE. But not with wrong $F$ substituted. |
| For one expression for $P$ in terms of $a$ only | M1 | Allow sign errors. Need 2 term expression. |
| $(15000a + 6000)\times 20 = (15000\times 0.5a + 6000)\times 25$ | A1 | Correct equation. |
| For solving for $a$ | M1 | Must get to '$a=$'. |
| Power $[= 200\,000\text{W}] = 200\text{kW}$, $a = \frac{4}{15}\ [\text{ms}^{-2}]$ | A1 | AG. OE awrt 0.267. Must show some working when they find $P$. |
**Alternative method (using given value $P = 200\text{kW}$):**
| Answer | Mark | Guidance |
|--------|------|----------|
| For use of $P = Fv$ | B1 | e.g. $200\,000 = 20F$ or $200\,000 = 25F$ OE. e.g. $F = \frac{200000}{20}[=10000]$ or $F = \frac{200000}{25}[=8000]$. |
| Attempt to use Newton's second law in at least one case | M1 | Must have 3 terms. Allow sign errors. Allow with $F$. Allow 200 in place of 200 000. |
| $\frac{200000}{20} - 6000 = 15000a$ and $\frac{200000}{25} - 6000 = 15000\left(\frac{1}{2}a\right)$ | A1 | For both. Allow $2a'$ and $a'$ here. |
| For solving for $a$ in both cases | M1 | |
| For showing that both equations lead to $a = \frac{4}{15}\ [\text{ms}^{-2}]$ | A1 | awrt 0.267. |
| | **5** | |
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| For attempt at resolving up hill | M1 | Or $\frac{200000}{v} - 6000 - 2618 = 0$. May see $\frac{200000}{8618}$. Must have correct number of terms. Allow sin/cos mix. Allow sign errors. Allow $g$ missing, but not a different acceleration. Do not allow $F$. |
| $\frac{200000}{v} - 6000 - 15000g\sin 1 = 0$ | | |
| Steady speed $= 23.2\ [\text{m s}^{-1}]$ | A1 | |
| | **2** | |
---4 A lorry of mass 15000 kg moves on a straight horizontal road in the direction from $A$ to $B$. It passes $A$ and $B$ with speeds $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The power of the lorry's engine is constant and there is a constant resistance to motion of magnitude 6000 N . The acceleration of the lorry at $B$ is 0.5 times the acceleration of the lorry at $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the power of the lorry's engine is 200 kW , and hence find the acceleration of the lorry when it is travelling at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
The lorry begins to ascend a straight hill inclined at $1 ^ { \circ }$ to the horizontal. It is given that the power of the lorry's engine and the resistance force do not change.
\item Find the steady speed up the hill that the lorry could maintain.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q4 [7]}}