| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Potential energy change calculation |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring standard application of work-energy principles. Parts (a)-(c) involve direct formula substitution (PE = mgh, work-energy equation, P = Fv), while part (d) requires one additional step of applying F = ma. All techniques are routine for M1 level with no novel problem-solving required. |
| Spec | 3.03d Newton's second law: 2D vectors6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = 30 \times 20\) | B1 | |
| PE change \(= 1600 \times g \times s \times 0.12\ [= 1600 \times g \times 20 \times 30 \times 0.12]\) | M1 | Attempt change in PE. May use angle \(= 6.9°\). Allow sin/cos error only. |
| Change in PE \(= 1152000\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1960000 = WD_{res} + \text{their PE}\ [1960000 = WD_{res} + 1152000]\ [WD_{res} = 808000\text{ J}]\) | M1 | Using work-energy, allow sign error. |
| \(R = WD_{res} \div 600\) | B1 | Using \(WD_{res} = R \times 600\). |
| Force resisting motion \(= R = 1350\) N to 3sf | A1 | Allow \(R = \dfrac{4040}{3}\) N. Allow \(R\) negative. |
| Alternative: \(DF - R - 1600g \times 0.12 = 0\) | M1 | \(R\) is the resisting force. |
| \(DF = \dfrac{196000}{20 \times 30}\ \left[= \dfrac{9800}{3}\right]\) | B1 | |
| Force resisting motion \(= R = \dfrac{4040}{3} = 1350\) N to 3sf | A1 | Allow \(R\) negative. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = \left(\dfrac{4040}{3} + 1600 \times g \times 0.12\right) \times 20\ \left[= \dfrac{196000}{3}\right]\) | M1 | For using \(P = DF \times v\). Allow use of their \(R\). |
| \(P = 65.3\) kW | A1 | |
| Alternative: \(P = \dfrac{1960000}{30}\) | M1 | For using \(P = \text{Work done} \div \text{Time}\). |
| \(P = 65.3\) kW | A1 | |
| Alternative: \(P = \dfrac{9800}{3} \times 20\) | M1 | For using \(P = DF \times v\). Allow use of their \(DF\). |
| \(P = 65.3\) kW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.85 \times \frac{196000}{3} = DF \times 20\) | B1 FT | \(P = DF \times v\), \(\left[DF = \frac{8330}{3}\right]\); FT on their \(P\) |
| \(DF - R - 1600g \times 0.12 = 1600a\), \(\left[\frac{8330}{3} - \frac{4040}{3} - 1920 = 1600a\right]\) | M1 | Newton's 2nd law, four terms, allow sin/cos error, their \(R\) and their \(DF\) |
| \(a = [-]0.306 \text{ ms}^{-2}\) | A1 | \(a = [-]\frac{490}{1600} = [-]\frac{49}{160}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(9800 = DF \times 20\) | B1 FT | Using reduction in power as cause of deceleration; \(9800 = 0.15 \times \text{their } P = DF \times v\) |
| \(DF = 1600d\), \(\left[\frac{9800}{20} = 1600d\right]\) | M1 | |
| \(a = [-]0.306 \text{ ms}^{-2}\) | A1 | \(a = [-]\frac{490}{1600} = [-]\frac{49}{160}\) |
| [3] |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = 30 \times 20$ | B1 | |
| PE change $= 1600 \times g \times s \times 0.12\ [= 1600 \times g \times 20 \times 30 \times 0.12]$ | M1 | Attempt change in PE. May use angle $= 6.9°$. Allow sin/cos error only. |
| Change in PE $= 1152000$ J | A1 | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1960000 = WD_{res} + \text{their PE}\ [1960000 = WD_{res} + 1152000]\ [WD_{res} = 808000\text{ J}]$ | M1 | Using work-energy, allow sign error. |
| $R = WD_{res} \div 600$ | B1 | Using $WD_{res} = R \times 600$. |
| Force resisting motion $= R = 1350$ N to 3sf | A1 | Allow $R = \dfrac{4040}{3}$ N. Allow $R$ negative. |
| **Alternative:** $DF - R - 1600g \times 0.12 = 0$ | M1 | $R$ is the resisting force. |
| $DF = \dfrac{196000}{20 \times 30}\ \left[= \dfrac{9800}{3}\right]$ | B1 | |
| Force resisting motion $= R = \dfrac{4040}{3} = 1350$ N to 3sf | A1 | Allow $R$ negative. |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = \left(\dfrac{4040}{3} + 1600 \times g \times 0.12\right) \times 20\ \left[= \dfrac{196000}{3}\right]$ | M1 | For using $P = DF \times v$. Allow use of their $R$. |
| $P = 65.3$ kW | A1 | |
| **Alternative:** $P = \dfrac{1960000}{30}$ | M1 | For using $P = \text{Work done} \div \text{Time}$. |
| $P = 65.3$ kW | A1 | |
| **Alternative:** $P = \dfrac{9800}{3} \times 20$ | M1 | For using $P = DF \times v$. Allow use of their $DF$. |
| $P = 65.3$ kW | A1 | |
## Question 5(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.85 \times \frac{196000}{3} = DF \times 20$ | **B1 FT** | $P = DF \times v$, $\left[DF = \frac{8330}{3}\right]$; FT on their $P$ |
| $DF - R - 1600g \times 0.12 = 1600a$, $\left[\frac{8330}{3} - \frac{4040}{3} - 1920 = 1600a\right]$ | **M1** | Newton's 2nd law, four terms, allow sin/cos error, their $R$ and their $DF$ |
| $a = [-]0.306 \text{ ms}^{-2}$ | **A1** | $a = [-]\frac{490}{1600} = [-]\frac{49}{160}$ |
**Alternative method for 5(d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $9800 = DF \times 20$ | **B1 FT** | Using reduction in power as cause of deceleration; $9800 = 0.15 \times \text{their } P = DF \times v$ |
| $DF = 1600d$, $\left[\frac{9800}{20} = 1600d\right]$ | **M1** | |
| $a = [-]0.306 \text{ ms}^{-2}$ | **A1** | $a = [-]\frac{490}{1600} = [-]\frac{49}{160}$ |
| | **[3]** | |
---5 A car of mass 1600 kg travels at constant speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a straight road inclined at an angle of $\sin ^ { - 1 } 0.12$ to the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Find the change in potential energy of the car in 30 s .
\item Given that the total work done by the engine of the car in this time is 1960 kJ , find the constant force resisting the motion.
\item Calculate, in kW , the power developed by the engine of the car.
\item Given that this power is suddenly decreased by $15 \%$, find the instantaneous deceleration of the car.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [11]}}