Standard +0.3 This is a standard M2 variable force problem requiring application of P=Fv and F=ma in two scenarios to form simultaneous equations. The method is routine (set up force equations, substitute P/v for driving force, solve system), though the two-scenario setup and algebraic manipulation add modest complexity beyond basic recall.
4. A truck of mass 900 kg is moving along a straight horizontal road with the engine of the truck working at a constant rate of \(P\) watts. The resistance to the motion of the truck is modelled as a constant force of magnitude \(R\) newtons.
At the instant when the speed of the truck is \(15 \mathrm {~ms} ^ { - 1 }\), the deceleration of the truck is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
Later the same truck is moving down a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 30 }\). The resistance to the motion of the truck is again modelled as a constant force of magnitude \(R\) newtons. The engine of the truck is again working at a constant rate of \(P\) watts.
At the instant when the speed of the truck is \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the acceleration of the truck is \(0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
Find the value of \(R\).
Formula with a speed substituted correctly. At least once
Equation for horizontal motion
M1
Dimensionally correct in \(P\) or \(F\). Condone sign errors. Need all terms
\(\frac{P}{15} - R = -0.2 \times 900 \quad \left(\frac{P}{15} - R = -180\right)\)
A1
Correct unsimplified equation in \(P\) and \(R\)
Equation for motion down hill
M1
Dimensionally correct in \(P\) or \(F_D\). Condone sign errors. Condone sin/cos confusion. Need all terms. M0 if using \(F(\text{down}) = F(\text{horizontal})\)
Unsimplified equation in \(F_D\) or \(P\) and \(R\) with at most one error
\(\left(\frac{P}{12} + 30g - R = 360\right) \quad \left(\frac{P}{12} = R + 66\right)\)
A1
Correct unsimplified equation in \((P\) and\()\) \(R\) with trig substituted. e.g. \(\frac{5}{4}(R-180) = 360 - 30g + R\)
Solve for \(R\)
DM1
Dependent on the 3 preceding M marks. Condone slips in the algebra
\(R = 1160\) or \(R = 1200\)
A1
3 sf or 2 sf only. NB answer follows use of 9.8, so final answer 1164 is A0. Clear use of 9.81 is a rubric infringement giving \((P=14742\) and\()\) \(R=1162.8\), scores maximum 7/8 (final A0)
Total [8]
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $F = \frac{P}{v}$ | M1 | Formula with a speed substituted correctly. At least once |
| Equation for horizontal motion | M1 | Dimensionally correct in $P$ or $F$. Condone sign errors. Need all terms |
| $\frac{P}{15} - R = -0.2 \times 900 \quad \left(\frac{P}{15} - R = -180\right)$ | A1 | Correct unsimplified equation in $P$ and $R$ |
| Equation for motion down hill | M1 | Dimensionally correct in $P$ or $F_D$. Condone sign errors. Condone sin/cos confusion. Need all terms. M0 if using $F(\text{down}) = F(\text{horizontal})$ |
| $F_D + 900g \times \sin\theta - R = 900 \times 0.4$ | A1 | Unsimplified equation in $F_D$ or $P$ and $R$ with at most one error |
| $\left(\frac{P}{12} + 30g - R = 360\right) \quad \left(\frac{P}{12} = R + 66\right)$ | A1 | Correct unsimplified equation in $(P$ and$)$ $R$ with trig substituted. e.g. $\frac{5}{4}(R-180) = 360 - 30g + R$ |
| Solve for $R$ | DM1 | Dependent on the 3 preceding M marks. Condone slips in the algebra |
| $R = 1160$ or $R = 1200$ | A1 | 3 sf or 2 sf only. NB answer follows use of 9.8, so final answer 1164 is A0. Clear use of 9.81 is a rubric infringement giving $(P=14742$ and$)$ $R=1162.8$, scores maximum 7/8 (final A0) |
| **Total [8]** | | |
---
4. A truck of mass 900 kg is moving along a straight horizontal road with the engine of the truck working at a constant rate of $P$ watts. The resistance to the motion of the truck is modelled as a constant force of magnitude $R$ newtons.\\
At the instant when the speed of the truck is $15 \mathrm {~ms} ^ { - 1 }$, the deceleration of the truck is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
Later the same truck is moving down a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 30 }$. The resistance to the motion of the truck is again modelled as a constant force of magnitude $R$ newtons. The engine of the truck is again working at a constant rate of $P$ watts.\\
At the instant when the speed of the truck is $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the acceleration of the truck is $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
Find the value of $R$.\\
\hfill \mbox{\textit{Edexcel M2 2022 Q4 [8]}}