| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Standard +0.3 This is a standard M2 mechanics problem involving power-force-velocity relationships and Newton's second law on an incline. Part (a) requires setting up power equation P=Fv with forces in equilibrium (constant speed), and part (b) applies F=ma after removing tension. The calculations are straightforward with clearly defined resistances and a simple angle (sin α = 1/20). Slightly easier than average due to the methodical setup and routine application of standard mechanics principles. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Constant speed: \(F = 200 + 800g\sin\alpha + 300 + 1800g\sin\alpha\ (= 1774)\) | M1, A2 | Complete method to equation in \(F\). Requires all terms including resolution of weights. Condone sign errors and sin/cos confusion. \(g\) missing from both weights is a single error. \(-1\) each error: A1A1 no errors, A1A0 one error, A0A0 two or more errors |
| \(40000 = Fv\ (= 1774v)\) | M1 | Use of \(P = Fv\). Allow with \(F\) or their \(F\). Independent of first M1 |
| \(v = 22.5\) | A1 | Accept 23. (maximum 3sf following use of 9.8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{"1774"} - 300 - 1800g\sin\alpha\ (= 592) = 1800a\) | M1, A2 | New equation of motion for the truck. Follow their 1774. Requires all terms including resolution of weights. Condone sign errors and sin/cos confusion. \(-1\) each error: A1A1 no errors, A1A0 one error, A0A0 two or more errors |
| \(F = ma\): \(a = \dfrac{F}{1800} = 0.32888... = 0.33\ (\text{m s}^{-2})\) | A1 | Accept 0.329 |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Constant speed: $F = 200 + 800g\sin\alpha + 300 + 1800g\sin\alpha\ (= 1774)$ | M1, A2 | Complete method to equation in $F$. Requires all terms including resolution of weights. Condone sign errors and sin/cos confusion. $g$ missing from both weights is a single error. $-1$ each error: A1A1 no errors, A1A0 one error, A0A0 two or more errors |
| $40000 = Fv\ (= 1774v)$ | M1 | Use of $P = Fv$. Allow with $F$ or their $F$. Independent of first M1 |
| $v = 22.5$ | A1 | Accept 23. (maximum 3sf following use of 9.8) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{"1774"} - 300 - 1800g\sin\alpha\ (= 592) = 1800a$ | M1, A2 | New equation of motion for the truck. Follow their 1774. Requires all terms including resolution of weights. Condone sign errors and sin/cos confusion. $-1$ each error: A1A1 no errors, A1A0 one error, A0A0 two or more errors |
| $F = ma$: $a = \dfrac{F}{1800} = 0.32888... = 0.33\ (\text{m s}^{-2})$ | A1 | Accept 0.329 |\begin{enumerate}
\item A truck of mass 1800 kg is towing a trailer of mass 800 kg up a straight road which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 1 } { 20 }$. The truck is connected to the trailer by a light inextensible rope which is parallel to the direction of motion of the truck. The resistances to motion of the truck and the trailer from non-gravitational forces are modelled as constant forces of magnitudes 300 N and 200 N respectively. The truck is moving at constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the engine of the truck is working at a rate of 40 kW .\\
(a) Find the value of $v$.
\end{enumerate}
As the truck is moving up the road the rope breaks.\\
(b) Find the acceleration of the truck immediately after the rope breaks.\\
\hfill \mbox{\textit{Edexcel M2 2014 Q4 [9]}}