| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard application of work-energy principles and power equations. Part (a) involves calculating work done against resistance and gravity over a known distance (constant speed means forces balance, then multiply by distance). Part (b) uses P=Fv where F comes from Newton's second law with the given acceleration. Both parts are routine applications of standard formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Resolving parallel to the plane: \(D = 900g\sin\theta + 800\) | M1 | Condone trig confusion |
| \(\frac{900}{25}g + 800 (= 1152.8)\) (N) | A1 | |
| Work done: Their \(D \times\) distance \(= 1152.8 \times 14 \times 10\) | M1 | Independent. For use of \(14 \times 10 \times\) their \(D\) |
| \(= 161392 = 161\) kJ (160) | A1 | Accept 161000 (J), 160000 (J). Ignore incorrect units. |
| Alternative using energy: | ||
| Work done \(= 900gd\sin\theta + 800d\) | M1A1 | Allow with incorrect \(d\) |
| Use of \(d = 14 \times 10\) | M1 | Independent – allow in an incorrect expression |
| \(= 161392 = 161\) kJ (160) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Equation of motion | M1 | All terms required. Condone trig confusion and sign errors. Allow with \(900a\) |
| \(D - 900g\sin\theta - 800 = 900 \times 0.7\) | A1 | Correct unsimplified with \(a = 0.7\) used. Accept with their 1152.8 arising from a 2 term expression in (a). |
| \((D - 1152.8 = 900 \times 0.7)\) | ||
| \(D = 1782.8\) (N) | ||
| Use of \(P = Fv\): \(P = 14 \times \frac{\text{their } D}{1000}\) | M1 | Independent. Treat missing 1000 as misread, so allow for \(14 \times \text{their } D\). Allow for \(\frac{1000P}{14}\) (or \(\frac{P}{14}\)) in their equation of motion |
| \(P = 25.0\) (25) | A1 | CAO |
# Question 1:
## Part 1a:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Resolving parallel to the plane: $D = 900g\sin\theta + 800$ | M1 | Condone trig confusion |
| $\frac{900}{25}g + 800 (= 1152.8)$ (N) | A1 | |
| Work done: Their $D \times$ distance $= 1152.8 \times 14 \times 10$ | M1 | Independent. For use of $14 \times 10 \times$ their $D$ |
| $= 161392 = 161$ kJ (160) | A1 | Accept 161000 (J), 160000 (J). Ignore incorrect units. |
| **Alternative using energy:** | | |
| Work done $= 900gd\sin\theta + 800d$ | M1A1 | Allow with incorrect $d$ |
| Use of $d = 14 \times 10$ | M1 | Independent – allow in an incorrect expression |
| $= 161392 = 161$ kJ (160) | A1 | |
## Part 1b:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Equation of motion | M1 | All terms required. Condone trig confusion and sign errors. Allow with $900a$ |
| $D - 900g\sin\theta - 800 = 900 \times 0.7$ | A1 | Correct unsimplified with $a = 0.7$ used. Accept with their 1152.8 arising from a 2 term expression in (a). |
| $(D - 1152.8 = 900 \times 0.7)$ | | |
| $D = 1782.8$ (N) | | |
| Use of $P = Fv$: $P = 14 \times \frac{\text{their } D}{1000}$ | M1 | Independent. Treat missing 1000 as misread, so allow for $14 \times \text{their } D$. Allow for $\frac{1000P}{14}$ (or $\frac{P}{14}$) in their equation of motion |
| $P = 25.0$ (25) | A1 | CAO |
---\begin{enumerate}
\item A car of mass 900 kg is travelling up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 25 }$. The car is travelling at a constant speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the resistance to motion from non-gravitational forces has a constant magnitude of 800 N . The car takes 10 seconds to travel from $A$ to $B$, where $A$ and $B$ are two points on the road.\\
(a) Find the work done by the engine of the car as the car travels from $A$ to $B$.
\end{enumerate}
When the car is at $B$ and travelling at a speed of $14 \mathrm {~ms} ^ { - 1 }$ the rate of working of the engine of the car is suddenly increased to $P \mathrm {~kW}$, resulting in an initial acceleration of the car of $0.7 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The resistance to motion from non-gravitational forces still has a constant magnitude of 800 N .\\
(b) Find the value of $P$.\\
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\hfill \mbox{\textit{Edexcel M2 2016 Q1 [8]}}