| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Standard +0.2 This is a straightforward M2 work-energy question requiring standard application of formulas. Part (a) uses work = force × distance with constant speed (so driving force = resistance + component down slope), and part (b) uses P = Fv with F = ma + resistance + component. Both parts are routine calculations with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02c Work by variable force: using integration6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving parallel to the plane | Condone trig confusion | M1 |
| \(D = 900g\sin\theta - 800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Their \(D \times\) distance \(= 1152.8 \times 14 \times 10\) | Independent. For use of \(14 \times 10 \times\) their \(D\) | M1 |
| \(= 161392 = 161\) kJ (160) | Accept 161000 (J), 160000 (J). Ignore incorrect units. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Work done \(= 900gd\sin\theta - 800d\) | Allow with incorrect \(d\) | M1A1 |
| Use of \(d = 14 \times 10\) | Independent – allow in an incorrect expression | M1 |
| \(= 161392 = 161\) kJ (160) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Equation of motion | All terms required. Condone trig confusion and sign errors. Allow with \(900a\) | M1 |
| \(D - 900g\sin\theta - 800 = 900 \times 0.7\) | Correct unsimplified with \(a = 0.7\) used. Accept with their 1152.8 arising from a 2 term expression in (a) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(P = Fv\): \(P = 14 \times \frac{\text{their } D}{1000}\) | Independent. Treat missing 1000 as misread, so allow for \(\frac{14 \times \text{their } D}{1000}\). Allow for \(\frac{P}{14}\) or \(\frac{P}{1000}\) in their equation of motion | M1 |
| \(P = 25.0\) (25) | cao | A1 |
# Question 1
## 1(a)
Resolving parallel to the plane | Condone trig confusion | M1
$D = 900g\sin\theta - 800$ | A1
$900g - 800 = 1152.8$ (N)
$25$
Work done:
Their $D \times$ distance $= 1152.8 \times 14 \times 10$ | Independent. For use of $14 \times 10 \times$ their $D$ | M1
$= 161392 = 161$ kJ (160) | Accept 161000 (J), 160000 (J). Ignore incorrect units. | A1
**Alternative using energy**
Work done $= 900gd\sin\theta - 800d$ | Allow with incorrect $d$ | M1A1
Use of $d = 14 \times 10$ | Independent – allow in an incorrect expression | M1
$= 161392 = 161$ kJ (160) | A1
**(4)**
## 1(b)
Equation of motion | All terms required. Condone trig confusion and sign errors. Allow with $900a$ | M1
$D - 900g\sin\theta - 800 = 900 \times 0.7$ | Correct unsimplified with $a = 0.7$ used. Accept with their 1152.8 arising from a 2 term expression in (a) | A1
$(D - 1152.8 = 900 \times 0.7)$
$D = 1782.8$ (N)
Use of $P = Fv$: $P = 14 \times \frac{\text{their } D}{1000}$ | Independent. Treat missing 1000 as misread, so allow for $\frac{14 \times \text{their } D}{1000}$. Allow for $\frac{P}{14}$ or $\frac{P}{1000}$ in their equation of motion | M1
$P = 25.0$ (25) | cao | A1
**(4)**
**(8 marks)**\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 2018 Q1 [8]}}