| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of P=Fv at constant speed (part a) and F=ma with power (part b). The calculations are routine with clearly given values and no conceptual subtlety, making it slightly easier than average but still requiring proper method. |
| Spec | 3.03d Newton's second law: 2D vectors6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Tractive force \(= \frac{25000}{20} = 1250\) N | B1 | Seen or implied |
| \(1250 = R + 600g\sin\theta\) | M1 | Equation in \(R\). Condone sign errors and sin/cos confusion |
| \(R = 1250 - 600g \times \frac{1}{16}\) (= 882.5) | A1ft | Correct unsimplified expression for \(R\). Allow with their 1250 |
| \(= 883\) or \(880\) N | A1 (4) | 2 or 3 s.f. 882.5 is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(TF = \frac{30000}{20} = 1500\) N | — | — |
| \(1500 - 600g \times \frac{1}{16} - R = 600a\) | M1 | Equation of motion. Must have all the terms. Condone sign errors and sin/cos confusion |
| A2 | -1 each error | |
| \(a = \frac{1500 - 600\times9.8\div16 - 882.5}{600}\) (= 0.4166…) | — | — |
| \(= 0.42\) or \(0.417\) m s\(^{-1}\) | A1 (4) | 2 or 3 s.f. \(\frac{5}{12}\) is A0 |
# Question 1:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Tractive force $= \frac{25000}{20} = 1250$ N | B1 | Seen or implied |
| $1250 = R + 600g\sin\theta$ | M1 | Equation in $R$. Condone sign errors and sin/cos confusion |
| $R = 1250 - 600g \times \frac{1}{16}$ (= 882.5) | A1ft | Correct unsimplified expression for $R$. Allow with their 1250 |
| $= 883$ or $880$ N | A1 (4) | 2 or 3 s.f. 882.5 is A0 |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $TF = \frac{30000}{20} = 1500$ N | — | — |
| $1500 - 600g \times \frac{1}{16} - R = 600a$ | M1 | Equation of motion. Must have all the terms. Condone sign errors and sin/cos confusion |
| | A2 | -1 each error |
| $a = \frac{1500 - 600\times9.8\div16 - 882.5}{600}$ (= 0.4166…) | — | — |
| $= 0.42$ or $0.417$ m s$^{-1}$ | A1 (4) | 2 or 3 s.f. $\frac{5}{12}$ is A0 |
**Total: [8]**
> $882.5$ and $\frac{5}{12}$ is A0 at the end of (a) and A1 at the end of (b) – penalise once only.
> Use of $9.81$ is an accuracy error – penalise at the end of the first part affected.
---\begin{enumerate}
\item A van of mass 600 kg is moving up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 16 }$. The resistance to motion of the van from non-gravitational forces has constant magnitude $R$ newtons. When the van is moving at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the van's engine is working at a constant rate of 25 kW .\\
(a) Find the value of $R$.
\end{enumerate}
The power developed by the van's engine is now increased to 30 kW . The resistance to motion from non-gravitational forces is unchanged. At the instant when the van is moving up the road at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the acceleration of the van is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(b) Find the value of $a$.\\
\hfill \mbox{\textit{Edexcel M2 2014 Q1 [8]}}