| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.3 This is a straightforward M3 variable force question requiring integration of F=ma where force depends on time. Students must apply Newton's second law, integrate to find velocity, use initial conditions to find the constant, then evaluate at t=1.5. The integration is routine (polynomial plus cosine), making this slightly easier than average but still requiring proper method. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.5a = 4 + \cos(\pi t)\) | B1 | |
| Integrating: \(0.5v = 4t + \frac{\sin(\pi t)}{\pi} (+C)\) | M1 A1 | |
| Using boundary values: \(3 = 4 + C \Rightarrow C = -1\) | M1 A1 | |
| When \(t = 1.5\): \(0.5v = 6 - \frac{1}{\pi} - 1\) | M1 | |
| \(v \approx 9.36 \text{ (m s}^{-1})\) | A1 | cao |
| Total | [7] |
# Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.5a = 4 + \cos(\pi t)$ | B1 | |
| Integrating: $0.5v = 4t + \frac{\sin(\pi t)}{\pi} (+C)$ | M1 A1 | |
| Using boundary values: $3 = 4 + C \Rightarrow C = -1$ | M1 A1 | |
| When $t = 1.5$: $0.5v = 6 - \frac{1}{\pi} - 1$ | M1 | |
| $v \approx 9.36 \text{ (m s}^{-1})$ | A1 | cao |
| **Total** | **[7]** | |
---\begin{enumerate}
\item A particle $P$ of mass 0.5 kg is moving along the positive $x$-axis. At time $t$ seconds, $P$ is moving under the action of a single force of magnitude $[ 4 + \cos ( \pi t ) ] \mathrm { N }$, directed away from the origin. When $t = 1$, the particle $P$ is moving away from the origin with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
Find the speed of $P$ when $t = 1.5$, giving your answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel M3 2010 Q1 [7]}}