| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| 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 to find velocity, then integration again to find displacement. The algebra is simple (linear force function, basic integration), and both parts follow a standard textbook procedure with no conceptual challenges beyond applying Newton's second law twice. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.03c Newton's second law: F=ma one dimension6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.6a = \left(3t + \frac{1}{2}\right)\), so \(\frac{dv}{dt} = \frac{5}{3}\left(3t + \frac{1}{2}\right)\) | M1 | Form equation of motion with acceleration as \(dv/dt\); must include mass. |
| \(v = \frac{5}{2}t^2 + \frac{5}{6}t \quad (+c)\); \((t=0, v=0 \Rightarrow c=0)\) | A1 (2) | Integrate and complete; correct expression even if no reference to constant. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{10}{3} = \frac{5}{2}t^2 + \frac{5}{6}t\) | M1 | Set expression equal to \(\frac{10}{3}\). |
| \(3t^2 + t - 4 = 0 \Rightarrow (3t+4)(t-1)=0\), so \(t=1\) at \(A\) | dM1A1 | Factorise/formula to reach \(t = ...\); correct value of \(t\). |
| \(s = \left[\frac{5t^3}{6} + \frac{5t^2}{12}\right]_0^1 = \frac{5}{4}\) m | M1, A1 (5) [7] | Integrate \(v\); lower limit 0, upper limit their \(t\). Correct answer \(\frac{5}{4}\) or 1.25 m. |
# Question 2:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.6a = \left(3t + \frac{1}{2}\right)$, so $\frac{dv}{dt} = \frac{5}{3}\left(3t + \frac{1}{2}\right)$ | M1 | Form equation of motion with acceleration as $dv/dt$; must include mass. |
| $v = \frac{5}{2}t^2 + \frac{5}{6}t \quad (+c)$; $(t=0, v=0 \Rightarrow c=0)$ | A1 (2) | Integrate and complete; correct expression even if no reference to constant. |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{10}{3} = \frac{5}{2}t^2 + \frac{5}{6}t$ | M1 | Set expression equal to $\frac{10}{3}$. |
| $3t^2 + t - 4 = 0 \Rightarrow (3t+4)(t-1)=0$, so $t=1$ at $A$ | dM1A1 | Factorise/formula to reach $t = ...$; correct value of $t$. |
| $s = \left[\frac{5t^3}{6} + \frac{5t^2}{12}\right]_0^1 = \frac{5}{4}$ m | M1, A1 (5) [7] | Integrate $v$; lower limit 0, upper limit their $t$. Correct answer $\frac{5}{4}$ or 1.25 m. |
---2. A particle $P$ of mass 0.6 kg is moving along the positive $x$-axis in the positive direction. The only force acting on $P$ acts in the direction of $x$ increasing and has magnitude $\left( 3 t + \frac { 1 } { 2 } \right) \mathrm { N }$, where $t$ seconds is the time after $P$ leaves the origin $O$.
When $t = 0 , P$ is at rest at $O$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $t$, for the velocity of $P$ at time $t$ seconds.
The particle passes through the point $A$ with speed $\frac { 10 } { 3 } \mathrm {~ms} ^ { - 1 }$.
\item Find the distance $O A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q2 [7]}}