| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.8 This M3 variable force question requires integrating F=ma with force as a function of time, then integrating velocity to find displacement. While the integration itself is straightforward (involving 1/(t+2)² and its antiderivative), students must correctly handle the negative force direction, apply initial conditions carefully, and work through two successive integrations—more demanding than standard M1/M2 questions but routine for M3 level. |
| Spec | 1.08h Integration by substitution3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(0.6a = -\frac{12}{(t+2)^2}\) | M1 | |
| \(0.6\int dv = -\int\frac{12}{(t+2)^2}\,dt\) | ||
| \(0.6v = \frac{12}{(t+2)} (+c)\) | M1depA1 | |
| \(t=0,\ v=15\): \(0.6\times15 = 6+c \Rightarrow c=3\) | M1dep | |
| \(v = \frac{20}{(t+2)}+5 = 5\left(\frac{4}{t+2}+1\right)\) | A1 | Given/shown result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{dx}{dt} = 5\left(\frac{4}{t+2}+1\right)\) | M1 | |
| \(x = \int 5\left(\frac{4}{t+2}+1\right)dt\) | ||
| \(x = 5\left(4\ln(t+2)+t\right)(+c')\) | M1depA1 | |
| \(t=0,\ x=0\): \(c' = -20\ln 2\) | ||
| \(t=5\): \(x = 5(4\ln 7+5)-20\ln 2 = 50.05... = 50.1\) or better | M1dep | |
| or \(20\ln\left(\frac{7}{2}\right)+25\) | A1 |
## Question 3:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $0.6a = -\frac{12}{(t+2)^2}$ | M1 | |
| $0.6\int dv = -\int\frac{12}{(t+2)^2}\,dt$ | | |
| $0.6v = \frac{12}{(t+2)} (+c)$ | M1depA1 | |
| $t=0,\ v=15$: $0.6\times15 = 6+c \Rightarrow c=3$ | M1dep | |
| $v = \frac{20}{(t+2)}+5 = 5\left(\frac{4}{t+2}+1\right)$ | A1 | Given/shown result |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dx}{dt} = 5\left(\frac{4}{t+2}+1\right)$ | M1 | |
| $x = \int 5\left(\frac{4}{t+2}+1\right)dt$ | | |
| $x = 5\left(4\ln(t+2)+t\right)(+c')$ | M1depA1 | |
| $t=0,\ x=0$: $c' = -20\ln 2$ | | |
| $t=5$: $x = 5(4\ln 7+5)-20\ln 2 = 50.05... = 50.1$ or better | M1dep | |
| or $20\ln\left(\frac{7}{2}\right)+25$ | A1 | |
---\begin{enumerate}
\item A particle $P$ of mass 0.6 kg is moving along the $x$-axis in the positive direction. At time $t = 0 , P$ passes through the origin $O$ with speed $15 \mathrm {~ms} ^ { - 1 }$. At time $t$ seconds the distance $O P$ is $x$ metres, the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the resultant force acting on $P$ has magnitude $\frac { 12 } { ( t + 2 ) ^ { 2 } }$ newtons. The resultant force is directed towards $O$.\\
(a) Show that $v = 5 \left( \frac { 4 } { t + 2 } + 1 \right)$.\\
(b) Find the value of $x$ when $t = 5$\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q3 [10]}}