| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | January |
| Marks | 12 |
| 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 where force depends on time, recognizing that velocity approaches a limit as the force term vanishes, then integrating velocity for displacement. While the integration itself is straightforward (standard form), the multi-step process (force→acceleration→velocity→displacement) and the conceptual understanding of limiting behavior elevate this above routine mechanics problems. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(800\frac{dv}{dt} = \frac{48000}{(t+2)^2}\) | M1 | |
| \(v = 60\int\frac{dt}{(t+2)^2} = \frac{-60}{(t+2)} (+c)\) | M1 A1 | |
| \(t=0, v=0 \Rightarrow c = 30\) | M1 A1 | |
| \(v = 30 - \frac{60}{(t+2)} \Rightarrow v \to 30\) as \(t \to \infty\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(s = \int v\,dt = 30t - 60\ln(t+2)\ (+c)\) | M1 A1 | |
| Substitute \(t=0\) and \(t=6\) | M1 | |
| \(s = 180 - 60\ln 8 - (-60\ln 2)\) | A1, A1 | |
| \(s \approx 96.8\) m | A1 | (6) |
# Question 5:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $800\frac{dv}{dt} = \frac{48000}{(t+2)^2}$ | M1 | |
| $v = 60\int\frac{dt}{(t+2)^2} = \frac{-60}{(t+2)} (+c)$ | M1 A1 | |
| $t=0, v=0 \Rightarrow c = 30$ | M1 A1 | |
| $v = 30 - \frac{60}{(t+2)} \Rightarrow v \to 30$ as $t \to \infty$ | A1 | (6) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $s = \int v\,dt = 30t - 60\ln(t+2)\ (+c)$ | M1 A1 | |
| Substitute $t=0$ and $t=6$ | M1 | |
| $s = 180 - 60\ln 8 - (-60\ln 2)$ | A1, A1 | |
| $s \approx 96.8$ m | A1 | (6) |
---5. A car of mass 800 kg moves along a horizontal straight road. At time $t$ seconds, the resultant force acting on the car has magnitude $\frac { 48000 } { ( t + 2 ) ^ { 2 } }$ newtons, acting in the direction of the motion of the car. When $t = 0$, the car is at rest.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the car approaches a limiting value as $t$ increases and find this value.
\item Find the distance moved by the car in the first 6 seconds of its motion.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q5 [12]}}