| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Distance traveled with variable force |
| Difficulty | Challenging +1.2 This is a standard M4 variable force question requiring Newton's second law with v dv/dx = a, separation of variables, and integration with logarithms. Part (a) follows a well-established method (show that result), and part (b) requires solving v dv/dt with partial fractions. While it involves multiple steps and M4-level calculus, the techniques are routine for this module with no novel problem-solving required. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mv\frac{dv}{dx} = -(a + bv^2)\) | M1A1 | Need equation linking speed and displacement |
| \(\int \frac{6v}{a + bv^2}dv = \int -1\,dx\) | M1 | Separating variables |
| \(\frac{3}{b}\ln(a + bv^2) = -x + (C)\) | A1 | Integrating |
| \(X = \frac{3}{b}\Big[\ln(a + bU^2) - \ln(a)\Big] = \frac{3}{b}\ln\left[1 + \frac{bU^2}{a}\right]\) ** | M1A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6\frac{dv}{dt} = -(12 + 3v^2)\) | M1 | Equation connecting \(v\) and \(t\) |
| \(\int \frac{-6}{12 + 3v^2}dv = \int 1\,dt\) | M1, A1 | Separate variables |
| \(\int_U^0 \frac{-2}{4 + v^2}dv = \int_0^U \frac{2}{4 + v^2}dv = T\) | M1 | |
| \(T = \frac{2}{2}\tan^{-1}\frac{U}{2} = \tan^{-1}\frac{U}{2} \ (\text{s})\) | A1 | (5) Total: 11 |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv\frac{dv}{dx} = -(a + bv^2)$ | M1A1 | Need equation linking speed and displacement |
| $\int \frac{6v}{a + bv^2}dv = \int -1\,dx$ | M1 | Separating variables |
| $\frac{3}{b}\ln(a + bv^2) = -x + (C)$ | A1 | Integrating |
| $X = \frac{3}{b}\Big[\ln(a + bU^2) - \ln(a)\Big] = \frac{3}{b}\ln\left[1 + \frac{bU^2}{a}\right]$ ** | M1A1 | (6) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6\frac{dv}{dt} = -(12 + 3v^2)$ | M1 | Equation connecting $v$ and $t$ |
| $\int \frac{-6}{12 + 3v^2}dv = \int 1\,dt$ | M1, A1 | Separate variables |
| $\int_U^0 \frac{-2}{4 + v^2}dv = \int_0^U \frac{2}{4 + v^2}dv = T$ | M1 | |
| $T = \frac{2}{2}\tan^{-1}\frac{U}{2} = \tan^{-1}\frac{U}{2} \ (\text{s})$ | A1 | (5) **Total: 11** |
---5. A particle $Q$ of mass 6 kg is moving along the $x$-axis. At time $t$ seconds the displacement of $Q$ from the origin $O$ is $x$ metres and the speed of $Q$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The particle moves under the action of a retarding force of magnitude ( $a + b v ^ { 2 }$ ) N, where $a$ and $b$ are positive constants. At time $t = 0 , Q$ is at $O$ and moving with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$-direction. The particle $Q$ comes to instantaneous rest at the point $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance $O X$ is
$$\frac { 3 } { b } \ln \left( 1 + \frac { b U ^ { 2 } } { a } \right) \mathrm { m }$$
Given that $a = 12$ and $b = 3$,
\item find, in terms of $U$, the time taken to move from $O$ to $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2011 Q5 [11]}}