| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Energy method with work done |
| Difficulty | Challenging +1.2 This is a standard M4/Further Mechanics question requiring work-energy methods and separation of variables. Part (i) is routine application of P=Fv and Newton's second law with chain rule. Part (ii) involves straightforward integration with partial fractions (though the cubic factorization is given by the form). Part (iii) requires separating variables and integrating, which is standard for this module. While it requires multiple techniques and careful algebra, these are well-practiced methods for M4 students with no novel insights required. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{P}{U} = kU^2\) | B1 | |
| \(\frac{P}{v} - kv^2 = mv\frac{dv}{dx}\) | M1 | For use of N2L with any expression for \(a\) (3 terms) |
| \(P\left(1 - \frac{v^3}{U^3}\right) = mv^2\frac{dv}{dx}\) | E1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\int dx = mU^3\int \frac{v^2 dv}{U^3 - v^3}\) | M1 | Separate variables – condone minor slips |
| M1 | Integrate – of the form \(Px = k\ln\ | U^3 - v^3\ |
| \(Px = -\frac{1}{3}mU^3\ln\ | U^3 - v^3\ | (+c)\) |
| \(x=0, v=\frac{1}{4}U\) to find \(c\) and substitute \(v=\frac{1}{2}U\) | M1 | Use initial condition – or for use of limits on definite integral |
| \(Px = -\frac{mU^3}{3}\left[\ln\left(U^3 - \frac{1}{8}U^3\right) - \ln\left(U^3 - \frac{1}{64}U^3\right)\right]\) | A1 | AEF |
| \(x = \frac{mU^3}{3P}\ln\left(\frac{9}{8}\right)\) so \(A = \frac{9}{8}\) | E1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\frac{P}{U^3}v^2 = m\frac{dv}{dt} \Rightarrow \int\frac{dv}{v^2} = -\frac{P}{mU^3}\int dt\) | B1 | Use of \(a = \frac{dv}{dt}\), power \(= 0\) and separate variables |
| \(-\frac{1}{v} = -\frac{P}{mU^3}t + c\), \(t=0, v=\frac{1}{2}U \Rightarrow c = -\frac{2}{U}\) | M1 | Attempt to integrate and use conditions to find \(+c\) |
| \(t = \frac{2mU^2}{P}\) | A1 | If left in terms of \(k\) eg \(t = \frac{2m}{kU}\) then B1M1A0 |
| [3] |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{P}{U} = kU^2$ | B1 | |
| $\frac{P}{v} - kv^2 = mv\frac{dv}{dx}$ | M1 | For use of N2L with any expression for $a$ (3 terms) |
| $P\left(1 - \frac{v^3}{U^3}\right) = mv^2\frac{dv}{dx}$ | E1 | |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\int dx = mU^3\int \frac{v^2 dv}{U^3 - v^3}$ | M1 | Separate variables – condone minor slips |
| | M1 | Integrate – of the form $Px = k\ln\|U^3 - v^3\|(+c)$ cst. $k$ |
| $Px = -\frac{1}{3}mU^3\ln\|U^3 - v^3\|(+c)$ | A1 | |
| $x=0, v=\frac{1}{4}U$ to find $c$ **and** substitute $v=\frac{1}{2}U$ | M1 | Use initial condition – or for use of limits on definite integral |
| $Px = -\frac{mU^3}{3}\left[\ln\left(U^3 - \frac{1}{8}U^3\right) - \ln\left(U^3 - \frac{1}{64}U^3\right)\right]$ | A1 | AEF |
| $x = \frac{mU^3}{3P}\ln\left(\frac{9}{8}\right)$ so $A = \frac{9}{8}$ | E1 | |
| **[6]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\frac{P}{U^3}v^2 = m\frac{dv}{dt} \Rightarrow \int\frac{dv}{v^2} = -\frac{P}{mU^3}\int dt$ | B1 | Use of $a = \frac{dv}{dt}$, power $= 0$ and separate variables |
| $-\frac{1}{v} = -\frac{P}{mU^3}t + c$, $t=0, v=\frac{1}{2}U \Rightarrow c = -\frac{2}{U}$ | M1 | Attempt to integrate and use conditions to find $+c$ |
| $t = \frac{2mU^2}{P}$ | A1 | If left in terms of $k$ eg $t = \frac{2m}{kU}$ then B1M1A0 |
| **[3]** | | |
---1 A car of mass $m$ moves horizontally in a straight line. At time $t$ the car is a distance $x$ from a point O and is moving away from O with speed $v$. There is a force of magnitude $k v ^ { 2 }$, where $k$ is a constant, resisting the motion of the car. The car's engine has a constant power $P$. The terminal speed of the car is $U$.\\
(i) Show that
$$m v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = P \left( 1 - \frac { v ^ { 3 } } { U ^ { 3 } } \right)$$
(ii) Show that the distance moved while the car accelerates from a speed of $\frac { 1 } { 4 } U$ to a speed of $\frac { 1 } { 2 } U$ is
$$\frac { m U ^ { 3 } } { 3 P } \ln A$$
stating the exact value of the constant $A$.
Once the car attains a speed of $\frac { 1 } { 2 } U$, no further power is supplied by the car's engine.\\
(iii) Find, in terms of $m , P$ and $U$, the time taken for the speed of the car to reduce from $\frac { 1 } { 2 } U$ to $\frac { 1 } { 4 } U$.
\hfill \mbox{\textit{OCR MEI M4 2016 Q1 [12]}}