| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| 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 variable force question requiring Newton's second law with chain rule (F=mv dv/dx), separation of variables, and integration - all routine techniques for this module. The work done calculation in part (ii) follows directly from the given result. While it involves multiple steps and careful algebra, it's a textbook application of well-practiced methods without requiring novel insight. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1200v\frac{\mathrm{d}v}{\mathrm{d}x} = 4000 - \frac{40}{49}v^2\) | M1 | For use of N2L with any expression for \(a\) – numerical/sign slips only |
| A1 | Correct, with any expression for \(a\) | |
| \(\int \frac{1470v}{4900 - v^2}\,\mathrm{d}v = \int \mathrm{d}x\) | M1* | Separate variables leading to \(A\ln(B - Dv^2) = Ex(+c)\) |
| \(-\frac{1470}{2}\ln\ | 4900 - v^2\ | = x + c\) |
| When \(x = 0\), \(v = 10\), so \(c = -735\ln 4800\) | M1dep* | Use correct initial condition leading to \(c = \ldots\) |
| \(v^2 = 4900 - 4800e^{-\frac{1}{735}x}\) | M1 | Using correct log laws (not trivialised) to make \(v^2\) the subject |
| E1 | CAO | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F = \frac{40}{49}v^2\) | ||
| so \(F = \frac{40}{49} \times 4900 - 4800e^{-\frac{1}{735}x}\) | B1 | |
| Work done \(= \int F dx\) | M1 | Integral of their \(F\) |
| \(= \frac{40}{49}\left(4900x + 3528000e^{-\frac{1}{735}x}\right)\) | A1 | CAO (AEF) |
| \(= \frac{40}{49}\left[4900x + 3528000e^{-\frac{1}{735}x}\right]_0^{100}\) | M1 | Integrated expression evaluated between both correct limits; integrated expression is of the form \(Ax + Be^{-\frac{1}{735}x}\) |
| \(= 33649.98057\ldots = 33.6\) kJ to 3 sf | A1 [5] | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total work done by driving force \(= 400000\) J | B1 | |
| Change in KE \(= \frac{1}{2}\times1200\times\left(4900 - 4800e^{-\frac{100}{735}}\right) - \frac{1}{2}\times1200\times10^2\) | M1 | |
| \(= 366350.0194\ldots\) | A1 | |
| Work done against resistance \(=\) total work done by driving force \(-\) change in kinetic energy | M1 | |
| So work done against resistance \(= 400000 - 366350.01\ldots = 33649.98\ldots = 33.6\) kJ to 3 sf | A1 | CAO |
## Question 1(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1200v\frac{\mathrm{d}v}{\mathrm{d}x} = 4000 - \frac{40}{49}v^2$ | M1 | For use of N2L with any expression for $a$ – numerical/sign slips only |
| | A1 | Correct, with any expression for $a$ |
| $\int \frac{1470v}{4900 - v^2}\,\mathrm{d}v = \int \mathrm{d}x$ | M1* | Separate variables leading to $A\ln(B - Dv^2) = Ex(+c)$ |
| $-\frac{1470}{2}\ln\|4900 - v^2\| = x + c$ | A1ft | Integrate, condone missing modulus signs and $+ c$. FT their DE |
| When $x = 0$, $v = 10$, so $c = -735\ln 4800$ | M1dep* | Use correct initial condition leading to $c = \ldots$ |
| $v^2 = 4900 - 4800e^{-\frac{1}{735}x}$ | M1 | Using correct log laws (not trivialised) to make $v^2$ the subject |
| | E1 | CAO |
| **[7]** | | |
## Question 1(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F = \frac{40}{49}v^2$ | | |
| so $F = \frac{40}{49} \times 4900 - 4800e^{-\frac{1}{735}x}$ | B1 | |
| Work done $= \int F dx$ | M1 | Integral of their $F$ |
| $= \frac{40}{49}\left(4900x + 3528000e^{-\frac{1}{735}x}\right)$ | A1 | CAO (AEF) |
| $= \frac{40}{49}\left[4900x + 3528000e^{-\frac{1}{735}x}\right]_0^{100}$ | M1 | Integrated expression evaluated between both correct limits; integrated expression is of the form $Ax + Be^{-\frac{1}{735}x}$ |
| $= 33649.98057\ldots = 33.6$ kJ to 3 sf | A1 [5] | CAO |
**OR:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Total work done by driving force $= 400000$ J | B1 | |
| Change in KE $= \frac{1}{2}\times1200\times\left(4900 - 4800e^{-\frac{100}{735}}\right) - \frac{1}{2}\times1200\times10^2$ | M1 | |
| $= 366350.0194\ldots$ | A1 | |
| Work done against resistance $=$ total work done by driving force $-$ change in kinetic energy | M1 | |
| So work done against resistance $= 400000 - 366350.01\ldots = 33649.98\ldots = 33.6$ kJ to 3 sf | A1 | CAO |
---1 A sports car of mass 1.2 tonnes is being tested on a horizontal, straight section of road. After $t \mathrm {~s}$, the car has travelled $x \mathrm {~m}$ from the starting line and its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The engine produces a driving force of 4000 N and the total resistance to the motion of the car is given by $\frac { 40 } { 49 } v ^ { 2 } \mathrm {~N}$. The car crosses the starting line with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Write down an equation of motion for the car and solve it to show that $v ^ { 2 } = 4900 - 4800 \mathrm { e } ^ { - \frac { 1 } { 735 } x }$.\\
(ii) Hence find the work done against the resistance to motion over the first 100 m beyond the starting line.
\hfill \mbox{\textit{OCR MEI M4 2014 Q1 [12]}}