| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable mass problems (mass increasing) |
| Difficulty | Challenging +1.8 This is a challenging M4 variable mass mechanics problem requiring derivation of equations of motion using F=dp/dt, solving differential equations with exponential and polynomial mass models, and comparing models. It demands strong calculus skills, physical insight into variable mass systems, and multi-step algebraic manipulation across four interconnected parts, placing it well above average difficulty but within reach of well-prepared Further Maths students. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((m+\delta m)(v+\delta v)-(mv+\delta m(0))=(m+\delta m)g\delta t\) | M1, A1 | Impulse = change in momentum; Condone lack of \(\delta m\) on rhs |
| \(m\frac{\delta v}{\delta t}+v\frac{\delta m}{\delta t}+\frac{\delta m}{\delta t}\frac{\delta v}{\delta t}\delta t=mg+g\frac{\delta m}{\delta t}\delta t\) | M1 | Form differential equation |
| \(mv\frac{\mathrm{d}v}{\mathrm{d}x}+v^2\frac{\mathrm{d}m}{\mathrm{d}x}=mg\) | E1 | Complete argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}}{\mathrm{d}t}(mv)=mg\) | M1A1 | |
| \(\frac{\mathrm{d}}{\mathrm{d}t}(mv)=mg \Rightarrow m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}=mg\) | A1 | |
| \(mv\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}m}{\mathrm{d}t} \Rightarrow mv\frac{\mathrm{d}v}{\mathrm{d}x}+v^2\frac{\mathrm{d}m}{\mathrm{d}x}=mg\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}m}{\mathrm{d}x}=m_0k_1e^{k_1x}\) | B1 | |
| \(v\frac{\mathrm{d}v}{\mathrm{d}x}=g-k_1v^2 \Rightarrow \int\frac{v\,\mathrm{d}v}{g-k_1v^2}=\int\mathrm{d}x\) | M1 | Substitute \(m\) and \(\frac{\mathrm{d}m}{\mathrm{d}t}\) and separate variables |
| M1 | Integrate of the form \(\lambda\ln(g-k_1v^2)=x(+c)\), cst. \(\lambda\) | |
| \(-\frac{1}{2k_1}\ln(g-k_1v^2)=x+c\) | A1 | Including \(+c\) |
| \(x=0, v=0 \Rightarrow c=-\frac{1}{2k_1}\ln g\) | M1 | Use conditions to obtain \(c\) and attempt to make \(v^2\) the subject |
| \(v^2=\frac{g}{k_1}(1-e^{-2k_1x})\) | E1 | www after correct integration |
| Terminal velocity is \(\sqrt{\frac{g}{k_1}}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}}{\mathrm{d}x}\left[v^2(1+k_2x)^2\right]=\) | M1 | Differentiate \(v^2(1+k_2x)^2\) using product rule |
| \(2v\frac{\mathrm{d}v}{\mathrm{d}x}(1+k_2x)^2+2k_2v^2(1+k_2x)\) | A1 A1 | A1 for each term |
| \(\frac{\mathrm{d}m}{\mathrm{d}x}=k_2m_0 \Rightarrow (1+k_2x)\frac{\mathrm{d}v}{\mathrm{d}x}+k_2v^2=(1+k_2x)g\) | B1 | Correct \(\frac{\mathrm{d}m}{\mathrm{d}x}\) and substitute both \(m\) and \(\frac{\mathrm{d}m}{\mathrm{d}x}\) into DE |
| \(\frac{\mathrm{d}}{\mathrm{d}x}\left[v^2(1+k_2x)^2\right]=2g(1+k_2x)^2\) | E1 | Working must be clear (dependent on all previous marks) |
| \(v^2(1+k_2x)^2=2g\int(1+k_2x)^2\,\mathrm{d}x\) | B1 | |
| \(=2g\left(\frac{(1+k_2x)^3}{3k_2}\right)(+c)\) | B1 | |
| \(v=0, x=0 \Rightarrow c=-\frac{2g}{3k_2}\) | M1 | Use conditions to find \(+c\) or limits from 0 to \(x\) on definite integral |
| \(v^2=\frac{2g}{3k_2}\left[1+k_2x-\frac{1}{(1+k_2x)^2}\right]\) | A1 | oe eg \(v^2=\frac{2gx\!\left(1+k_2x+\frac{1}{3}k_2^2x^2\right)}{(1+k_2x)^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| First model: \(x=\frac{1}{k_1}\ln 2\) | B1 | |
| Second model: \(x=\frac{1}{k_2}\) | B1 | |
| \(\frac{2g}{3k_2}\!\left(1+k_2\!\left(\frac{1}{k_2}\right)\right)-\frac{2g}{3k_2\!\left(1+k_2\!\left(\frac{1}{k_2}\right)\right)^2}=\frac{g}{k_1}\!\left(1-e^{-2k_1\!\left(\frac{1}{k_1}\ln2\right)}\right)\) | M1 | Substitute both \(x\) values and equate |
| \(\frac{k_1}{k_2}=\frac{9}{14}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| First model: \(e^{k_1x}=2\) | B1 | |
| Second model: \(k_2x=1\) | B1 | |
| \(\frac{3g}{4k_1}=\frac{7g}{6k_2}\) | M1 | Substituting for both and equating; \(v^2=\frac{3g}{4k_1}\) (1st model) and \(v^2=\frac{7g}{6k_2}\) (2nd model) |
| \(\frac{k_1}{k_2}=\frac{9}{14}\) | A1 |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(m+\delta m)(v+\delta v)-(mv+\delta m(0))=(m+\delta m)g\delta t$ | M1, A1 | Impulse = change in momentum; Condone lack of $\delta m$ on rhs |
| $m\frac{\delta v}{\delta t}+v\frac{\delta m}{\delta t}+\frac{\delta m}{\delta t}\frac{\delta v}{\delta t}\delta t=mg+g\frac{\delta m}{\delta t}\delta t$ | M1 | Form differential equation |
| $mv\frac{\mathrm{d}v}{\mathrm{d}x}+v^2\frac{\mathrm{d}m}{\mathrm{d}x}=mg$ | E1 | Complete argument |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}}{\mathrm{d}t}(mv)=mg$ | M1A1 | |
| $\frac{\mathrm{d}}{\mathrm{d}t}(mv)=mg \Rightarrow m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}=mg$ | A1 | |
| $mv\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}m}{\mathrm{d}t} \Rightarrow mv\frac{\mathrm{d}v}{\mathrm{d}x}+v^2\frac{\mathrm{d}m}{\mathrm{d}x}=mg$ | E1 | |
---
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}m}{\mathrm{d}x}=m_0k_1e^{k_1x}$ | B1 | |
| $v\frac{\mathrm{d}v}{\mathrm{d}x}=g-k_1v^2 \Rightarrow \int\frac{v\,\mathrm{d}v}{g-k_1v^2}=\int\mathrm{d}x$ | M1 | Substitute $m$ and $\frac{\mathrm{d}m}{\mathrm{d}t}$ and separate variables |
| | M1 | Integrate of the form $\lambda\ln(g-k_1v^2)=x(+c)$, cst. $\lambda$ |
| $-\frac{1}{2k_1}\ln(g-k_1v^2)=x+c$ | A1 | Including $+c$ |
| $x=0, v=0 \Rightarrow c=-\frac{1}{2k_1}\ln g$ | M1 | Use conditions to obtain $c$ and attempt to make $v^2$ the subject |
| $v^2=\frac{g}{k_1}(1-e^{-2k_1x})$ | E1 | www after correct integration |
| Terminal velocity is $\sqrt{\frac{g}{k_1}}$ | B1 | |
---
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}}{\mathrm{d}x}\left[v^2(1+k_2x)^2\right]=$ | M1 | Differentiate $v^2(1+k_2x)^2$ using product rule |
| $2v\frac{\mathrm{d}v}{\mathrm{d}x}(1+k_2x)^2+2k_2v^2(1+k_2x)$ | A1 A1 | A1 for each term |
| $\frac{\mathrm{d}m}{\mathrm{d}x}=k_2m_0 \Rightarrow (1+k_2x)\frac{\mathrm{d}v}{\mathrm{d}x}+k_2v^2=(1+k_2x)g$ | B1 | Correct $\frac{\mathrm{d}m}{\mathrm{d}x}$ and substitute both $m$ and $\frac{\mathrm{d}m}{\mathrm{d}x}$ into DE |
| $\frac{\mathrm{d}}{\mathrm{d}x}\left[v^2(1+k_2x)^2\right]=2g(1+k_2x)^2$ | E1 | Working must be clear (dependent on all previous marks) |
| $v^2(1+k_2x)^2=2g\int(1+k_2x)^2\,\mathrm{d}x$ | B1 | |
| $=2g\left(\frac{(1+k_2x)^3}{3k_2}\right)(+c)$ | B1 | |
| $v=0, x=0 \Rightarrow c=-\frac{2g}{3k_2}$ | M1 | Use conditions to find $+c$ or limits from 0 to $x$ on definite integral |
| $v^2=\frac{2g}{3k_2}\left[1+k_2x-\frac{1}{(1+k_2x)^2}\right]$ | A1 | oe eg $v^2=\frac{2gx\!\left(1+k_2x+\frac{1}{3}k_2^2x^2\right)}{(1+k_2x)^2}$ |
---
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| First model: $x=\frac{1}{k_1}\ln 2$ | B1 | |
| Second model: $x=\frac{1}{k_2}$ | B1 | |
| $\frac{2g}{3k_2}\!\left(1+k_2\!\left(\frac{1}{k_2}\right)\right)-\frac{2g}{3k_2\!\left(1+k_2\!\left(\frac{1}{k_2}\right)\right)^2}=\frac{g}{k_1}\!\left(1-e^{-2k_1\!\left(\frac{1}{k_1}\ln2\right)}\right)$ | M1 | Substitute both $x$ values and equate |
| $\frac{k_1}{k_2}=\frac{9}{14}$ | A1 | |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| First model: $e^{k_1x}=2$ | B1 | |
| Second model: $k_2x=1$ | B1 | |
| $\frac{3g}{4k_1}=\frac{7g}{6k_2}$ | M1 | Substituting for both and equating; $v^2=\frac{3g}{4k_1}$ (1st model) and $v^2=\frac{7g}{6k_2}$ (2nd model) |
| $\frac{k_1}{k_2}=\frac{9}{14}$ | A1 | |4 A raindrop falls from rest through a stationary cloud. The raindrop has mass $m$ and speed $v$ when it has fallen a distance $x$. You may assume that resistances to motion are negligible.\\
(i) Derive the equation of motion
$$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } + v ^ { 2 } \frac { \mathrm {~d} m } { \mathrm {~d} x } = m g .$$
Initially the mass of the raindrop is $m _ { 0 }$. Two different models for the mass of the raindrop are suggested.\\
In the first model $m = m _ { 0 } \mathrm { e } ^ { k _ { 1 } x }$, where $k _ { 1 }$ is a positive constant.\\
(ii) Show that
$$v ^ { 2 } = \frac { g } { k _ { 1 } } \left( 1 - \mathrm { e } ^ { - 2 k _ { 1 } x } \right) ,$$
and hence state, in terms of $g$ and $k _ { 1 }$, the terminal velocity of the raindrop according to this first model.
In the second model $m = m _ { 0 } \left( 1 + k _ { 2 } x \right)$, where $k _ { 2 }$ is a positive constant.\\
(iii) By considering the expression obtained from differentiating $v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 }$ with respect to $x$, show that, for the second model, the equation of motion in part (i) may be written as
$$\frac { d } { d x } \left[ v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 } \right] = 2 g \left( 1 + k _ { 2 } x \right) ^ { 2 } .$$
Hence find an expression for $v ^ { 2 }$ in terms of $g , k _ { 2 }$ and $x$.\\
(iv) Suppose that the models give the same value for the speed of the raindrop at the instant when it has doubled its initial mass. Find the exact value of $\frac { k _ { 1 } } { k _ { 2 } }$, giving your answer in the form $\frac { a } { b }$ where $a$ and $b$\\
are integers. are integers.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR MEI M4 2016 Q4 [24]}}