| 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 | Given acceleration function find velocity |
| Difficulty | Challenging +1.8 This is a challenging variable mass mechanics problem requiring derivation of equation of motion with mass loss, integration involving logarithms, and optimization. While the setup is clearly guided and the mathematical techniques are A-level standard (separation of variables, logarithmic integration), the conceptual demand of handling variable mass systems and the multi-step nature with several marks pushes this well above average difficulty for M4 material. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| mass of counterweight \(= m_1 - \lambda t\) | B1 | Condone use of \(\delta t\) |
| \((m_1 - \lambda t)\frac{dv}{dt} = (m_1 - \lambda t)g - T\) | M1* | Equation of motion of the counterweight — their \((m_1-\lambda t)\) must be three terms; condone sign errors (condone use of \(m_1\) for \((m_1-\lambda t)\) for this M mark only) |
| \(m_2\frac{dv}{dt} = T - m_2 g\) | M1* | Equation of motion of the bag — 3 terms; condone sign errors |
| \((m_1 - \lambda t + m_2)\frac{dv}{dt} = (m_1 - \lambda t - m_2)g\) | M1 dep* | Eliminate \(T\); must be using \((m_1 - \lambda t)\) |
| CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((m_1-\lambda t)g - m_2 g = (m_1 - \lambda t + m_2)\frac{dv}{dt}\) | M1 | 3 terms shown without \(\frac{dv}{dt}\) — condone sign errors; 3 terms shown with \(\frac{dv}{dt}\) — condone sign errors |
| A1 A1 | Each side must have scored both M marks | |
| \(\int dv = g\int\frac{m_1 - \lambda t - m_2}{m_1 - \lambda t + m_2}dt\) | M1 | Separating the variables and integrating to obtain \(v = gt + A\ln\) (must include \(m_1, m_2\) and \(\lambda t\)) |
| \(v = g\int\left(1 - \frac{2m_2}{m_1 - \lambda t + m_2}\right)dt\) | ||
| \(= gt + \frac{2m_2 g}{\lambda}\ln\ | m_1 - \lambda t + m_2\ | + c\) |
| at \(t=0, v=0\), so \(c = -\frac{2m_2 g}{\lambda}\ln\ | m_1 + m_2\ | \) |
| \(v = gt + \frac{2m_2 g}{\lambda}\ln\left\ | \frac{m_1 - \lambda t + m_2}{m_1 + m_2}\right\ | \) |
| \(= gt + \frac{2m_2 g}{\lambda}\ln\left\ | 1 - \frac{\lambda t}{m_1 + m_2}\right\ | \) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = \frac{1}{10}m_1\) | B1 | Seen |
| \(t = \frac{m_1 - m_2}{\lambda} = 2\) seconds | M1* | Setting their \(\frac{dv}{dt} = 0\) or other argument and obtaining \(t\) |
| M1 dep* | Substitute their \(t\) and \(\lambda = \frac{1}{10}m_1\) into \(v\) | |
| \(v = 1.13162\ldots = 1.13\) m s\(^{-1}\) | A1 [4] | \(2g + 16g\ln(8/9)\) |
## Question 2(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| mass of counterweight $= m_1 - \lambda t$ | B1 | Condone use of $\delta t$ |
| $(m_1 - \lambda t)\frac{dv}{dt} = (m_1 - \lambda t)g - T$ | M1* | Equation of motion of the counterweight — their $(m_1-\lambda t)$ must be three terms; condone sign errors (condone use of $m_1$ for $(m_1-\lambda t)$ for this M mark only) |
| $m_2\frac{dv}{dt} = T - m_2 g$ | M1* | Equation of motion of the bag — 3 terms; condone sign errors |
| $(m_1 - \lambda t + m_2)\frac{dv}{dt} = (m_1 - \lambda t - m_2)g$ | M1 dep* | Eliminate $T$; must be using $(m_1 - \lambda t)$ |
| | | CAO |
**OR (first five marks):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(m_1-\lambda t)g - m_2 g = (m_1 - \lambda t + m_2)\frac{dv}{dt}$ | M1 | 3 terms shown without $\frac{dv}{dt}$ — condone sign errors; 3 terms shown with $\frac{dv}{dt}$ — condone sign errors |
| | A1 A1 | Each side must have scored both M marks |
| $\int dv = g\int\frac{m_1 - \lambda t - m_2}{m_1 - \lambda t + m_2}dt$ | M1 | Separating the variables and integrating to obtain $v = gt + A\ln$ (must include $m_1, m_2$ and $\lambda t$) |
| $v = g\int\left(1 - \frac{2m_2}{m_1 - \lambda t + m_2}\right)dt$ | | |
| $= gt + \frac{2m_2 g}{\lambda}\ln\|m_1 - \lambda t + m_2\| + c$ | A1 | Condone missing modulus signs throughout |
| at $t=0, v=0$, so $c = -\frac{2m_2 g}{\lambda}\ln\|m_1 + m_2\|$ | | |
| $v = gt + \frac{2m_2 g}{\lambda}\ln\left\|\frac{m_1 - \lambda t + m_2}{m_1 + m_2}\right\|$ | | |
| $= gt + \frac{2m_2 g}{\lambda}\ln\left\|1 - \frac{\lambda t}{m_1 + m_2}\right\|$ | E1 [8] | |
---
## Question 2(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = \frac{1}{10}m_1$ | B1 | Seen |
| $t = \frac{m_1 - m_2}{\lambda} = 2$ seconds | M1* | Setting their $\frac{dv}{dt} = 0$ or other argument and obtaining $t$ |
| | M1 dep* | Substitute their $t$ and $\lambda = \frac{1}{10}m_1$ into $v$ |
| $v = 1.13162\ldots = 1.13$ m s$^{-1}$ | A1 [4] | $2g + 16g\ln(8/9)$ |
---2 On a building site, a pulley system is used for moving small amounts of material up to roof level. A light pulley, which can rotate freely, is attached with its axis horizontal to the top of some scaffolding. A light inextensible rope hangs over the pulley with a counterweight of mass $m _ { 1 } \mathrm {~kg}$ attached to one end. Attached to the other end of the rope is a bag of negligible mass into which $m _ { 2 } \mathrm {~kg}$ of roof tiles are placed, where $m _ { 2 } < m _ { 1 }$. This situation is shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c3ac9277-d34d-4d0e-9f9b-d0bce8c741af-2_554_711_1098_678}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Initially the system is held at rest with the rope taut, the counterweight at the top of the scaffolding and the bag of tiles on the ground. When the counterweight is released, the bag ascends towards the top of the scaffolding. At time $t \mathrm {~s}$ the velocity of the counterweight is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ downwards.
The counterweight is made from a bag of negligible mass filled with sand. At the moment the counterweight is released, this bag is accidentally ripped and after this time the sand drops out at a constant rate of $\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }$.\\
(i) Find the equation of motion for the counterweight while it still contains sand, and hence show that
$$v = g t + \frac { 2 g m _ { 2 } } { \lambda } \ln \left( 1 - \frac { \lambda t } { m _ { 1 } + m _ { 2 } } \right) .$$
(ii) Given that the sand would run out after 10 seconds and that $m _ { 2 } = \frac { 4 } { 5 } m _ { 1 }$, find the maximum velocity attained by the counterweight towards the ground. You may assume that the scaffolding is sufficiently high that the counterweight does not hit the ground before this velocity is reached.
\hfill \mbox{\textit{OCR MEI M4 2014 Q2 [12]}}