| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring Newton's second law with resistance proportional to v, forming a first-order linear differential equation. The solution method (separating variables or integrating factor) is routine for this module, and part (ii) is straightforward substitution. Slightly above average difficulty due to the inclined plane setup and differential equation solving, but follows a well-practiced template. |
| 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/Working | Marks | Guidance |
| \([\text{mg}\sin\alpha - 0.2\text{mv} = \text{ma}]\) | M1 | For using Newton's second law |
| \(5\frac{dv}{dt} = 28 - v\) | A1 | AG |
| \(\left[\int\frac{5}{28-v}dv = \int dt\right]\) | M1 | For separating variables and integrating |
| \((C) - 5\ln(28-v) = t\) | A1 | |
| M1 | For using \(v = 0\) when \(t = 0\); ft for \(\ln[(28-v)/28] = t/A\) from | |
| \(\ln[(28-v)/28] = -t/5\) | A1ft | \(C + A\ln(28-v) = t\) previously |
| \([28 - v = 28e^{-t/5}]\) | M1 | For expressing \(v\) in terms of \(t\); ft for \(v = 28(1 - e^{v/A})\) from |
| \(v = 28(1 - e^{-t/5})\) | A1ft | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For using \(a = (28 - v(t))/5\) or \(a = d(28 - 28e^{-t/5})dt\) and substituting \(t = 10\) | ||
| \([a = 28e^{-2}/5]\) | M1 | ft from incorrect \(v\) in the form \(a + be^{ct}\) \((b \neq 0)\); Accept \(5.6/e^2\) |
| Acceleration is \(0.758\text{ ms}^{-2}\) | A1ft | 2 |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[\text{mg}\sin\alpha - 0.2\text{mv} = \text{ma}]$ | M1 | For using Newton's second law |
| $5\frac{dv}{dt} = 28 - v$ | A1 | AG |
| $\left[\int\frac{5}{28-v}dv = \int dt\right]$ | M1 | For separating variables and integrating |
| $(C) - 5\ln(28-v) = t$ | A1 | |
| | M1 | For using $v = 0$ when $t = 0$; ft for $\ln[(28-v)/28] = t/A$ from |
| $\ln[(28-v)/28] = -t/5$ | A1ft | $C + A\ln(28-v) = t$ previously |
| $[28 - v = 28e^{-t/5}]$ | M1 | For expressing $v$ in terms of $t$; ft for $v = 28(1 - e^{v/A})$ from |
| $v = 28(1 - e^{-t/5})$ | A1ft | 8 | $\ln[(28-v)/28] = t/A$ previously |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| | | For using $a = (28 - v(t))/5$ or $a = d(28 - 28e^{-t/5})dt$ and substituting $t = 10$ |
| $[a = 28e^{-2}/5]$ | M1 | ft from incorrect $v$ in the form $a + be^{ct}$ $(b \neq 0)$; Accept $5.6/e^2$ |
| Acceleration is $0.758\text{ ms}^{-2}$ | A1ft | 2 | |
---4 A particle $P$ of mass $m \mathrm {~kg}$ is held at rest at a point $O$ on a fixed plane inclined at an angle $\sin ^ { - 1 } \left( \frac { 4 } { 7 } \right)$ to the horizontal. $P$ is released and moves down the plane. The total resistance acting on $P$ is $0.2 m v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ at time $t \mathrm {~s}$ after leaving $O$.\\
(i) Show that $5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = 28 - v$ and hence find an expression for $v$ in terms of $t$.\\
(ii) Find the acceleration of $P$ when $t = 10$.
\hfill \mbox{\textit{OCR M3 2008 Q4 [10]}}