| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on velocity v |
| Difficulty | Challenging +1.2 This is a multi-part Further Mechanics question requiring impulse-momentum (straightforward), forming and solving a separable differential equation with v dv/dx substitution (standard FM technique), and motion description. While it involves several steps and FM content, the techniques are all standard textbook methods with clear signposting. The differential equation solving is routine for FM students, making this above average but not exceptionally challenging. |
| Spec | 6.03f Impulse-momentum: relation6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | I = mv – mu = 2.5(18 – –5) = 57.5 |
| [1] | 1.1 | |
| (b) | F =ma |
| Answer | Marks |
|---|---|
| dx x | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 2.2a | Using F = ma to express a |
| Answer | Marks |
|---|---|
| (c) | 𝑑𝑣 2𝑣 𝑑𝑣 2 |
| Answer | Marks |
|---|---|
| c = –6 => v = 24x – 6x2 | B1FT |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 3.4 | Correctly determining the |
| Answer | Marks |
|---|---|
| find c | FT their value of a provided 𝑎 ≠ 0 |
| Answer | Marks |
|---|---|
| (d) | • As t → , x → 4 OR P moves from its |
| Answer | Marks |
|---|---|
| when x = 2 | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.4 |
Question 6:
6 | (a) | I = mv – mu = 2.5(18 – –5) = 57.5 | B1
[1] | 1.1
(b) | F =ma
5v2 dv
−60v=2.5v
x dx
dv 2v
= −24 so a = 2 and b = –24
dx x | M1
A1
[2] | 3.3
2.2a | Using F = ma to express a
differential form of the
acceleration in terms of the given
forces with the signs correct
(c) | 𝑑𝑣 2𝑣 𝑑𝑣 2
= −24⇒ − 𝑣 =−24
𝑑𝑥 𝑥 𝑑𝑥 𝑥
2
⇒IF=𝑒∫− 𝑑𝑥 =𝑒−2𝑙𝑛𝑥 =𝑥−2
𝑥
x − 2 d v − 2 x − 3 v = d ( v x − 2 ) = − 2 4 x − 2
d x d x
v x − 2 = − 2 4 x − 2 d x = 2 4 x − 1 + c
x = 1, v = 18 => 181–2 = 241–1 + c =>
c = –6 => v = 24x – 6x2 | B1FT
M1*
M1dep*
A1
[4] | 1.1
1.1
1.1
3.4 | Correctly determining the
integrating factor 𝑥−𝑎 𝑎 ≠ 0
Multiplying by the IF and writing
LHS as an exact derivative.
Integrating BS with a constant of
integration
Substituting initial conditions to
find c | FT their value of a provided 𝑎 ≠ 0
IF must be in the form 𝑥𝑘
or using a definite integral with
suitable limits
or using the limits correctly in a
definite integral with suitable limits
(d) | • As t → , x → 4 OR P moves from its
initial position to the point on the x-axis
where x = 4
• P initially accelerates and then
decelerates (to 0)
• P reaches its max speed of 24ms-1 or
reaches max speed when 𝑡 =
1
ln3 (0.0458) or reaches max speed
24
when x = 2 | B1
B1
B1
[3] | 2.4
2.4
2.2a6 A particle $P$ of mass 2.5 kg is free to move along the $x$-axis. When its displacement from the origin is $x \mathrm {~m}$ its velocity is $v \mathrm {~ms} ^ { - 1 }$.
At time $t = 0$ seconds, $P$ is at the point where $x = 1$ and is travelling in the negative $x$-direction with speed $5 \mathrm {~ms} ^ { - 1 }$.
At this time an impulse of $I$ Ns is applied to $P$ in the positive $x$-direction so that $P$ moves in the positive $x$-direction with speed $18 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $I$.
Subsequently, whenever $P$ is in motion, two forces act on it. The first force acts in the positive $x$-direction and has magnitude $\frac { 5 v ^ { 2 } } { x } N$. The second force acts in the negative $x$-direction and has magnitude 60 vN .
\item Show that the motion of $P$ can be modelled by the differential equation $\frac { \mathrm { dV } } { \mathrm { dx } } = \frac { \mathrm { aV } } { \mathrm { x } } + \mathrm { b }$ where $a$ and $b$ are constants whose values should be determined.
\item By solving the differential equation derived in part (b) find an expression for $v$ in terms of $x$.
You are given that $\mathrm { x } = \frac { 4 } { 3 \mathrm { e } ^ { - 24 \mathrm { t } } + 1 }$ when $t \geqslant 0$.
\item Describe in detail the motion of $P$ when $t \geqslant 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2022 Q6 [10]}}