| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on time t |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring integration of a time-dependent force through multiple steps (finding v, then x, then solving for a specific condition involving the limiting velocity). While the integrations themselves are standard, the multi-stage problem-solving, work with limits, and final percentage calculation make it moderately challenging but still within typical Further Maths scope. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (t+1)2 dt dt 3(t+1)2 | B1 | |
| [1] | 3.1b | AG |
| 5 | (b) | k⌠ 1 −k |
| Answer | Marks |
|---|---|
| 1+t | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Separating variables correctly |
| Answer | Marks |
|---|---|
| 1+t | May use + c instead of u |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | dx 4 |
| Answer | Marks |
|---|---|
| t =0, x=1⇒c=1so x=4t−4ln(1+t)+1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | For integrating their ‘v’ to reach |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | 95% of v =0.95×4=3.8 |
| Answer | Marks |
|---|---|
| 1+t | B1 |
| Answer | Marks |
|---|---|
| A1 | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | Setting their v to their 3.8 in the |
| Answer | Marks |
|---|---|
| 76−4ln20m or awrt 64 m | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| 1.1 | Substituting their t into the |
Question 5:
5 | (a) | 1
F ∝
(t+1)2
k dv dv k
∴F = =ma=3 ⇒ =
(t+1)2 dt dt 3(t+1)2 | B1
[1] | 3.1b | AG
5 | (b) | k⌠ 1 −k
∴v= dx= +u
3⌡(1+t)2 3(1+t)
t = 0, v = 0 ⇒ k = 3u
−k
t = 1, v = 2 ⇒ 2= +u
3(1+1)
4
⇒u=4, k =12⇒v=4− oe
1+t | M1
M1
M1
A1
[4] | 3.1b
3.1b
3.1b
1.1 | Separating variables correctly
C
and integrating to ; award if
1+t
“+ u” missing
Substituting initial values to
determine a relationship between
k and u.
Substituting t = 1 to determine a
second relationship between k
and u oe.
4t
eg v=
1+t | May use + c instead of u
NB The units of k are Ns2 or kgm
but these are not required.
5 | (c) | dx 4
=4− ⇒x=4t−4ln(1+t)+c
dt 1+t
t =0, x=1⇒c=1so x=4t−4ln(1+t)+1 | M1
A1
[2] | 1.1
1.1 | For integrating their ‘v’ to reach
an expression involving
oe
Can be awarded even if n𝑘𝑘ol n“+(1 c”+
𝑡𝑡)
5 | (d) | 95% of v =0.95×4=3.8
T
4
v=3.8⇒3.8=4−
1+t
4
⇒0.2= ⇒1+t =20⇒t =19
1+t | B1
M1
A1 | 2.2a
3.1b
1.1 | Setting their v to their 3.8 in the
appropriate equation
so x=4×19−4ln(1+19)+1
x=77−4ln20 so distance moved is
76−4ln20m or awrt 64 m | M1
A1
[5] | 1.1
1.1 | Substituting their t into the
equation for x5 A particle $P$ of mass 3 kg moves on the $x$-axis under the action of a single force acting in the positive $x$-direction. At time $t \mathrm {~s}$, where $t \geqslant 0$, the displacement of $P$ is $x \mathrm {~m}$ and its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The magnitude of the force acting is inversely proportional to $( t + 1 ) ^ { 2 }$.
Initially $P$ is at rest at the point where $x = 1$. When $t = 1 , v = 2$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { dv } } { \mathrm { dt } } = \frac { \mathrm { k } } { 3 ( \mathrm { t } + 1 ) ^ { 2 } }$ where $k$ is a constant.
\item Find an expression for $v$ in terms of $t$.
\item Find an expression for $x$ in terms of $t$.
As $t$ increases, $v$ approaches a limiting value, $\mathrm { V } _ { \mathrm { T } }$.
\item Determine how far $P$ is from its initial position at the instant when $v$ is $95 \%$ of $\mathrm { V } _ { \mathrm { T } }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q5 [12]}}