| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Power from force and speed |
| Difficulty | Challenging +1.2 This Further Mechanics question involves hyperbolic functions, work-energy theorem, and power calculations. While it requires familiarity with hyperbolic identities and vector mechanics, the solution path is relatively standard: compute KE at two times for part (a), find when velocity is parallel to i for part (b), then calculate FΒ·v. The main challenge is algebraic manipulation of hyperbolic functions rather than novel problem-solving insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | (i) |
| Attempt to find v at t = 0 | Alternative using calculus: |
| Answer | Marks | Guidance |
|---|---|---|
| = 6 0 i β 1 8 9 j | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| v 2 ( ln 2 ) = 6 0 2 + ( β 1 8 9 ) 2 = 3 9 3 2 1 | M1 | 1.1 |
| Answer | Marks |
|---|---|
| may see KE = 58981.5J | Attempts the dot product: |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | 1.1 |
| Answer | Marks |
|---|---|
| 2 | Integrates and applies limits: |
| Answer | Marks | Guidance |
|---|---|---|
| awrt β17000 J | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | It is negative because over the interval P ends | |
| up moving slower than when it started | B1 | 2.4 |
| Answer | Marks |
|---|---|
| motion of the particleβ | Or β(the overall effect of F over |
| Answer | Marks |
|---|---|
| (b) | P moving parallel to x-axis => v = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| => cosh2t = 257/32 | M1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 1.39) | A1 | 2.3 |
| Answer | Marks | Guidance |
|---|---|---|
| May see t=0.5arcosh (257/32) | 1.386β¦ | |
| t = ln4 =>v=(32sinh(2ln4))i=255i | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| d t | M1* | 3.1b |
| Answer | Marks |
|---|---|
| (may just see the i-component) | May be seen in part (a), must be |
| Answer | Marks | Guidance |
|---|---|---|
| x | M1dep | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| awrt 393000 W | A1 | 1.1 |
Question 6:
6 | (a) | (i) | v(0)=(32sinh0)i+(32cosh0β257)j=β225j | M1 | 3.1b | DR required (determine)
Attempt to find v at t = 0 | Alternative using calculus:
64πππ β2π‘
π = ( ) so π = 3π =
64π ππβ2π‘
πππ β2π‘
192( ) seen used in F.v
π ππβ2π‘
v ( ln 2 ) = ( 3 2 s in h ( 2 ln 2 ) ) i + ( 3 2 c o s h ( 2 ln 2 ) β 2 5 7 ) j
= 6 0 i β 1 8 9 j | M1 | 1.1 | Attempt to find v at t = ln2 | Uses ππ·=β«πΉβπ£ ππ‘
ππ2 πππ β2π‘ 32π ππβ2π‘
ππ·=192β« ( )β( )ππ‘
π ππβ2π‘ 32πππ β2π‘β257
π‘=0
v 2 ( ln 2 ) = 6 0 2 + ( β 1 8 9 ) 2 = 3 9 3 2 1 | M1 | 1.1 | Attempt to use v.v to find v2 at
t = ln2 (v = 198.295...)
may see KE = 58981.5J | Attempts the dot product:
ππ2
=192β« 32π ππβ4π‘β257π ππβ2π‘ ππ‘
0
1 ( )
W D = ο΄ 3 3 9 3 2 1 β ( β 2 2 5 ) 2 J
2 | M1 | 1.1 | Using WD = change in KE
1
= Ξ m v .v
2 | Integrates and applies limits:
257 ππ2
=192[8πππ β4π‘β πππ β2π‘]
2
0
awrt β17000 J | A1 | 1.1 | -16956 J ISW if replaced with
magnitude
[5]
(ii) | It is negative because over the interval P ends
up moving slower than when it started | B1 | 2.4 | Condone eg βBecause P is
slowing downβ, βparticle is
deceleratingβ, βP is opposing the
motion of the particleβ | Or β(the overall effect of F over
this interval is such that) the force
F is acting in the opposite direction
to the motion of Pβ
βThe particle does work against the
forceβ
Allow βF is a resistive forceβ
[1]
(b) | P moving parallel to x-axis => v = 0
y
=> cosh2t = 257/32 | M1 | 2.2a | Deriving equation for t for P to
be moving parallel to x-axis
=> 2t = ο±ln16 but t > 0 => t = ln4 (or awrt
1.39) | A1 | 2.3 | May go straight to t = ln 4 (or 2t
= ln 16).
May see t=0.5arcosh (257/32) | 1.386β¦
t = ln4 =>v=(32sinh(2ln4))i=255i | A1 | 1.1 | or just v . If shown, v must be 0
x y
d v
F = m a = 3 = (1 9 2 c o s h 2 t ) i + (1 9 2 s i n h 2 t ) j
d t | M1* | 3.1b | Attempt to differentiate v
(implied by either (64cosh2t)i or
(64sinh2t)j) and multiply by m
(may just see the i-component) | May be seen in part (a), must be
used in part (b) to gain this method
mark
t = ln4 => F = 1 9 2 c o s h ( 2 ln 4 ) = 1 5 4 2
x | M1dep | 1.1 | Attempting to find (i-
component) of F at their time.
Ignore attempt to find F
y
P = F.v => Power is 1542ο΄255 =
awrt 393000 W | A1 | 1.1 | 3 x 514 x 255=393210
[6]A particle $P$ of mass $3$ kg is moving on a smooth horizontal surface under the influence of a variable horizontal force $\mathbf{F}$ N. At time $t$ seconds, where $t \geqslant 0$, the velocity of $P$, $\mathbf{v}$ m s$^{-1}$, is given by
$$\mathbf{v} = (32\sinh(2t))\mathbf{i} + (32\cosh(2t) - 257)\mathbf{j}.$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item By considering kinetic energy, determine the work done by $\mathbf{F}$ over the interval $0 \leqslant t \leqslant \ln 2$. [5]
\item Explain the significance of the sign of the answer to part (a)(i). [1]
\end{enumerate}
\item Determine the rate at which $\mathbf{F}$ is working at the instant when $P$ is moving parallel to the $\mathbf{i}$-direction. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2023 Q6 [12]}}