| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question testing standard work-energy formulas. Part (a) requires computing work as F·s (dot product), part (b) uses power = F·v, and part (c) applies Hooke's law and EPE formula. All are direct applications of memorized formulas with simple vector arithmetic—no problem-solving insight needed. Slightly above average difficulty only because it's Further Maths content. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Displacement = (5 – –1)i + (12 – 6)j m |
| (7i – 2j).(6i + 6j) = 42 + –12 = 30 so 30 J | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Subtracting the position vectors |
| Answer | Marks |
|---|---|
| (b) | P = F.v = (7i – 2j).(–i – 5j) |
| = –7 + 10 = 3 so 3 W | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Use of P = F.v |
| (c) | OB = (52 + 122) = 13 so extension is 5m |
| Answer | Marks |
|---|---|
| (28) = 37.5 so 37.5 J | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Calculating OB and hence finding |
| the extension | If M0M1A0 SC1 for T = 1.46N and |
Question 2:
2 | (a) | Displacement = (5 – –1)i + (12 – 6)j m
(7i – 2j).(6i + 6j) = 42 + –12 = 30 so 30 J | M1
A1
[2] | 1.1
1.1 | Subtracting the position vectors
of the points to find the
displacement 6i + 6j m
(b) | P = F.v = (7i – 2j).(–i – 5j)
= –7 + 10 = 3 so 3 W | M1
A1
[2] | 1.1
1.1 | Use of P = F.v
(c) | OB = (52 + 122) = 13 so extension is 5m
T = 24 (“OB”-8) / 8 (= 15) or
EPE = 24 (“OB”-8) 2 / (28) (= 37.5)
T = 24 5 / 8 = 15 so 15 N and EPE = 24 52 /
(28) = 37.5 so 37.5 J | M1
M1
A1
[3] | 3.1b
1.1
1.1 | Calculating OB and hence finding
the extension | If M0M1A0 SC1 for T = 1.46N and
EPE = 0.353J2 The coordinates of two points, $A$ and $B$, are $( - 1,6 )$ and $( 5,12 )$ respectively, where the units of the coordinate axes are metres. A particle $P$ moves from $A$ to $B$ under the action of several forces. The force $\mathbf { F } = 7 \mathbf { i } - 2 \mathbf { j } \mathbf { N }$ is one of the forces acting on $P$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the work done by $\mathbf { F }$ on $P$ as $P$ moves from $A$ to $B$.
At the instant when $P$ reaches $B$ its velocity is $- \mathbf { i } - 5 \mathbf { j } \mathrm {~ms} ^ { - 1 }$.
\item Find the power generated by $\mathbf { F }$ at the instant that $P$ reaches $B$.
One end of a light elastic string was attached to the origin of the coordinate system and the other to $P$ when $P$ was at $A$, before it moved to $B$. The natural length of the string is 8 m and its modulus of elasticity is 24 N .
\item At the instant that $P$ reaches $B$, find the following.
\begin{itemize}
\item The tension in the string
\item The elastic potential energy stored in the string
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2022 Q2 [7]}}