| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| 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 requiring standard application of work = force · displacement, average power = work/time, and the work-energy theorem. While it involves vectors and multiple parts, each step follows directly from formula application with no conceptual challenges or novel problem-solving required. Being Further Maths raises the baseline slightly, but the mechanics are routine. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 2 50 |
| Answer | Marks |
|---|---|
| 2×50 + 10×140 = 1500 J | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | Using WD = F.x | |
| 1 | (b) | 1500/5 = 300 W |
| [1] | 1.1a | Their 1500 |
| 1 | (c) | 1 1 |
| Answer | Marks |
|---|---|
| 50 ms–1 | M1 |
| A1 | 1.1 |
| 1.1 | Correct use of work-energy |
| Answer | Marks | Guidance |
|---|---|---|
| Not ± | Can use their WD for M mark | |
| Alternative method | M1 | complete method involving |
| Answer | Marks |
|---|---|
| v =50 ms–1 | A1 |
Question 1:
1 | (a) | 2 50
.
10 140
2×50 + 10×140 = 1500 J | M1
A1
[2] | 1.1
1.1 | Using WD = F.x
1 | (b) | 1500/5 = 300 W | B1ft
[1] | 1.1a | Their 1500 | Must be a scalar value
1 | (c) | 1 1
×1.25v2 = ×1.25×102 +1500
2 2
50 ms–1 | M1
A1 | 1.1
1.1 | Correct use of work-energy
principle
Not ± | Can use their WD for M mark
Alternative method | M1 | complete method involving
constant acceleration
formula(e)
50 8 6
=5u+ 1. 5 .52 ⇒u=
140 2 8 8
6 8 14
Then v= + 5 .5=
8 8 48
v =50 ms–1 | A1
[2]
M1
complete method involving
constant acceleration
formula(e)1 A force of $\binom { 2 } { 10 } \mathrm {~N}$ is the only horizontal force acting on a particle $P$ of mass 1.25 kg as it moves in a horizontal plane. Initially $P$ is at the origin, $O$, and 5 seconds later it is at the point $A ( 50,140 )$. The units of the coordinate system are metres.
\begin{enumerate}[label=(\alph*)]
\item Calculate the work done by the force during these 5 seconds.
\item Calculate the average power generated by the force during these 5 seconds.
The speed of $P$ at $O$ is $10 \mathrm {~ms} ^ { - 1 }$.
\item Calculate the speed of $P$ at $A$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2020 Q1 [5]}}