| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question requiring application of F=ma with vector forces, followed by integration to find velocity. Part (a) is simple division by mass; part (b) involves integrating trigonometric functions (cos 2t and sin t) and finding speed from velocity components—all standard techniques with no novel insight required. Slightly above average difficulty due to being Further Maths content and requiring careful integration, but still a routine textbook exercise. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | F=(−2i+6j )+ ( 2cos(2t) i+4sintj ) |
| a= ( cos(2t)−1 ) i+(2sint+3) j (ms- 2) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Find the resultant of the two forces by vector addition |
| Answer | Marks |
|---|---|
| (b) | 1 |
| Answer | Marks |
|---|---|
| 9.15 (ms- 1) | M1* |
| Answer | Marks |
|---|---|
| [5] | 2.1 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt to integrate - at least two (of their four) terms |
Question 5:
5 | (a) | F=(−2i+6j )+ ( 2cos(2t) i+4sintj )
a= ( cos(2t)−1 ) i+(2sint+3) j (ms- 2) | M1
A1
[2] | 1.1
1.1 | Find the resultant of the two forces by vector addition
Using F=mato correctly find a
Must be written as a vector but ISW if magnitude
considered. Brackets must be present (oe) for this
mark
(b) | 1
v= sin(2t)−ti+(−2cost+3t) j(+c)
2
t =0,v=0⇒c=...
2
1
v = sin4−2 +(8−2cos2)2
2
9.15 (ms- 1) | M1*
A1
M1dep*
M1
A1
[5] | 2.1
1.1
3.4
1.1
1.1 | Attempt to integrate - at least two (of their four) terms
correct – must be a vector expression for a
Correct integration +c not required for this mark
Uses correct initial conditions to determine c (if
correct c=2j) – just stating c = 0 because v = 0
when t = 0 is M0
Substitute t = 2 and attempt to calculate the speed
(dependent on first M mark only)
9.1469232… (an answer of 6.31477… implies M1
(working in degrees))At time $t$ seconds, where $t \geq 0$, a particle P of mass 2 kg is moving on a smooth horizontal surface. The particle moves under the action of a constant horizontal force of $(-2\mathbf{i} + 6\mathbf{j})$ N and a variable horizontal force of $(2\cos 2t \mathbf{i} + 4\sin t \mathbf{j})$ N. The acceleration of P at time $t$ seconds is $\mathbf{a}$ m s$^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf{a}$ in terms of $t$. [2]
\end{enumerate}
The particle P is at rest when $t = 0$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine the speed of P at the instant when $t = 2$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q5 [7]}}