| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done by constant force - vector setup |
| Difficulty | Standard +0.3 This is a straightforward application of the work-energy principle with standard mechanics bookwork. Part (a) requires applying KE change = work done over the full path (routine). Part (b) needs a second equation from O to A (standard technique). Part (c) checks consistency at point B, requiring students to recognize negative KE is impossible—slightly more thoughtful but still a standard check. The multi-part structure and need to set up two equations adds modest complexity, but all techniques are standard A-level mechanics with no novel insight required. |
| Spec | 6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | 1⋅2⋅32 +40T −60R= 1⋅2⋅152 |
| 2 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒40T −60R=216 ⇒20T −30R=108 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (b) | Either 1⋅2⋅32 +18T −25R= 1⋅2⋅112 ⇒18T −25R=112 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| T =16.5 | A1 | 1.1 |
| R=7.4 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | If the particle can move as described from O to C, then T and | |
| R must have the values found in part (b). | M1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| where E joules is the kinetic energy of the particle at B. | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| which is not possible. | A1 | 2.2a |
Question 4:
4 | (a) | 1⋅2⋅32 +40T −60R= 1⋅2⋅152
2 2 | M1 | 3.3 | Attempt at WEP: all terms present;
allow one slip in coefficients of T
and R.
⇒40T −60R=216 ⇒20T −30R=108 | A1 | 1.1 | AG
[2]
(b) | Either 1⋅2⋅32 +18T −25R= 1⋅2⋅112 ⇒18T −25R=112
2 2
or 1⋅2⋅112 +22T −35R= 1⋅2⋅152 ⇒22T −35R=104
2 2 | M1 | 3.4 | WEP: all terms present; allow one
slip in coefficients of T and R.
T =16.5 | A1 | 1.1
R=7.4 | A1 | 1.1
[3]
(c) | If the particle can move as described from O to C, then T and
R must have the values found in part (b). | M1 | 2.1 | Seen or implied.
But then 1⋅2⋅112 +6(16.5)−12(7.4)=2(9.8)⋅10+E,
2
where E joules is the kinetic energy of the particle at B. | M1 | 3.4 | Attempt at WEP must take into
account GPE to find KE at B. This
calculation can done using B and
any of O, A and C; the calculation
in the MS is based on A.
⇒E =−64.8
which is not possible. | A1 | 2.2a | Explicit recognition of a
contradiction required.
[3]4 The diagram shows the path of a particle P of mass 2 kg as it moves from the origin O to C via A and B . The lengths of the sections $\mathrm { OA } , \mathrm { AB }$ and BC are given in the diagram. The units of the axes are metres.\\
\includegraphics[max width=\textwidth, alt={}, center]{5c1cfe41-d7a2-4f69-ae79-67d9f023c246-4_670_1322_404_246}
P , starting from O , moves along the path indicated in the diagram to C under the action of a constant force of magnitude $T \mathrm {~N}$ acting in the positive $x$-direction. As P moves, it does $R \mathrm {~J}$ of work for every metre travelled against resistances to motion.
It is given that
\begin{itemize}
\item the speed of P at O is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item the speed of P at A is $11 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,
\item the speed of P at C is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{itemize}
You should assume that both $x$ - and $y$-axes lie in a horizontal plane.
\begin{enumerate}[label=(\alph*)]
\item By considering the entire path of P from O to C , show that
$$20 \mathrm {~T} - 30 \mathrm { R } = 108 .$$
\item By formulating a second equation, determine the values of $T$ and $R$.
\item It is now given that the $x$-axis is horizontal, and the $y$-axis is directed vertically upwards. By considering the kinetic energy of P at B , show that the motion as described above is impossible.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2021 Q4 [8]}}