OCR MEI Further Mechanics Minor 2021 November — Question 3 5 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2021
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeConstant speed up/down incline
DifficultyStandard +0.8 This is a multi-stage mechanics problem requiring understanding of power at constant speed (P=Fv), Newton's second law for connected particles, and the relationship between power and motion when mass changes. It requires setting up equations for two different equilibrium states and solving simultaneously, which is more sophisticated than standard A-level mechanics but still follows established methods without requiring novel insight.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

3 The diagram shows an electric winch raising two crates A and B , with masses 40 kg and 25 kg , respectively. The cable connecting the winch to A , and the cable connecting A to B may both be modelled as light and inextensible. Furthermore, it can be assumed that there are no resistances to motion. \includegraphics[max width=\textwidth, alt={}, center]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-4_499_300_447_246} Throughout the entire motion, the power \(P \mathrm {~W}\) developed by the winch is constant.
Crates A and B are both being raised at a constant speed \(\nu \mathrm { m } \mathrm { s } ^ { - 1 }\) when the cable connecting A and B breaks. After the cable between A and B breaks, crate A continues to be raised by the winch. Crate A now accelerates until it reaches a new constant speed of \(( v + 3 ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Determine
  • the value of \(v\),
  • the value of \(P\).

Question 3:
3
AnswerMarks Guidance
P=65gvB1 3.1b
P=40g(v+3)B1 1.1
65gv=40g(v+3)M1 3.4
P – with at least one correct
equation
AnswerMarks Guidance
−1A1 1.1
𝑣𝑣 =4.8 𝑚𝑚𝑠𝑠A1 1.1
this case)3057.6
𝑃𝑃 =3060 𝑊𝑊[5]
Question 3:
3
P=65gv | B1 | 3.1b | Use of P=Fv (either one)
P=40g(v+3) | B1 | 1.1
65gv=40g(v+3) | M1 | 3.4 | Equating their two expressions for
P – with at least one correct
equation
−1 | A1 | 1.1
𝑣𝑣 =4.8 𝑚𝑚𝑠𝑠 | A1 | 1.1 | or 3.06 kW (but must state kW in
this case) | 3057.6
𝑃𝑃 =3060 𝑊𝑊 | [5]
3 The diagram shows an electric winch raising two crates A and B , with masses 40 kg and 25 kg , respectively. The cable connecting the winch to A , and the cable connecting A to B may both be modelled as light and inextensible. Furthermore, it can be assumed that there are no resistances to motion.\\
\includegraphics[max width=\textwidth, alt={}, center]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-4_499_300_447_246}

Throughout the entire motion, the power $P \mathrm {~W}$ developed by the winch is constant.\\
Crates A and B are both being raised at a constant speed $\nu \mathrm { m } \mathrm { s } ^ { - 1 }$ when the cable connecting A and B breaks. After the cable between A and B breaks, crate A continues to be raised by the winch. Crate A now accelerates until it reaches a new constant speed of $( v + 3 ) \mathrm { m } \mathrm { s } ^ { - 1 }$.

Determine

\begin{itemize}
  \item the value of $v$,
  \item the value of $P$.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q3 [5]}}