CAIE M1 2012 November — Question 1 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeParticle on smooth curved surface
DifficultyStandard +0.3 This is a straightforward energy conservation problem requiring students to apply work-energy principles in two parts. Part (i) involves a direct calculation using gravitational PE change and the work-energy theorem. Part (ii) requires setting up an equation with the given ratio of work done, but the method is standard. The geometry is given explicitly (no need to derive heights), and the problem follows a familiar textbook pattern for M1 level.
Spec6.02d Mechanical energy: KE and PE concepts

1 \includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-2_366_780_251_680} \(A B C D\) is a semi-circular cross-section, in a vertical plane, of the inner surface of half a hollow cylinder of radius 2.5 m which is fixed with its axis horizontal. \(A D\) is horizontal, \(B\) is the lowest point of the cross-section and \(C\) is at a height of 1.8 m above the level of \(B\) (see diagram). A particle \(P\) of mass 0.8 kg is released from rest at \(A\) and comes to instantaneous rest at \(C\).
  1. Find the work done on \(P\) by the resistance to motion while \(P\) travels from \(A\) to \(C\). The work done on \(P\) by the resistance to motion while \(P\) travels from \(A\) to \(B\) is 0.6 times the work done while \(P\) travels from \(A\) to \(C\).
  2. Find the speed of \(P\) when it passes through \(B\).

AnswerMarks Guidance
(i) PE loss = \(0.8g \times (2.5 - 1.8)\) (= 5.6J)B1
Work done is 5.6 JB1 2 marks total
(ii) \(\frac{1}{2} \times 0.8v^2 = 0.8g \times 2.5 - 0.6 \times 5.6\)A1ft
Speed at \(B\) is 6.45 m s\(^{-1}\)A1 3 marks total
**(i)** PE loss = $0.8g \times (2.5 - 1.8)$ (= 5.6J) | B1 | 
Work done is 5.6 J | B1 | 2 marks total

**(ii)** $\frac{1}{2} \times 0.8v^2 = 0.8g \times 2.5 - 0.6 \times 5.6$ | A1ft |
Speed at $B$ is 6.45 m s$^{-1}$ | A1 | 3 marks total | For using KE gain = PE loss – WD against resistance

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\includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-2_366_780_251_680}\\
$A B C D$ is a semi-circular cross-section, in a vertical plane, of the inner surface of half a hollow cylinder of radius 2.5 m which is fixed with its axis horizontal. $A D$ is horizontal, $B$ is the lowest point of the cross-section and $C$ is at a height of 1.8 m above the level of $B$ (see diagram). A particle $P$ of mass 0.8 kg is released from rest at $A$ and comes to instantaneous rest at $C$.\\
(i) Find the work done on $P$ by the resistance to motion while $P$ travels from $A$ to $C$.

The work done on $P$ by the resistance to motion while $P$ travels from $A$ to $B$ is 0.6 times the work done while $P$ travels from $A$ to $C$.\\
(ii) Find the speed of $P$ when it passes through $B$.

\hfill \mbox{\textit{CAIE M1 2012 Q1 [5]}}