CAIE M1 2009 November — Question 2 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2009
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeParticle on smooth curved surface
DifficultyModerate -0.3 This is a straightforward application of conservation of energy with clearly defined heights and speeds. Part (i) requires a single energy equation between A and E, while part (ii) requires recognizing that maximum speed occurs at the lowest point C. Both parts use standard SUVAT/energy principles with no geometric complications or novel problem-solving required.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle

2 \includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-2_609_967_536_589} A smooth narrow tube \(A E\) has two straight parts, \(A B\) and \(D E\), and a curved part \(B C D\). The part \(A B\) is vertical with \(A\) above \(B\), and \(D E\) is horizontal. \(C\) is the lowest point of the tube and is 0.65 m below the level of \(D E\). A particle is released from rest at \(A\) and travels through the tube, leaving it at \(E\) with speed \(6 \mathrm {~ms} ^ { - 1 }\) (see diagram). Find
  1. the height of \(A\) above the level of \(D E\),
  2. the maximum speed of the particle.

AnswerMarks Guidance
(i) \([\text{mgh} = \frac{1}{2}m\dot{v}^2]\)M1 For using PE loss = KE gain
Height is \(1.8\text{m}\)A1 2
(ii) \(\frac{1}{2}m\dot{v}^2 = \text{mg}(1.8 + 0.65)\) or \(\frac{1}{2}v^2 = \frac{1}{2}m\dot{v}^2 = \text{mg} \times 0.65\)B1ft
Maximum speed is \(7\text{ms}^{-1}\)B1 2
(i) $[\text{mgh} = \frac{1}{2}m\dot{v}^2]$ | M1 | For using PE loss = KE gain
Height is $1.8\text{m}$ | A1 | 2 |

(ii) $\frac{1}{2}m\dot{v}^2 = \text{mg}(1.8 + 0.65)$ or $\frac{1}{2}v^2 = \frac{1}{2}m\dot{v}^2 = \text{mg} \times 0.65$ | B1ft |
Maximum speed is $7\text{ms}^{-1}$ | B1 | 2 |
2\\
\includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-2_609_967_536_589}

A smooth narrow tube $A E$ has two straight parts, $A B$ and $D E$, and a curved part $B C D$. The part $A B$ is vertical with $A$ above $B$, and $D E$ is horizontal. $C$ is the lowest point of the tube and is 0.65 m below the level of $D E$. A particle is released from rest at $A$ and travels through the tube, leaving it at $E$ with speed $6 \mathrm {~ms} ^ { - 1 }$ (see diagram). Find\\
(i) the height of $A$ above the level of $D E$,\\
(ii) the maximum speed of the particle.

\hfill \mbox{\textit{CAIE M1 2009 Q2 [4]}}