Question 3 - A-Level Maths

Edexcel M5 2014 June — Question 3 8 marks

Exam Board	Edexcel
Module	M5 (Mechanics 5)
Year	2014
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Oblique and successive collisions
Type	Coalescing particles collision
Difficulty	Challenging +1.8 This is a challenging M5 mechanics problem requiring conservation of angular momentum about the hinge, coefficient of restitution for the collision, and calculation of the moment of inertia for a rectangular lamina rotating about an edge. It combines multiple advanced mechanics concepts in a non-trivial way, placing it well above average difficulty but not at the extreme end for Further Maths mechanics.
Spec	6.03k Newton's experimental law: direct impact

3. A uniform rectangular lamina \(A B C D\), where \(A B = a\) and \(B C = 2 a\), has mass \(2 m\). The lamina is free to rotate about its edge \(A B\), which is fixed and vertical. The lamina is at rest when it is struck at \(C\) by a particle \(P\) of mass \(m\). The particle \(P\) is moving horizontally with speed \(U\) in a direction which is perpendicular to the lamina. The coefficient of restitution between \(P\) and the lamina is 0.5 Find the angular speed of the lamina immediately after the impact.
(8)

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Question 3:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Moment of inertia of lamina about \(AB\): \(I = \frac{2m}{3}(2a)^2 \times \frac{1}{3} = \frac{8ma^2}{3}\)	M1 A1	\(I_{AB} = \frac{1}{3}(2m)(2a)^2 = \frac{8ma^2}{3}\)
Point \(C\) is at distance \(2a\) from axis \(AB\)	B1
Angular impulse = linear impulse \(\times\) distance: \(J \cdot 2a = \frac{8ma^2}{3}\omega\)	M1	Angular momentum equation
Restitution at \(C\): \(v_C - 0 = 0.5(U - 0)\) wait — separation/approach	M1	Applying Newton's law of restitution
Velocity of \(C\) after \(= 2a\omega\); velocity of \(P\) after \(= v\)
Restitution: \(2a\omega - v = 0.5U\)	A1
Impulse-momentum on \(P\): \(mU - mv = J\), so \(J = m(U-v)\)	M1	Linear impulse on particle
\(m(U-v)(2a) = \frac{8ma^2}{3}\omega\)
Solving: \(2a\omega - v = 0.5U\) and \(2m(U-v)a = \frac{8ma^2}{3}\omega\)
From restitution: \(v = 2a\omega - 0.5U\)
Substituting: \(2ma(U - 2a\omega + 0.5U) = \frac{8ma^2}{3}\omega\)
\(2a(1.5U - 2a\omega) = \frac{8a^2}{3}\omega\)
\(3aU = 4a^2\omega + \frac{8a^2}{3}\omega = \frac{20a^2\omega}{3}\)	A1
\(\omega = \frac{9U}{20a}\)	A1	cao

# Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of inertia of lamina about $AB$: $I = \frac{2m}{3}(2a)^2 \times \frac{1}{3} = \frac{8ma^2}{3}$ | M1 A1 | $I_{AB} = \frac{1}{3}(2m)(2a)^2 = \frac{8ma^2}{3}$ |
| Point $C$ is at distance $2a$ from axis $AB$ | B1 | |
| Angular impulse = linear impulse $\times$ distance: $J \cdot 2a = \frac{8ma^2}{3}\omega$ | M1 | Angular momentum equation |
| Restitution at $C$: $v_C - 0 = 0.5(U - 0)$ wait — separation/approach | M1 | Applying Newton's law of restitution |
| Velocity of $C$ after $= 2a\omega$; velocity of $P$ after $= v$ | | |
| Restitution: $2a\omega - v = 0.5U$ | A1 | |
| Impulse-momentum on $P$: $mU - mv = J$, so $J = m(U-v)$ | M1 | Linear impulse on particle |
| $m(U-v)(2a) = \frac{8ma^2}{3}\omega$ | | |
| Solving: $2a\omega - v = 0.5U$ and $2m(U-v)a = \frac{8ma^2}{3}\omega$ | | |
| From restitution: $v = 2a\omega - 0.5U$ | | |
| Substituting: $2ma(U - 2a\omega + 0.5U) = \frac{8ma^2}{3}\omega$ | | |
| $2a(1.5U - 2a\omega) = \frac{8a^2}{3}\omega$ | | |
| $3aU = 4a^2\omega + \frac{8a^2}{3}\omega = \frac{20a^2\omega}{3}$ | A1 | |
| $\omega = \frac{9U}{20a}$ | A1 | cao |

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3. A uniform rectangular lamina $A B C D$, where $A B = a$ and $B C = 2 a$, has mass $2 m$. The lamina is free to rotate about its edge $A B$, which is fixed and vertical. The lamina is at rest when it is struck at $C$ by a particle $P$ of mass $m$. The particle $P$ is moving horizontally with speed $U$ in a direction which is perpendicular to the lamina. The coefficient of restitution between $P$ and the lamina is 0.5

Find the angular speed of the lamina immediately after the impact.\\
(8)\\

\hfill \mbox{\textit{Edexcel M5 2014 Q3 [8]}}

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