M5 June 2016 Q7
7. A uniform square lamina \(PQRS\), of mass \(m\) and side \(2a\), is free to rotate about a fixed smooth horizontal axis which passes through \(P\) and \(Q\). The lamina hangs at rest in a vertical plane with \(SR\) below \(PQ\) and is given a horizontal impulse of magnitude \(J\) at the midpoint of \(SR\). The impulse is perpendicular to \(SR\).
| Scheme | Marks |
|---|---|
| \(J \cdot 2a = \dfrac{4}{3}ma^2\omega\) | M1 A1 |
| \(\omega = \dfrac{3J}{2ma}\) | A1 |
| (3) |
Notes
First M1 for moment of impulse = Gain in Angular Momentum with all terms dimensionally correct
First A1 for a correct equation
Second A1 for correct answer
| Scheme | Marks |
|---|---|
| \(mga\sin 30^\circ = -\dfrac{4}{3}ma^2\ddot{\theta}\) | M1 A1 |
| \(|\ddot{\theta}| = \dfrac{3g}{8a}\) | A1 |
| (3) |
Notes
First M1 for taking moments about the axis with all terms dimensionally correct
First A1 for a correct equation (\(\theta\) need not be substituted)
Second A1 for a correct positive answer
Alternative:
First M1 for differentiating an energy equation with all terms dimensionally correct
First A1 for a correct equation
Second A1 for a correct positive answer
| Scheme | Marks |
|---|---|
| \(X - mg\cos 60^\circ = ma\ddot{\theta}\) | M1 A1 A1 |
| \(X - \dfrac{1}{2}mg = m\left(-\dfrac{3g}{8a}\right)a\) | M1 |
| \(X = \dfrac{mg}{8}\) | A1 |
| (5) | |
| (11 marks) |
Notes
First M1 for resolving perpendicular to the lamina, with correct no. of terms, with all terms dimensionally correct
First and Second A1’s for a correct equation (\(\theta\) need not be substituted)
Second M1 for substituting for (must be dimensionally correct) \(\ddot{\theta}\)
Third A1 for answer