5 \includegraphics[max width=\textwidth, alt={}, center]{f8dd2aee-4ed5-4588-aa03-5dd56d9e7529-3_378_625_1272_758} A light elastic band, of total natural length \(a\) and modulus of elasticity \(\frac { 1 } { 2 } m g\), is stretched over two small smooth pins fixed at the same horizontal level and at a distance \(a\) apart. A particle of mass \(m\) is attached to the lower part of the band and when the particle is in equilibrium the sloping parts of the band each make an angle \(\beta\) with the vertical (see diagram). Express the tension in the band in terms of \(m , g\) and \(\beta\), and hence show that \(\beta = \frac { 1 } { 4 } \pi\). The particle is given a velocity of magnitude \(\sqrt { } ( a g )\) vertically downwards. At time \(t\) the displacement of the particle from its equilibrium position is \(x\). Show that, neglecting air resistance, $$\ddot { x } = - \frac { 2 g } { a } x .$$ Show that the particle passes through the level of the pins in the subsequent motion, and find the time taken to reach this level for the first time.

## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{1}{2}mg(a/\sin\beta)/a = mg/2\sin\beta$ | M1, A1 | Find extension and apply Hooke's Law |
| $mg = 2T\cos\beta\; [T = mg/2\cos\beta]$ | B1 | Resolve vertically for particle |
| $\sin\beta = \cos\beta,\; \beta = \frac{1}{4}\pi$ **A.G.** | B1 | Equate two expressions for $T$ |
| $m\,d^2x/dt^2 = mg - 2T'\cos\beta' = mg - (mg/\sin\beta')\cos\beta'$ | M1 A1 | Use Newton's Law for vertical motion (A.E.F.) |
| $d^2x/dt^2 = g - g(\frac{1}{2}a + x)/\frac{1}{2}a = -2gx/a$ **A.G.** | A1 | Simplify |
| $A = \sqrt{ag}\cdot\sqrt{2g/a} = a/\sqrt{2}$ | M1 A1 | Use $v = A\omega$ to find amplitude $A$ |
| $a/\sqrt{2} > a/2$ | A1 | Hence show particle reaches pins |
| $t = (\sin^{-1}(-\frac{1}{2}a/A))/\omega$ or $\frac{1}{2}T + (\sin^{-1}(\frac{1}{2}a/A))/\omega = (\sin^{-1}(-1/\sqrt{2}))/\sqrt{2g/a}$ or $(\pi + \sin^{-1}(1/\sqrt{2}))/\sqrt{2g/a}$ | M1, A1 | Use $x = A\sin\omega t$ to find time $t$ |
| $= (5\pi/4)\sqrt{a/2g}$ or $0.878\sqrt{a}$ | A1 | Simplify (A.E.F.) |

**Total: 13 marks**

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\includegraphics[max width=\textwidth, alt={}, center]{f8dd2aee-4ed5-4588-aa03-5dd56d9e7529-3_378_625_1272_758}

A light elastic band, of total natural length $a$ and modulus of elasticity $\frac { 1 } { 2 } m g$, is stretched over two small smooth pins fixed at the same horizontal level and at a distance $a$ apart. A particle of mass $m$ is attached to the lower part of the band and when the particle is in equilibrium the sloping parts of the band each make an angle $\beta$ with the vertical (see diagram). Express the tension in the band in terms of $m , g$ and $\beta$, and hence show that $\beta = \frac { 1 } { 4 } \pi$.

The particle is given a velocity of magnitude $\sqrt { } ( a g )$ vertically downwards. At time $t$ the displacement of the particle from its equilibrium position is $x$. Show that, neglecting air resistance,

$$\ddot { x } = - \frac { 2 g } { a } x .$$

Show that the particle passes through the level of the pins in the subsequent motion, and find the time taken to reach this level for the first time.

\hfill \mbox{\textit{CAIE FP2 2010 Q5 [13]}}