Challenging +1.2 This is a standard Further Maths mechanics problem requiring application of Newton's second law to find k, then energy conservation to find maximum height. The multi-step nature and elastic string context elevate it above average, but the methods are well-practiced techniques from the syllabus with clear signposting of what to find.
6 One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(k\), is attached to a particle \(P\) of mass \(m\). The other end of the string is attached to a fixed point \(Q\). The particle \(P\) is projected vertically upwards from \(Q\). When \(P\) is moving upwards and at a distance \(\frac { 4 } { 3 } a\) directly above \(Q\), it has a speed \(\sqrt { 2 g a }\). At this point, its acceleration is \(\frac { 7 } { 3 } g\) downwards.
Show that \(\mathrm { k } = 4 \mathrm { mg }\) and find in terms of \(a\) the greatest height above \(Q\) reached by \(P\).
\(x = \frac{1}{3}a\) so greatest height is \(\frac{5}{3}a\)
A1
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T + mg = m \cdot \frac{7}{3}g$ | M1 | |
| With $T = k\dfrac{\frac{a}{3}}{a}$ giving $k = 4mg$ | A1 | AG |
| Let greatest height above $Q$ be $\frac{4}{3}a + x$; Gain in GPE $= mgx$ and Loss in KE $= \frac{1}{2}m \cdot 2ga$ | B1 | The length being found may be expressed as the total extension of the string or the greatest height above $Q$. GPE and KE |
| Gain in EPE $= \frac{1}{2} \cdot \frac{4mg}{a}\left(\left(x+\frac{a}{3}\right)^2 - \left(\frac{a}{3}\right)^2\right)$ | B1 | EPE Note: initial EPE $= \frac{2mga}{9}$ |
| $\frac{4mg}{2a}\left(x^2 + \frac{2ax}{3} + \frac{a^2}{9} - \frac{a^2}{9}\right) + mgx = mga$ | M1 A1 | Energy equation, correct number of terms |
| $2x^2 + \frac{7ax}{3} - a^2 = 0$ | M1 | Simplify to quadratic |
| $x = \frac{1}{3}a$ so greatest height is $\frac{5}{3}a$ | A1 | |
6 One end of a light elastic string, of natural length $a$ and modulus of elasticity $k$, is attached to a particle $P$ of mass $m$. The other end of the string is attached to a fixed point $Q$. The particle $P$ is projected vertically upwards from $Q$. When $P$ is moving upwards and at a distance $\frac { 4 } { 3 } a$ directly above $Q$, it has a speed $\sqrt { 2 g a }$. At this point, its acceleration is $\frac { 7 } { 3 } g$ downwards.
Show that $\mathrm { k } = 4 \mathrm { mg }$ and find in terms of $a$ the greatest height above $Q$ reached by $P$.\\
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q6 [8]}}