| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with hole removed |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard composite body techniques. Part (a) involves basic Hooke's law and Newton's second law with simple arithmetic. Part (b) uses energy conservation with elastic potential energy. The lamina problem uses the standard method of subtracting a triangle from a square, with clearly defined dimensions and no geometric complications. All steps are routine applications of standard formulas, making this easier than average for Further Maths. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T - mg = m \cdot a\) | M1 | |
| \(T = 5mg\), \(\frac{1}{2}a / a = \frac{5}{2}mg\) | M1 | |
| \(a = \frac{3}{2}g\) (upwards) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gain in KE \(= \frac{1}{2}mv^2\); Gain in GPE \(= \frac{1}{2}mga\) | B1 | |
| Loss in EPE \(= \frac{1}{2}\dfrac{5mg\cdot\left(\frac{1}{2}a\right)^2}{a}\) | B1 | |
| \(\frac{1}{2}mv^2 + \frac{1}{2}mga = \frac{1}{2}\dfrac{5mg\cdot\left(\frac{1}{2}a\right)^2}{a}\) \(\Rightarrow \frac{1}{2}mv^2 + \frac{1}{2}mga = \frac{5}{8}mga\) | M1 | |
| \(v = \frac{1}{2}\sqrt{ga}\) | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T - mg = m \cdot a$ | M1 | |
| $T = 5mg$, $\frac{1}{2}a / a = \frac{5}{2}mg$ | M1 | |
| $a = \frac{3}{2}g$ (upwards) | A1 | AG |
---
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gain in KE $= \frac{1}{2}mv^2$; Gain in GPE $= \frac{1}{2}mga$ | B1 | |
| Loss in EPE $= \frac{1}{2}\dfrac{5mg\cdot\left(\frac{1}{2}a\right)^2}{a}$ | B1 | |
| $\frac{1}{2}mv^2 + \frac{1}{2}mga = \frac{1}{2}\dfrac{5mg\cdot\left(\frac{1}{2}a\right)^2}{a}$ $\Rightarrow \frac{1}{2}mv^2 + \frac{1}{2}mga = \frac{5}{8}mga$ | M1 | |
| $v = \frac{1}{2}\sqrt{ga}$ | A1 | |
---3 One end of a light elastic spring, of natural length $a$ and modulus of elasticity 5 mg , is attached to a fixed point $A$. The other end of the spring is attached to a particle $P$ of mass $m$. The spring hangs with $P$ vertically below $A$. The particle $P$ is released from rest in the position where the extension of the spring is $\frac { 1 } { 2 } a$.
\begin{enumerate}[label=(\alph*)]
\item Show that the initial acceleration of $P$ is $\frac { 3 } { 2 } g$ upwards.
\item Find the speed of $P$ when the spring first returns to its natural length.\\
\includegraphics[max width=\textwidth, alt={}, center]{7251b13f-1fae-4138-80ea-e6b8091c94ab-08_581_659_267_708}
A uniform square lamina $A B C D$ has sides of length 10 cm . The point $E$ is on $B C$ with $E C = 7.5 \mathrm {~cm}$, and the point $F$ is on $D C$ with $\mathrm { CF } = \mathrm { xcm }$. The triangle $E F C$ is removed from $A B C D$ (see diagram). The centre of mass of the resulting shape $A B E F D$ is a distance $\bar { x } \mathrm {~cm}$ from $C B$ and a distance $\bar { y } \mathrm {~cm}$ from CD.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [7]}}