| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Year | 2023 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Collision/impulse during SHM |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics question requiring derivation of a second-order differential equation from Newton's second law with damping, solving it (critical damping case), and applying boundary conditions. However, the steps are fairly standard for FM students who have studied damped SHM: F=ma setup, recognizing critical damping from the discriminant, and applying initial conditions. The multi-part structure and need to connect physical parameters to mathematical form elevate it above average, but it follows a well-established template for damped oscillator problems. |
| Spec | 3.03l Newton's third law: extend to situations requiring force resolution4.10e Second order non-homogeneous: complementary + particular integral4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 6 | y |
| Answer | Marks | Guidance |
|---|---|---|
| High/Low | Time | Depth |
| Answer | Marks | Guidance |
|---|---|---|
| Low Tide | 5 a.m. | 2 |
| High Tide | 11 a.m. | 10 |
| Low Tide | 5 p.m. | 2 |
| High Tide | 11 p.m. | 10 |
Question 6:
6 | y
3x
Tidal Times
High/Low | Time | Depth
(metres)
Low Tide | 5 a.m. | 2
High Tide | 11 a.m. | 10
Low Tide | 5 p.m. | 2
High Tide | 11 p.m. | 106. The diagram on the left shows a train of mass 50 tonnes approaching a buffer at the end of a straight horizontal railway track. The buffer is designed to prevent the train from running off the end of the track. The buffer may be modelled as a light horizontal spring $A B$, as shown in the diagram on the right, which is fixed at the end $A$. The train strikes the buffer so that $P$ makes contact with $B$ at $t = 0$ seconds.
While $P$ is in contact with $B$, an additional resistive force of $250000 v \mathrm {~N}$ will oppose the motion of the train, where $v \mathrm {~ms} ^ { - 1 }$ is the speed of the train at time $t$ seconds. The spring has natural length 1 m and modulus of elasticity 312500 N . At time $t$ seconds, the compression of the spring is $x$ metres.\\
\includegraphics[max width=\textwidth, alt={}, center]{d7f600c5-af4a-4708-bfd9-92b37a95c634-7_358_1506_824_283}
\begin{enumerate}[label=(\alph*)]
\item Show that, while $P$ is in contact with $B$, $x$ satisfies the differential equation
$$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 20 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 0$$
\item Given that, when $P$ first makes contact with $B$, the speed of the train is $U \mathrm {~ms} ^ { - 1 }$, find an expression for $x$ in terms of $U$ and $t$.
\item When the train comes to rest, the compression of the buffer is 0.3 m . Determine the speed of the train when it strikes the buffer.
\item State which type of damping is described by the motion of $P$. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 2023 Q6 [16]}}