| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Topic | Momentum and Collisions |
| Type | String becomes taut problem |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring geometric insight (finding the angle when string becomes taut from the ½d spacing), conservation of momentum in 2D with impulse resolution, and energy calculations. The 'show that' format with a specific non-obvious result and the large-m limiting case analysis elevate this beyond standard A-level mechanics to Further Maths territory, requiring multi-step reasoning and physical insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation |
4 Two particles $A$ and $B$, of masses $m \mathrm {~kg}$ and 1 kg respectively, are connected by a light inextensible string of length $d \mathrm {~m}$ and placed at rest on a smooth horizontal plane a distance of $\frac { 1 } { 2 } d \mathrm {~m}$ apart. $B$ is then projected horizontally with speed $v \mathrm {~ms} ^ { - 1 }$ in a direction perpendicular to $A B$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at the instant that the string becomes taut, the magnitude of the instantaneous impulse in the string, $I \mathrm { Ns }$, is given by $I = \frac { \sqrt { 3 } m v } { 2 ( 1 + m ) }$.
\item Find, in terms of $m$ and $v$, the kinetic energy of $B$ at the instant after the string becomes taut. Give your answer as a single algebraic fraction.
\item In the case where $m$ is very large, describe, with justification, the approximate motion of $B$ after the string becomes taut.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q4 [9]}}