| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Standard +0.3 This is a standard Atwood machine problem with straightforward application of Newton's second law, followed by routine kinematics and coefficient of restitution calculations. All three parts use well-practiced mechanics techniques with no novel insight required, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Applying Newton's law to the particles, where \(a\) is the acceleration of \(A\) and \(T\) is the tension in the string: \(2a = T - 2g; 3a = 3g - T\) | M1 A1 | |
| Adding and solving: \(a = 0.2g\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(v^2 = u^2 + 2as\) (\(u = 0, s = 1\)), the common speed of \(A\) and \(B\) is: \(\sqrt{\frac{2g}{5}}\) (= 2 ms\(^{-1}\)) | M1 A1 | |
| Using \(v^2 = u^2 + 2as\) for the motion of \(A\), with \(a = -g, u = \sqrt{\frac{2g}{5}}\), gives a total height of 2.2 m above the ground | B1 | [3] |
| If \(e\) is the coefficient of restitution, we use above formula with \(v = 0\), \(a\) replaced by \(-g, s = 0.05, u = e\sqrt{\frac{2g}{5}}\) | M1 | |
| Obtain \(e = 0.5\) | A1 | [2] |
### (i)
Applying Newton's law to the particles, where $a$ is the acceleration of $A$ and $T$ is the tension in the string: $2a = T - 2g; 3a = 3g - T$ | M1 A1 |
Adding and solving: $a = 0.2g$ | A1 | [3]
### (ii)
Using $v^2 = u^2 + 2as$ ($u = 0, s = 1$), the common speed of $A$ and $B$ is: $\sqrt{\frac{2g}{5}}$ (= 2 ms$^{-1}$) | M1 A1 |
Using $v^2 = u^2 + 2as$ for the motion of $A$, with $a = -g, u = \sqrt{\frac{2g}{5}}$, gives a total height of 2.2 m above the ground | B1 | [3]
If $e$ is the coefficient of restitution, we use above formula with $v = 0$, $a$ replaced by $-g, s = 0.05, u = e\sqrt{\frac{2g}{5}}$ | M1 |
Obtain $e = 0.5$ | A1 | [2]A light inextensible string passes over a fixed smooth light pulley. Particles $A$ and $B$, of masses 2 kg and 3 kg respectively, are attached to the ends so that the portions of the string below the axis of the pulley are vertical (see diagram). The centre of the horizontal axis of the pulley is 4 m above ground level.
\includegraphics{figure_13}
The particles are released from rest 1 m above ground level with the string taut.
\begin{enumerate}[label=(\roman*)]
\item Determine the acceleration of both particles prior to the impact of $B$ with the ground. [3]
\item Determine the greatest height attained by $A$ above ground level. [3]
\item If $B$ rebounds after impact to a first maximum height of 0.05 m, determine the coefficient of restitution between $B$ and the ground. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q13 [8]}}