| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with straightforward application of Newton's second law to connected particles. Parts (a)-(c) involve routine F=ma calculations with given acceleration. Part (d) requires using SUVAT equations and energy/kinematics after impact, which is a common M1 scenario but adds modest complexity through multi-stage motion analysis. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{15}{4}mg\) ✱ | M1 A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 3\) | M1 A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| The tensions in the two parts of the string are the same | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = 2s_1 + s_2 = 3.969 \approx 4.0\) (m) | M1 A1 M1 A1 M1 A1 A1 | (7) [14] |
**Part (a):**
N2L A: $5mg - T = 5m \times \frac{1}{4}g$
$T = \frac{15}{4}mg$ ✱ | M1 A1 A1 | (3) | cso
**Part (b):**
N2L B: $T - kmg = km \times \frac{1}{4}g$
$k = 3$ | M1 A1 A1 | (3)
**Part (c):**
The tensions in the two parts of the string are the same | B1 | (1)
**Part (d):**
Distance of A above ground: $s_1 = \frac{1}{2} \times \frac{1}{4}g \times 1.2^2 = 0.18g$ ($\approx$ 1.764)
Speed on reaching ground: $v = \frac{1}{4}g \times 1.2 = 0.3g$ ($\approx$ 2.94)
For B under gravity: $(0.3g)^2 = 2gs_2 \Rightarrow s_2 = \frac{(0.3)^2g}{2} \approx 0.441)$
$S = 2s_1 + s_2 = 3.969 \approx 4.0$ (m) | M1 A1 M1 A1 M1 A1 A1 | (7) [14]\includegraphics{figure_4}
Two particles $A$ and $B$ have masses $5m$ and $km$ respectively, where $k < 5$. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut, the hanging parts of the string vertical and with $A$ and $B$ at the same height above a horizontal plane, as shown in Figure 4. The system is released from rest. After release, $A$ descends with acceleration $\frac{1}{4}g$.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string as $A$ descends is $\frac{15}{4}mg$. [3]
\item Find the value of $k$. [3]
\item State how you have used the information that the pulley is smooth. [1]
\end{enumerate}
After descending for 1.2 s, the particle $A$ reaches the plane. It is immediately brought to rest by the impact with the plane. The initial distance between $B$ and the pulley is such that, in the subsequent motion, $B$ does not reach the pulley.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the greatest height reached by $B$ above the plane. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2010 Q6 [14]}}