| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with connected particles requiring straightforward application of Newton's second law and kinematics. Part (a) uses SUVAT with given values, parts (b-c) involve routine force equations for the two-particle system, and part (e) requires tracking motion after impact—all standard M1 techniques with no novel insight required. Slightly easier than average due to clear structure and guided steps. |
| 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 pulleys6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(s = ut + \frac{1}{2}at^2\) ⟹ \(3.15 = \frac{1}{4}a \times 2^2\) | M1 A1 | (3 marks) |
| \(a = 2.8\) (ms⁻²) ⭐ | cso A1 | |
| (b) N2L for \(P\): \(0.5g - T = 0.5 \times 2.8\) | M1 A1 | (3 marks) |
| \(T = 3.5\) (N) | A1 | |
| (c) N2L for \(Q\): \(T - mg = 2.8m\) | M1 A1 | (4 marks) |
| \(m = \frac{3.5}{12.6} = \frac{5}{18}\) ⭐ | cso DM1 A1 | |
| (d) The acceleration of \(P\) is equal to the acceleration of \(Q\). | B1 | (1 mark) |
| (e) \(v = u + at\) ⟹ \(v = 2.8 \times 1.5\) | M1 A1 | (6 marks) |
| Answer | Marks |
|---|---|
| \(v = u + at\) ⟹ \(4.2 = -4.2 + 9.8t\) | DM1 A1 |
| \(t = \frac{6}{7}, 0.86, 0.857\) (s) | DM1 A1 |
| [17] |
**(a)** $s = ut + \frac{1}{2}at^2$ ⟹ $3.15 = \frac{1}{4}a \times 2^2$ | M1 A1 | (3 marks)
$a = 2.8$ (ms⁻²) ⭐ | cso A1 |
**(b)** N2L for $P$: $0.5g - T = 0.5 \times 2.8$ | M1 A1 | (3 marks)
$T = 3.5$ (N) | A1 |
**(c)** N2L for $Q$: $T - mg = 2.8m$ | M1 A1 | (4 marks)
$m = \frac{3.5}{12.6} = \frac{5}{18}$ ⭐ | cso DM1 A1 |
**(d)** The acceleration of $P$ is equal to the acceleration of $Q$. | B1 | (1 mark)
**(e)** $v = u + at$ ⟹ $v = 2.8 \times 1.5$ | M1 A1 | (6 marks)
(or $v^2 = u^2 + 2as$ ⟹ $v^2 = 2 \times 2.8 \times 3.15$)
($v^2 = 17.64, v = 4.2$)
$v = u + at$ ⟹ $4.2 = -4.2 + 9.8t$ | DM1 A1 |
$t = \frac{6}{7}, 0.86, 0.857$ (s) | DM1 A1 |
| | [17] |6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\includegraphics[alt={},max width=\textwidth]{5b5d70b1-1eb6-461f-9277-5912b914f443-10_572_586_299_696}
\end{center}
\end{figure}
Two particles $P$ and $Q$ have mass 0.5 kg and $m \mathrm {~kg}$ respectively, where $m < 0.5$. The particles are connected by a light inextensible string which passes over a smooth, fixed pulley. Initially $P$ is 3.15 m above horizontal ground. The particles are released from rest with the string taut and the hanging parts of the string vertical, as shown in Figure 4. After $P$ has been descending for 1.5 s , it strikes the ground. Particle $P$ reaches the ground before $Q$ has reached the pulley.
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of $P$ as it descends is $2.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the tension in the string as $P$ descends.
\item Show that $m = \frac { 5 } { 18 }$.
\item State how you have used the information that the string is inextensible.
When $P$ strikes the ground, $P$ does not rebound and the string becomes slack. Particle $Q$ then moves freely under gravity, without reaching the pulley, until the string becomes taut again.
\item Find the time between the instant when $P$ strikes the ground and the instant when the string becomes taut again.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2007 Q6 [17]}}