| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - inclined plane involved |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with connected particles. Parts (a)-(d) involve routine application of F=ma to both particles and using the constraint that accelerations are equal, which is textbook material. Parts (e)-(f) add a collision element requiring use of SUVAT equations, but the problem-solving is straightforward and methodical. Slightly easier than average due to the step-by-step scaffolding and standard techniques throughout. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| (a) N2L for \(Q\): \(2g - T = 2a\) | M1 A1 | N2L for \(P\): \(T - 3g \sin 30° = 3a\) |
| (b) \(2g - 3g \sin 30° = 5a\) → \(a = 0.98\) (m s\(^{-2}\)) | M1 A1 | cso |
| (c) \(T = 2(g - a)\) → \(≈ 18\) (N) | M1 A1 | accept 17.6 |
| (d) The (magnitudes of the) accelerations of \(P\) and \(Q\) are equal | B1 | 1 mark |
| (e) \(v^2 = u^2 + 2as ⇒ v^2 = 2 \times 0.98 \times 0.8\) (\(= 1.568\)) → \(v ≈ 1.3\) (m s\(^{-1}\)) | M1 A1 | accept 1.25 |
| (f) N2L for \(P\): \(-3g \sin 30° = 3a\) → \(a = (-) \frac{1}{2} g\) | M1 A1 | \(s = ut + \frac{1}{2}at^2 ⇒ 0 = \sqrt{1.568}t - \frac{1}{2} \times 4.9t^2\) or equivalent |
**(a)** N2L for $Q$: $2g - T = 2a$ | M1 A1 | N2L for $P$: $T - 3g \sin 30° = 3a$ | M1 A1 | 4 marks
**(b)** $2g - 3g \sin 30° = 5a$ → $a = 0.98$ (m s$^{-2}$) | M1 A1 | cso | 2 marks
**(c)** $T = 2(g - a)$ → $≈ 18$ (N) | M1 A1 | accept 17.6 | 2 marks
**(d)** The (magnitudes of the) accelerations of $P$ and $Q$ are equal | B1 | 1 mark
**(e)** $v^2 = u^2 + 2as ⇒ v^2 = 2 \times 0.98 \times 0.8$ ($= 1.568$) → $v ≈ 1.3$ (m s$^{-1}$) | M1 A1 | accept 1.25 | 2 marks
**(f)** N2L for $P$: $-3g \sin 30° = 3a$ → $a = (-) \frac{1}{2} g$ | M1 A1 | $s = ut + \frac{1}{2}at^2 ⇒ 0 = \sqrt{1.568}t - \frac{1}{2} \times 4.9t^2$ or equivalent | M1 A1 | → $t = 0.51$ (s) | A1 | accept 0.511 | 5 marks | 16 marks total
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**General Note:** A maximum of one mark can be lost for giving too great accuracy.\includegraphics{figure_4}
Figure 4 shows two particles $P$ and $Q$, of mass 3 kg and 2 kg respectively, connected by a light inextensible string. Initially $P$ is held at rest on a fixed smooth plane inclined at 30° to the horizontal. The string passes over a small smooth light pulley $A$ fixed at the top of the plane. The part of the string from $P$ to $A$ is parallel to a line of greatest slope of the plane. The particle $Q$ hangs freely below $A$. The system is released from rest with the string taut.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation of motion for $P$ and an equation of motion for $Q$. [4]
\item Hence show that the acceleration of $Q$ is 0.98 m s$^{-2}$. [2]
\item Find the tension in the string. [2]
\item State where in your calculations you have used the information that the string is inextensible. [1]
\end{enumerate}
On release, $Q$ is at a height of 0.8 m above the ground. When $Q$ reaches the ground, it is brought to rest immediately by the impact with the ground and does not rebound. The initial distance of $P$ from $A$ is such that in the subsequent motion $P$ does not reach $A$. Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item the speed of $Q$ as it reaches the ground, [2]
\item the time between the instant when $Q$ reaches the ground and the instant when the string becomes taut again. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2007 Q7 [16]}}