| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Three or more connected particles |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with connected particles requiring Newton's second law applied to multiple bodies. While it involves three masses and multiple parts, the techniques are routine: resolve forces on the inclined plane (using given sin α), apply F=ma to the system, then individual particles, and finally vector addition for the pulley force. The given sin α = 3/5 simplifies calculations. This is slightly easier than average because it's a textbook application of standard methods with no novel insight required. |
| Spec | 3.03i Normal reaction force3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03m Equilibrium: sum of resolved forces = 03.03o Advanced connected particles: and pulleys3.03p Resultant forces: using vectors3.03q Dynamics: motion under force in plane |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T - 5g\sin\alpha = 5a\) | M1 A1 | |
| \(15g - T = 15a\) | M1 A1 | |
| solving for \(a\): \(a = 0.6g\) | M1 A1 | |
| solving for \(T\): \(T = 6g\) | M1 A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| For \(Q\): \(5g - N = 5a\) | M1 A1 | |
| \(N = 2g\) | A1 f.t. | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(F = 2T\cos\!\left(\frac{90° - \alpha}{2}\right)\) | M1 A2 | |
| \(= 12g\cos 26.56°\) | A1 f.t. | |
| \(= 105\ \text{N}\) | A1 | (5) [16] |
## Question 7:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T - 5g\sin\alpha = 5a$ | M1 A1 | |
| $15g - T = 15a$ | M1 A1 | |
| solving for $a$: $a = 0.6g$ | M1 A1 | |
| solving for $T$: $T = 6g$ | M1 A1 | **(8)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| For $Q$: $5g - N = 5a$ | M1 A1 | |
| $N = 2g$ | A1 f.t. | **(3)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $F = 2T\cos\!\left(\frac{90° - \alpha}{2}\right)$ | M1 A2 | |
| $= 12g\cos 26.56°$ | A1 f.t. | |
| $= 105\ \text{N}$ | A1 | **(5) [16]** |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86bb11a4-b409-49b1-bffb-d0e3727d345c-11_495_892_301_523}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
One end of a light inextensible string is attached to a block $P$ of mass 5 kg . The block $P$ is held at rest on a smooth fixed plane which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 3 } { 5 }$. The string lies along a line of greatest slope of the plane and passes over a smooth light pulley which is fixed at the top of the plane. The other end of the string is attached to a light scale pan which carries two blocks $Q$ and $R$, with block $Q$ on top of block $R$, as shown in Figure 3. The mass of block $Q$ is 5 kg and the mass of block $R$ is 10 kg . The scale pan hangs at rest and the system is released from rest. By modelling the blocks as particles, ignoring air resistance and assuming the motion is uninterrupted, find
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the acceleration of the scale pan,
\item the tension in the string,
\end{enumerate}\item the magnitude of the force exerted on block $Q$ by block $R$,
\item the magnitude of the force exerted on the pulley by the string.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2009 Q7 [16]}}