| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Force on pulley from string |
| Difficulty | Moderate -0.3 This is a standard M1 pulley problem requiring Newton's second law applied to a two-particle system. Parts (a) and (b) are routine textbook exercises (set up equations for each mass, solve simultaneously), while part (c) requires vector addition of tensions but follows a well-practiced method. Slightly easier than average due to the straightforward setup and lack of complications. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| For \(A\): \(T = 2ma\) | B1 |
| For \(B\): \(3mg - T = 3ma\) | M1 A1 |
| \(3mg = 5ma\) | DM1 |
| \(\frac{3g}{5} = a\) | A1 |
| \((5.9\) or \(5.88\) m s\(^{-2})\) | |
| (5) |
| Answer | Marks |
|---|---|
| \(T = 6mg/5; 12m; 11.8m\) | B1 |
| (1) |
| Answer | Marks |
|---|---|
| \(F = \sqrt{T^2 + T^2}\) | M1 A1 ft |
| \(F = \frac{6mg\sqrt{2}}{5}\) ;1.7mg (or better);16.6m;17m | A1 |
| Direction clearly marked on a diagram, with an arrow, and \(45°\) (oe) marked | B1 |
| (4) | |
| [10] |
## Part (a)
For $A$: $T = 2ma$ | B1
For $B$: $3mg - T = 3ma$ | M1 A1
$3mg = 5ma$ | DM1
$\frac{3g}{5} = a$ | A1
$(5.9$ or $5.88$ m s$^{-2})$ |
| (5)
## Part (b)
$T = 6mg/5; 12m; 11.8m$ | B1
| (1)
## Part (c)
$F = \sqrt{T^2 + T^2}$ | M1 A1 ft
$F = \frac{6mg\sqrt{2}}{5}$ ;1.7mg (or better);16.6m;17m | A1
Direction clearly marked on a diagram, with an arrow, and $45°$ (oe) marked | B1
| (4)
| [10]
**Notes for Question 8:**
**Q8(a)**
- B1 for $T = 2ma$
- First M1 for resolving vertically (up or down) for $B$, with correct no. of terms. (allow omission of $m$, provided 3 is there)
- First A1 for a correct equation.
- Second M1, dependent on first M1, for eliminating $T$, to give an equation in $a$ only.
- Second A1 for 0.6g, 5.88 or 5.9.
- N.B. 'Whole system' equation: $3mg = 5ma$ earns first 4 marks but any error loses all 4.
**Q8(b)**
- B1 for $\frac{6mg}{5}$ ,11.8m, 12m
**Q8(c)**
- M1 $\sqrt{(T^2 + T^2)}$ or $\frac{T}{\sin 45°}$ or $\frac{T}{\cos 45°}$ or $2T\cos 45°$ or $2T\sin 45°$ (allow if $m$ omitted)
- (M0 for $T \sin 45°$)
- First A1 ft on their $T$.
- Second A1 cao for $\frac{6mg\sqrt{2}}{5}$ oe, 1.7mg (or better),16.6m,17m
- B1 for the direction clearly shown on a diagram with an arrow and $45°$ marked.
---\includegraphics{figure_2}
Two particles $A$ and $B$ have masses $2m$ and $3m$ respectively. The particles are attached to the ends of a light inextensible string. Particle $A$ is held at rest on a smooth horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. Particle $B$ hangs at rest vertically below the pulley with the string taut, as shown in Figure 2. Particle $A$ is released from rest. Assuming that $A$ has not reached the pulley, find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $B$, [5]
\item the tension in the string, [1]
\item the magnitude and direction of the force exerted on the pulley by the string. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q8 [10]}}