| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.8 This is a straightforward two-particle pulley problem requiring standard application of Newton's second law to each particle separately, then solving simultaneous equations. The setup is clearly described, forces are given explicitly, and the method is routine for M1 students. Easier than average due to its direct approach and lack of complicating factors. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| For applying \(F = ma\) | M1 | Requires at least \(ma\), \(T\) and air resistance in linear combination in at least one equation |
| At least one equation with not more than one error | A1 | |
| \(0.2g + T - 0.4 = 0.2a\) | A1 | |
| \(0.3g - T - 0.25 = 0.3a\) | A1 | Total: 4 |
| SR: \(0.2g - T - 0.4 = 0.2a\) and \(0.3g + T - 0.25 = 0.3a\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| For obtaining equation in \(T\) or \(a\) only | M1 | Either by eliminating \(a\) or \(T\) from equations in (i), or by applying \(F=ma\) to the complete system |
| \(0.5g - 0.65 = 0.5a\) or \(5T - 0.7 = 0\) | A1ft | For correct equation in \(a\) only or \(T\) only; ft opposite direction of \(T\) only |
| \(a = 8.5\) and \(T = 0.14\) (positive only) | A1 | Total: 3 |
# Question 2:
## Part (i)
| For applying $F = ma$ | M1 | Requires at least $ma$, $T$ and air resistance in linear combination in at least one equation |
| At least one equation with not more than one error | A1 | |
| $0.2g + T - 0.4 = 0.2a$ | A1 | |
| $0.3g - T - 0.25 = 0.3a$ | A1 | **Total: 4** |
| SR: $0.2g - T - 0.4 = 0.2a$ **and** $0.3g + T - 0.25 = 0.3a$ | B1 | |
## Part (ii)
| For obtaining equation in $T$ or $a$ only | M1 | Either by eliminating $a$ or $T$ from equations in (i), or by applying $F=ma$ to the complete system |
| $0.5g - 0.65 = 0.5a$ or $5T - 0.7 = 0$ | A1ft | For correct equation in $a$ only or $T$ only; ft opposite direction of $T$ only |
| $a = 8.5$ **and** $T = 0.14$ (positive only) | A1 | **Total: 3** |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-2_643_289_1475_927}
Particles $A$ and $B$, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. Particle $A$ is held at rest at a fixed point and $B$ hangs vertically below $A$. Particle $A$ is now released. As the particles fall the air resistance acting on $A$ is 0.4 N and the air resistance acting on $B$ is 0.25 N (see diagram). The downward acceleration of each of the particles is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and the tension in the string is $T \mathrm {~N}$.\\
(i) Write down two equations in $a$ and $T$ obtained by applying Newton's second law to $A$ and to $B$.\\
(ii) Find the values of $a$ and $T$.
\hfill \mbox{\textit{OCR M1 2005 Q2 [7]}}