| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| 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. Part (a) requires routine application of F=ma to both particles with given answer to show. Parts (b)-(d) involve straightforward kinematics after Q hits ground. All techniques are standard textbook exercises requiring no novel insight, though the multi-part nature and careful tracking of motion phases makes it slightly above average difficulty for M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3g - T = 3a\) | M1 | Eqn of motion for \(Q\): must have the correct terms but condone sign errors |
| A1 | Correct equation | |
| \(T - 2g\cos60 = 2a \quad (T - g = 2a)\) | M1 | Eqn of motion for \(P\): must have the correct terms but condone sign errors. Weight must be resolved |
| A1 | Correct equation | |
| Allow M1A1 for \(3g - 2g\cos60 = 5a\) in place of either of these two equations | ||
| \(2g = 5a \quad a = \frac{2g}{5}\) ✱ | DM1 | Use an exact method to solve for \(a\) (i.e. not the equation solver on their calculator). Dependent on the first 2 M marks or the M for the combined equation |
| A1 | Given answer derived correctly from exact working | |
| \(T = 2 \times \frac{2g}{5} + g = \frac{9g}{5}\) | M1 | Use given acceleration to solve for \(T\) |
| A1(8) | Accept 18 or 17.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 2 \times \frac{2g}{5} \times 0.6 = \frac{2.4g}{5}\) | M1 | Use the given acceleration to find the speed |
| \(v = \frac{2}{5}\sqrt{3g}\) oe involving \(g\) | A1(2) | Accept 2.2 or 2.17 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| String slack: accel of \(P\) (up plane) \(= -g\cos60 = -\frac{1}{2}g\) | B1 | |
| \(0 = \frac{2.4g}{5} - gs\) | M1 | Use of \(v^2 = u^2 + 2as\) or equivalent for their acceleration \(\neq \frac{2g}{5}\) |
| \(s = \frac{2.4g}{5} \times \frac{1}{g} = \frac{2.4}{5} = 0.48\) | A1 | |
| Total dist \(= 1.08\) m | A1ft(4) | \(0.6 +\) their \(0.48\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = \frac{2}{5}\sqrt{3g} - \frac{g}{2}t \quad (0 = 2.17 - 4.9t)\) | M1 | Use of \(v = u + at\) or equivalent with their acceleration \(\neq \frac{2g}{5}\) to find \(t\) |
| \(t = \frac{4\sqrt{3g}}{5g} = 0.4426...\) | ||
| \(= 0.44\) or \(0.443\) | A1(2) | only |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3g - T = 3a$ | M1 | Eqn of motion for $Q$: must have the correct terms but condone sign errors |
| | A1 | Correct equation |
| $T - 2g\cos60 = 2a \quad (T - g = 2a)$ | M1 | Eqn of motion for $P$: must have the correct terms but condone sign errors. Weight must be resolved |
| | A1 | Correct equation |
| Allow M1A1 for $3g - 2g\cos60 = 5a$ in place of either of these two equations | | |
| $2g = 5a \quad a = \frac{2g}{5}$ ✱ | DM1 | Use an exact method to solve for $a$ (i.e. not the equation solver on their calculator). Dependent on the first 2 M marks or the M for the combined equation |
| | A1 | Given answer derived correctly from exact working |
| $T = 2 \times \frac{2g}{5} + g = \frac{9g}{5}$ | M1 | Use given acceleration to solve for $T$ |
| | A1(8) | Accept 18 or 17.6 |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 2 \times \frac{2g}{5} \times 0.6 = \frac{2.4g}{5}$ | M1 | Use the given acceleration to find the speed |
| $v = \frac{2}{5}\sqrt{3g}$ oe involving $g$ | A1(2) | Accept 2.2 or 2.17 |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| String slack: accel of $P$ (up plane) $= -g\cos60 = -\frac{1}{2}g$ | B1 | |
| $0 = \frac{2.4g}{5} - gs$ | M1 | Use of $v^2 = u^2 + 2as$ or equivalent for their acceleration $\neq \frac{2g}{5}$ |
| $s = \frac{2.4g}{5} \times \frac{1}{g} = \frac{2.4}{5} = 0.48$ | A1 | |
| Total dist $= 1.08$ m | A1ft(4) | $0.6 +$ their $0.48$ |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = \frac{2}{5}\sqrt{3g} - \frac{g}{2}t \quad (0 = 2.17 - 4.9t)$ | M1 | Use of $v = u + at$ or equivalent with their acceleration $\neq \frac{2g}{5}$ to find $t$ |
| $t = \frac{4\sqrt{3g}}{5g} = 0.4426...$ | | |
| $= 0.44$ or $0.443$ | A1(2) | only |
**Total: [16]**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{edcc4603-f006-4c4f-a4e5-063cab41da98-12_486_1257_230_347}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two particles $P$ and $Q$, of mass 2 kg and 3 kg respectively, are connected by a light inextensible string. Initially $P$ is held at rest on a fixed smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The string passes over a small smooth fixed pulley at the top of the plane. The particle $Q$ hangs freely below the pulley and 0.6 m above the ground, as shown in Figure 3. The part of the string from $P$ to the pulley is parallel to a line of greatest slope of the plane. The system is released from rest with the string taut.
For the motion before $Q$ hits the ground,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item show that the acceleration of $Q$ is $\frac { 2 g } { 5 }$,
\item find the tension in the string.
On hitting the ground $Q$ is immediately brought to rest by the impact.
\end{enumerate}\item Find the speed of $P$ at the instant when $Q$ hits the ground.
In its subsequent motion $P$ does not reach the pulley.
\item Find the total distance moved up the plane by $P$ before it comes to instantaneous rest.
\item Find the length of time between $Q$ hitting the ground and $P$ first coming to instantaneous rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q7 [16]}}