| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| 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 students to find acceleration using F=ma for the system, then calculate tension, and finally find the force on the pulley (2T). The method is routine and well-practiced, though it requires careful application of multiple steps. Part (b) is straightforward recall. Slightly easier than average due to being a textbook-standard question type. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5mg - T = 5ma\) OR \(5mg - T = -5ma\) | M1 A1 | Correct number of terms; condone sign errors (M0 if \(m\)'s missing); correct equation |
| \(T - 3mg = 3ma\) OR \(T - 3mg = -3ma\) | M1 A1 | Correct number of terms; condone sign errors (M0 if \(m\)'s missing); correct equation |
| Solve for \(T\) | DM1 | Dependent on previous two M marks, must be in terms of \(m\) |
| \(T = \frac{15mg}{4}\) oe (allow unsimplified and not in terms of \(mg\) at this stage) | A1 | Correct expression for \(T\) |
| Force on pulley \(= 2T\) | M1 | Correct method |
| \(\frac{15mg}{2}\) oe (must be a single positive term) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The tension is the same on both sides of the pulley. | B1 | Any equivalent statement. B0 if any incorrect extras; B0 if pulley not mentioned |
# Question 7:
## Part 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5mg - T = 5ma$ **OR** $5mg - T = -5ma$ | M1 A1 | Correct number of terms; condone sign errors (M0 if $m$'s missing); correct equation |
| $T - 3mg = 3ma$ **OR** $T - 3mg = -3ma$ | M1 A1 | Correct number of terms; condone sign errors (M0 if $m$'s missing); correct equation |
| Solve for $T$ | DM1 | Dependent on previous two M marks, must be in terms of $m$ |
| $T = \frac{15mg}{4}$ oe (allow unsimplified and not in terms of $mg$ at this stage) | A1 | Correct expression for $T$ |
| Force on pulley $= 2T$ | M1 | Correct method |
| $\frac{15mg}{2}$ oe (must be a single positive term) | A1 | Correct answer |
## Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The tension is the same on both sides of the pulley. | B1 | Any equivalent statement. B0 if any incorrect extras; B0 if pulley not mentioned |
---7.
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\caption{Figure 4}
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One end of a light inextensible string is attached to a particle $A$ of mass $5 m$. The other end of the string is attached to a particle $B$ of mass $3 m$. The string passes over a small, smooth, light fixed pulley. Particle $A$ is held at rest with the string taut and the hanging parts of the string vertical, as shown in Figure 4.
Particle A is released.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the force exerted on the pulley by the string while $A$ is falling and before $B$ hits the pulley.
\item State how, in your solution to part (a), you have used the fact that the pulley is smooth.
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\hfill \mbox{\textit{Edexcel M1 2020 Q7 [9]}}