| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Moderate -0.3 This is a standard M1 connected particles problem requiring Newton's second law applied to both masses and solving simultaneous equations. Part (a) tests understanding of constraint (same acceleration magnitude), part (b) is routine application of F=ma to find k, and part (c) uses basic kinematics (v²=u²+2as). While multi-part with 11 marks total, each step follows textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| (a) e.g. string is inextensible so \(B\) moves down same dist. \(A\) moves up | B1 | |
| \(\therefore \text{acceleration of } B \text{ is } \frac{1}{4}g \text{ m s}^{-2} \text{ downwards}\) | B1 | |
| (b) \(\text{eqn. of motion for } A: kmg - T = kma\) (1) | M1 | |
| \(\text{eqn. of motion for } B: T - mg = ma\) (2) | M1 | |
| \((1) + (2) \text{ gives } kmg - mg = kma + ma\) | M1 A1 | |
| \(k(g - a) = g + a \cdot k = \frac{g+a}{g-a} = \frac{+g}{-} \therefore k = \frac{5}{3}\) | M1 A1 | |
| (c) \(u = 0, s = 0.5, a = \frac{1}{4}g \text{ use } v^2 = u^2 + 2as\) | M1 | |
| \(v^2 = 0 + 2(0.25g)(0.5) = 2.45 \therefore v = 1.57 \text{ m s}^{-1}\) (3sf) | M1 A1 | (11 marks) |
(a) e.g. string is inextensible so $B$ moves down same dist. $A$ moves up | B1 |
$\therefore \text{acceleration of } B \text{ is } \frac{1}{4}g \text{ m s}^{-2} \text{ downwards}$ | B1 |
(b) $\text{eqn. of motion for } A: kmg - T = kma$ (1) | M1 |
$\text{eqn. of motion for } B: T - mg = ma$ (2) | M1 |
$(1) + (2) \text{ gives } kmg - mg = kma + ma$ | M1 A1 |
$k(g - a) = g + a \cdot k = \frac{g+a}{g-a} = \frac{+g}{-} \therefore k = \frac{5}{3}$ | M1 A1 |
(c) $u = 0, s = 0.5, a = \frac{1}{4}g \text{ use } v^2 = u^2 + 2as$ | M1 |
$v^2 = 0 + 2(0.25g)(0.5) = 2.45 \therefore v = 1.57 \text{ m s}^{-1}$ (3sf) | M1 A1 | (11 marks)\includegraphics{figure_3}
Figure 3 shows two particles $A$ and $B$ of masses $m$ and $km$ respectively, connected by a light inextensible string which passes over a smooth fixed pulley.
When the system is released from rest with both particles 0.5 m above the ground, particle $A$ moves vertically upwards with acceleration $\frac{1}{4} g \text{ m s}^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Write down, with a brief justification, the magnitude and direction of the acceleration of $B$. [2 marks]
\item Find the value of $k$. [6 marks]
Given that $A$ does not hit the pulley,
\item calculate, correct to 3 significant figures, the speed with which $B$ hits the ground. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q5 [11]}}