| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with a string-breaking complication. Part (i) requires routine application of Newton's second law to a connected system. Parts (ii) and (iii) involve straightforward kinematics (SUVAT equations) applied separately to each particle after the string breaks. The multi-step nature and need to track two particles adds modest complexity, but all techniques are standard M1 material with no novel insight required. |
| Spec | 3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | ||
| \(0.55g - T = 0.55a\) and \(T - 0.45g = 0.45a\) or \(a = [(0.55 - 0.45)(0.55 + 0.45)]g\) | M1 | |
| A1 | ||
| Acceleration is 1ms\(^{-2}\) | A1 | 3 |
| (ii) (a) | ||
| Height of P is 3m and height of Q is 7m | A1ft | 2 |
| (b) Speed is 2ms\(^{-1}\) | B1ft | 1 |
| (iii) \([3 = 2t_p + 5t_p^2, 7 = -2t_Q + 5t_Q^2]\) | M1 | For using \(s = ut + \frac{1}{2}gt^2\) for P or for Q (NB a = g) |
| \(t_p = 0.6\) | A1 | |
| Accept \(t_Q = 0.2 + 1.2\) following consideration of upward and downward motion under gravity of Q separately AG | ||
| \(t_Q = 1.4\) | A1 | |
| \(Q\) is 0.8s later than P | A1 | 4 |
(i) | | | For using Newton's second law to P or Q, or for using $a = \frac{M - m}{M + m}g$
$0.55g - T = 0.55a$ and $T - 0.45g = 0.45a$ or $a = [(0.55 - 0.45)(0.55 + 0.45)]g$ | M1 |
| A1 |
Acceleration is 1ms$^{-2}$ | A1 | 3 |
(ii) (a) | | | For using $s = 5 - \frac{1}{2}a2^2$ for P or $s = 5 + \frac{1}{2}a2^2$ for Q
Height of P is 3m and height of Q is 7m | A1ft | 2 | ft 5 – 2a and 5 + 2a
(b) Speed is 2ms$^{-1}$ | B1ft | 1 | ft 2a
(iii) $[3 = 2t_p + 5t_p^2, 7 = -2t_Q + 5t_Q^2]$ | M1 | For using $s = ut + \frac{1}{2}gt^2$ for P or for Q (NB a = g)
$t_p = 0.6$ | A1 |
| | | Accept $t_Q = 0.2 + 1.2$ following consideration of upward and downward motion under gravity of Q separately AG
$t_Q = 1.4$ | A1 |
$Q$ is 0.8s later than P | A1 | 4 |6\\
\includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-4_712_529_264_810}
Particles $P$ and $Q$, of masses 0.55 kg and 0.45 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. The particles are held at rest with the string taut and its straight parts vertical. Both particles are at a height of 5 m above the ground (see diagram). The system is released.\\
(i) Find the acceleration with which $P$ starts to move.
The string breaks after 2 s and in the subsequent motion $P$ and $Q$ move vertically under gravity.\\
(ii) At the instant that the string breaks, find
\begin{enumerate}[label=(\alph*)]
\item the height above the ground of $P$ and of $Q$,
\item the speed of the particles.\\
(iii) Show that $Q$ reaches the ground 0.8 s later than $P$.\\
$7 \quad$ A particle $P$ starts from rest at the point $A$ at time $t = 0$, where $t$ is in seconds, and moves in a straight line with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 10 s . For $10 \leqslant t \leqslant 20 , P$ continues to move along the line with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = \frac { 800 } { t ^ { 2 } } - 2$. Find
\begin{enumerate}[label=(\roman*)]
\item the speed of $P$ when $t = 10$, and the value of $a$,
\item the value of $t$ for which the acceleration of $P$ is $- a \mathrm {~m} \mathrm {~s} ^ { - 2 }$,
\item the displacement of $P$ from $A$ when $t = 20$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q6 [10]}}