| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with two straightforward parts: (i) requires setting up Newton's second law for a smooth surface system and using kinematics (s=ut+½at²), (ii) adds friction which is a routine modification. Both parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T = 0.8a\) for \(A\); \(2 - T = 0.2a\) for \(B\); \(0.2g = (0.2 + 0.8)a\) system\(]\) | M1 | For applying Newton's 2nd law either to particle \(A\) or particle \(B\) or to the system |
| M1 | For applying N2L to a second particle (if needed) and solving for \(a\) | |
| \([a = 2]\) | A1 | |
| \([2.5 = \frac{1}{2} \times 2 \times t^2]\) | M1 | A complete method for finding \(t\) using \(s = ut + \frac{1}{2}at^2\) |
| \(t = 1.58\) s | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.2 \times g \times 2.5\) or \(\frac{1}{2}(0.2 + 0.8)v^2]\) | M1 | Finding PE loss or KE gain (system) |
| \([0.2 \times g \times 2.5 = \frac{1}{2}(0.2 + 0.8)v^2]\) | M1 | Using PE loss \(=\) KE gain and find \(v\) |
| \([v^2 = 10]\) | A1 | |
| \([2.5 = \frac{1}{2}(0 + \sqrt{10})t]\) | M1 | For using \(s = \frac{1}{2}(u+v)t\) |
| \(t = 1.58\) s | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T = 0.8a \quad 2-T = 0.2a \rightarrow T = 1.6\text{ N}]\) | M1 | Apply N2 to \(A\) and \(B\) and solve for \(T\) |
| \([T \times 2.5 = \frac{1}{2}(0.8)v^2]\) | M1 | Use WD by \(T\) = KE gain by \(A\), find \(v\) |
| \([v^2 = 10]\) | A1 | |
| \([2.5 = \frac{1}{2}(0 + \sqrt{10})t]\) | M1 | Using \(s = \frac{1}{2}(u+v)t\) |
| \(t = 1.58\text{ s}\) | A1 | Allow \(t = \frac{1}{2}\sqrt{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N = 8\) and \(F = 0.1 \times N = 0.8\) | B1 | |
| \(T - 0.8 = 0.8a\) and \(2 - T = 0.2a\) or \(0.2g - 0.8 = (0.2+0.8)a\) | M1 | For applying N2 to both particles or to the system and solving for \(a\) |
| \(a = 1.2\) | A1 | |
| \(v^2 = 0 + 2 \times 1.2 \times 2.5\) | M1 | For using \(v^2 = u^2 + 2as\) |
| \(v = \sqrt{6} = 2.45\text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N = 8\) and \(F = 0.1 \times N = 0.8\) | B1 | |
| \([0.2 \times g \times 2.5 = \frac{1}{2}(0.8+0.2)v^2 + 0.8 \times 2.5]\) | M1 | Apply work/energy to system: PE loss = KE gain + WD against resistance |
| A1 | Correct Work/Energy equation | |
| M1 | For solving for \(v\) | |
| \(v = \sqrt{6} = 2.45\text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N = 8\) and \(F = 0.1 \times N = 0.8\) | B1 | |
| \(T - 0.8 = 0.8a\) and \(2 - T = 0.2a\) | M1 | Use N2 for \(A\) and \(B\) and solve for \(T\) |
| \(T = 1.76\text{ N}\) | A1 | |
| \([T \times 2.5 = 0.8 \times 2.5 + \frac{1}{2}(0.8)v^2]\) | M1 | Apply Work/Energy equation to \(A\) |
| \(v = \sqrt{6} = 2.45\text{ ms}^{-1}\) | A1 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T = 0.8a$ for $A$; $2 - T = 0.2a$ for $B$; $0.2g = (0.2 + 0.8)a$ system$]$ | M1 | For applying Newton's 2nd law either to particle $A$ or particle $B$ or to the system |
| | M1 | For applying N2L to a second particle (if needed) and solving for $a$ |
| $[a = 2]$ | A1 | |
| $[2.5 = \frac{1}{2} \times 2 \times t^2]$ | M1 | A complete method for finding $t$ using $s = ut + \frac{1}{2}at^2$ |
| $t = 1.58$ s | A1 | 5 | Allow $t = \dfrac{1}{2}\sqrt{10}$ |
**First Alternative Method for 6(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.2 \times g \times 2.5$ or $\frac{1}{2}(0.2 + 0.8)v^2]$ | M1 | Finding PE loss or KE gain (system) |
| $[0.2 \times g \times 2.5 = \frac{1}{2}(0.2 + 0.8)v^2]$ | M1 | Using PE loss $=$ KE gain and find $v$ |
| $[v^2 = 10]$ | A1 | |
| $[2.5 = \frac{1}{2}(0 + \sqrt{10})t]$ | M1 | For using $s = \frac{1}{2}(u+v)t$ |
| $t = 1.58$ s | A1 | 5 | Allow $t = \dfrac{1}{2}\sqrt{10}$ |
# Question 6:
## Part (i) — Second Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T = 0.8a \quad 2-T = 0.2a \rightarrow T = 1.6\text{ N}]$ | M1 | Apply N2 to $A$ and $B$ and solve for $T$ |
| $[T \times 2.5 = \frac{1}{2}(0.8)v^2]$ | M1 | Use WD by $T$ = KE gain by $A$, find $v$ |
| $[v^2 = 10]$ | A1 | |
| $[2.5 = \frac{1}{2}(0 + \sqrt{10})t]$ | M1 | Using $s = \frac{1}{2}(u+v)t$ |
| $t = 1.58\text{ s}$ | A1 | Allow $t = \frac{1}{2}\sqrt{10}$ |
**Total: 5 marks**
## Part (ii) — Main Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 8$ and $F = 0.1 \times N = 0.8$ | B1 | |
| $T - 0.8 = 0.8a$ and $2 - T = 0.2a$ or $0.2g - 0.8 = (0.2+0.8)a$ | M1 | For applying N2 to both particles or to the system and solving for $a$ |
| $a = 1.2$ | A1 | |
| $v^2 = 0 + 2 \times 1.2 \times 2.5$ | M1 | For using $v^2 = u^2 + 2as$ |
| $v = \sqrt{6} = 2.45\text{ ms}^{-1}$ | A1 | |
**Total: 5 marks**
## Part (ii) — First Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 8$ and $F = 0.1 \times N = 0.8$ | B1 | |
| $[0.2 \times g \times 2.5 = \frac{1}{2}(0.8+0.2)v^2 + 0.8 \times 2.5]$ | M1 | Apply work/energy to system: PE loss = KE gain + WD against resistance |
| | A1 | Correct Work/Energy equation |
| | M1 | For solving for $v$ |
| $v = \sqrt{6} = 2.45\text{ ms}^{-1}$ | A1 | |
**Total: 5 marks**
## Part (ii) — Second Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 8$ and $F = 0.1 \times N = 0.8$ | B1 | |
| $T - 0.8 = 0.8a$ and $2 - T = 0.2a$ | M1 | Use N2 for $A$ and $B$ and solve for $T$ |
| $T = 1.76\text{ N}$ | A1 | |
| $[T \times 2.5 = 0.8 \times 2.5 + \frac{1}{2}(0.8)v^2]$ | M1 | Apply Work/Energy equation to $A$ |
| $v = \sqrt{6} = 2.45\text{ ms}^{-1}$ | A1 | |
**Total: 5 marks**
---6 Two particles $A$ and $B$, of masses 0.8 kg and 0.2 kg respectively, are connected by a light inextensible string. Particle $A$ is placed on a horizontal surface. The string passes over a small smooth pulley $P$ fixed at the edge of the surface, and $B$ hangs freely. The horizontal section of the string, $A P$, is of length 2.5 m . The particles are released from rest with both sections of the string taut.\\
(i) Given that the surface is smooth, find the time taken for $A$ to reach the pulley.\\
(ii) Given instead that the surface is rough and the coefficient of friction between $A$ and the surface is 0.1 , find the speed of $A$ immediately before it reaches the pulley.\\
$7 \quad$ A particle $P$ moves in a straight line. The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ is given by
$$\begin{array} { l l }
v = 5 t ( t - 2 ) & \text { for } 0 \leqslant t \leqslant 4 \\
v = k & \text { for } 4 \leqslant t \leqslant 14 \\
v = 68 - 2 t & \text { for } 14 \leqslant t \leqslant 20
\end{array}$$
where $k$ is a constant.\\
(i) Find $k$.\\
(ii) Sketch the velocity-time graph for $0 \leqslant t \leqslant 20$.\\
(iii) Find the set of values of $t$ for which the acceleration of $P$ is positive.\\
(iv) Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 20$.
\hfill \mbox{\textit{CAIE M1 2016 Q6 [10]}}