| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard kinematic equations on an inclined plane. Part (i) is a simple show-that involving relative motion with constant velocities. Parts (ii) and (iii) require setting up equations with acceleration due to gravity component, but follow standard M1 procedures without requiring novel insight or complex problem-solving. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03g Gravitational acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Acceleration down (or up) the plane is the same for both particles | B1 | Stated or implied |
| \(d = (1.3t + \frac{1}{2}at^2) - (-1.3t + \frac{1}{2}at^2)\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) (either particle) |
| \(d = 2.6t\) | A1 | \(a\) must be the same for both \(s_A\) and \(s_B\) |
| A1 | 4 | AG |
| (ii) \(\sin \alpha = \frac{1.6(2 \times 2.5)}{2}\) | B1 | |
| \(\alpha = 32/13\) ms\(^{-2}\) (2.46) | M1 | For using \(a = g\sin \alpha\) |
| A1 | 3 | |
| (iii) \(\alpha = -32/13\) ms\(^{-2}\) (2.46) | M1 | For using \(0 = u + at\) for \(Q\) to find \(t\) |
| \(0 = -1.3 + 2.46t\) (0.528125) | M1 | For using this value of \(t\) in \(d_P = ut + \frac{1}{2}at^2\) |
| \(d_P = 1.3(0.528...) + \frac{1}{2} \times 2.46(0.528...)^2\) | or \(s_Q = 1.3^3(2 \times 2.46)\) or \(s_Q = \frac{1}{2} \times 2.46t\) | |
| or \(s_Q = 2.6(0.528...) - 0.343...\) | ||
| \(= 1.373... - 0.343...\) | M1 | For using \(0 = u + at\) for \(Q\) to find \(t\) |
| For using this value of \(t\) in \(d_P = ut + \frac{1}{2}at^2\) or in \(s_Q = \frac{1}{2} \times 2.46t\) and \(s_P = 2.6t - s_Q\) | ||
| Distance travelled is \(1.03\) m | A1 | 3 |
**(i)** Acceleration down (or up) the plane is the same for both particles | B1 | Stated or implied |
$d = (1.3t + \frac{1}{2}at^2) - (-1.3t + \frac{1}{2}at^2)$ | M1 | For using $s = ut + \frac{1}{2}at^2$ (either particle) |
$d = 2.6t$ | A1 | $a$ must be the same for both $s_A$ and $s_B$ |
| A1 | 4 | AG |
**(ii)** $\sin \alpha = \frac{1.6(2 \times 2.5)}{2}$ | B1 | |
$\alpha = 32/13$ ms$^{-2}$ (2.46) | M1 | For using $a = g\sin \alpha$ |
| A1 | 3 |
**(iii)** $\alpha = -32/13$ ms$^{-2}$ (2.46) | M1 | For using $0 = u + at$ for $Q$ to find $t$ |
$0 = -1.3 + 2.46t$ (0.528125) | M1 | For using this value of $t$ in $d_P = ut + \frac{1}{2}at^2$ |
$d_P = 1.3(0.528...) + \frac{1}{2} \times 2.46(0.528...)^2$ | | or $s_Q = 1.3^3(2 \times 2.46)$ or $s_Q = \frac{1}{2} \times 2.46t$ |
or $s_Q = 2.6(0.528...) - 0.343...$ | | |
$= 1.373... - 0.343...$ | M1 | For using $0 = u + at$ for $Q$ to find $t$ |
| | For using this value of $t$ in $d_P = ut + \frac{1}{2}at^2$ or in $s_Q = \frac{1}{2} \times 2.46t$ and $s_P = 2.6t - s_Q$ |
Distance travelled is $1.03$ m | A1 | 3 |7 Two particles $P$ and $Q$ move on a line of greatest slope of a smooth inclined plane. The particles start at the same instant and from the same point, each with speed $1.3 \mathrm {~ms} ^ { - 1 }$. Initially $P$ moves down the plane and $Q$ moves up the plane. The distance between the particles $t$ seconds after they start to move is $d \mathrm {~m}$.\\
(i) Show that $d = 2.6 t$.
When $t = 2.5$ the difference in the vertical height of the particles is 1.6 m . Find\\
(ii) the acceleration of the particles down the plane,\\
(iii) the distance travelled by $P$ when $Q$ is at its highest point.
\hfill \mbox{\textit{CAIE M1 2006 Q7 [10]}}