| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Standard +0.3 This is a straightforward two-part SUVAT problem on an inclined plane. Part (i) requires basic application of v=u+at and connecting acceleration to g sin θ. Part (ii) involves setting up position equations for both particles and solving a simple equation. All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2a = 3.5\) | M1 | For using \(v = 0 + at\) |
| Acceleration is \(1.75\text{ ms}^{-2}\) | A1 | |
| \(1.75 = g\sin\alpha\) or \(0.5\times3.5^2 = gh\); \(s = 0.5\times3.5\times2\) and \(\sin\alpha = h/s\) | M1 | For using \(a = g\sin\alpha\), or \(\frac{1}{2}mv^2 = mgh\), \(s = \frac{1}{2}vt\) and \(\sin\alpha = h/s\) |
| Angle is \(10.1°\) or \(0.176^c\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_P = \frac{1}{2}a\cdot2^2 + \{(a\cdot2)t + \frac{1}{2}at^2\}\) or \(s_P = \frac{1}{2}a(t+2)^2\) | M1 | For constructing an expression in \(t\) for \(s_P\) |
| \(s_P - s_Q = \frac{1}{2}a\cdot2^2 + (a\cdot2)t + \frac{1}{2}at^2 - \frac{1}{2}at^2\) | M1 | For constructing an expression in \(t\) for \(s_P - s_Q\) |
| \(2\times1.75 + 2\times1.75t\) | A1 | Correct expression for \(s_P - s_Q\) |
| \(4.9 = 2a + 2at\) | M1 | For using \(s_P - s_Q = 4.9\) to construct equation in \(t\) |
| \(t = 0.4\) | A1 | |
| [5] |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2a = 3.5$ | M1 | For using $v = 0 + at$ |
| Acceleration is $1.75\text{ ms}^{-2}$ | A1 | |
| $1.75 = g\sin\alpha$ or $0.5\times3.5^2 = gh$; $s = 0.5\times3.5\times2$ and $\sin\alpha = h/s$ | M1 | For using $a = g\sin\alpha$, or $\frac{1}{2}mv^2 = mgh$, $s = \frac{1}{2}vt$ and $\sin\alpha = h/s$ |
| Angle is $10.1°$ or $0.176^c$ | A1 | |
| **[4]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_P = \frac{1}{2}a\cdot2^2 + \{(a\cdot2)t + \frac{1}{2}at^2\}$ or $s_P = \frac{1}{2}a(t+2)^2$ | M1 | For constructing an expression in $t$ for $s_P$ |
| $s_P - s_Q = \frac{1}{2}a\cdot2^2 + (a\cdot2)t + \frac{1}{2}at^2 - \frac{1}{2}at^2$ | M1 | For constructing an expression in $t$ for $s_P - s_Q$ |
| $2\times1.75 + 2\times1.75t$ | A1 | Correct expression for $s_P - s_Q$ |
| $4.9 = 2a + 2at$ | M1 | For using $s_P - s_Q = 4.9$ to construct equation in $t$ |
| $t = 0.4$ | A1 | |
| **[5]** | | |
---6 Particles $P$ and $Q$ move on a line of greatest slope of a smooth inclined plane. $P$ is released from rest at a point $O$ on the line and 2 s later passes through the point $A$ with speed $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the acceleration of $P$ and the angle of inclination of the plane.
At the instant that $P$ passes through $A$ the particle $Q$ is released from rest at $O$. At time $t$ s after $Q$ is released from $O$, the particles $P$ and $Q$ are 4.9 m apart.\\
(ii) Find the value of $t$.
\hfill \mbox{\textit{CAIE M1 2010 Q6 [9]}}