| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Moderate -0.3 This is a standard two-particle SUVAT problem requiring straightforward application of kinematic equations. Part (i) involves setting up an equation relating the speeds using v = u + at, and part (ii) requires finding when the sum of distances equals 51m using s = ut + ½at². Both parts are routine mechanics exercises with no conceptual challenges beyond basic SUVAT manipulation. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(v = u + at\) and \(v_P = 1.8v_Q\) | |
| \(5 + 4t = 1.8(3 + 2t)\) or | A1 | |
| \(1.8v_Q = 5 + 4t\) and \(v_Q = 3 + 2t\) or | ||
| \(v_P = 5 + 4t\) and \(\frac{5}{9}v_P = 3 + 2t\) | ||
| \(t = 1\) or \(v_Q = 5\) or correct equation in \(v_P\) only \([\text{eg } (10/9 - 1)v_P = 6 - 5]\) | A1 | |
| Speed of \(P = 9 \text{ ms}^{-1}\) | A1ft | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(s = ut + \frac{1}{2}at^2\) and \(s_P + s_Q = 51\) | |
| \(5t + \frac{1}{2} \cdot 4t^2 + 3t + \frac{1}{2} \cdot 2t^2 = 51\) | A1 | |
| \(3t^2 + 8t - 51 = 0 \rightarrow (3t + 17)(t - 3)\) | M1 | For attempting to solve the resulting quadratic equation |
| Time is \(3 \text{ s}\) | A1 | 4 marks total |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $v = u + at$ and $v_P = 1.8v_Q$ |
| $5 + 4t = 1.8(3 + 2t)$ or | A1 | |
| $1.8v_Q = 5 + 4t$ and $v_Q = 3 + 2t$ or | | |
| $v_P = 5 + 4t$ and $\frac{5}{9}v_P = 3 + 2t$ | | |
| $t = 1$ or $v_Q = 5$ or correct equation in $v_P$ only $[\text{eg } (10/9 - 1)v_P = 6 - 5]$ | A1 | |
| Speed of $P = 9 \text{ ms}^{-1}$ | A1ft | **4 marks total** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $s = ut + \frac{1}{2}at^2$ and $s_P + s_Q = 51$ |
| $5t + \frac{1}{2} \cdot 4t^2 + 3t + \frac{1}{2} \cdot 2t^2 = 51$ | A1 | |
| $3t^2 + 8t - 51 = 0 \rightarrow (3t + 17)(t - 3)$ | M1 | For attempting to solve the resulting quadratic equation |
| Time is $3 \text{ s}$ | A1 | **4 marks total** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{38ece0f6-1c29-4e7a-9d66-16c3e2b695f9-3_240_862_274_644}
Particles $P$ and $Q$ start from points $A$ and $B$ respectively, at the same instant, and move towards each other in a horizontal straight line. The initial speeds of $P$ and $Q$ are $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The accelerations of $P$ and $Q$ are constant and equal to $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ respectively (see diagram).\\
(i) Find the speed of $P$ at the instant when the speed of $P$ is 1.8 times the speed of $Q$.\\
(ii) Given that $A B = 51 \mathrm {~m}$, find the time taken from the start until $P$ and $Q$ meet.
\hfill \mbox{\textit{CAIE M1 2004 Q5 [8]}}