| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.8 This question requires integration of a time-dependent acceleration (0.016t) to find velocity and position, then solving simultaneous equations using conditions at t=20. While the individual calculus steps are standard A-level mechanics, coordinating multiple particles with different acceleration types and matching conditions at a specific time requires careful systematic work beyond routine exercises. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| End speed \(= 1.3 + 0.1 \times 20\) | B1 | |
| \(v_Q(t) = 0.008t^2 + v_Q(0)\) | B1 | |
| \([3.3 = 0.008 \times 20^2 + v_Q(0)]\) | M1 | For substituting end speed and \(t = 20\) |
| Speed of \(Q\) when \(t = 0\) is \(0.1\) ms\(^{-1}\) | A1 | [Total: 4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(AO = (3.3^2 - 1.3^2) \div (2 \times 0.1)\) or \(20 \times \frac{1}{2}(1.3 + 3.3) \ [= 46]\) | B1 | or \(AO = 1.3(20) + \frac{1}{2}(0.1) \times 20^2\) |
| Distance \(OB = 0.008 \times 20^3 \div 3 + 0.1 \times 20 \ [= 70/3 = 23.3]\) | B1 | |
| Distance \(AB\) is \(69.3\) m | B1 | [Total: 3] |
## Question 4 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| End speed $= 1.3 + 0.1 \times 20$ | B1 | |
| $v_Q(t) = 0.008t^2 + v_Q(0)$ | B1 | |
| $[3.3 = 0.008 \times 20^2 + v_Q(0)]$ | M1 | For substituting end speed and $t = 20$ |
| Speed of $Q$ when $t = 0$ is $0.1$ ms$^{-1}$ | A1 | **[Total: 4]** |
## Question 4 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $AO = (3.3^2 - 1.3^2) \div (2 \times 0.1)$ or $20 \times \frac{1}{2}(1.3 + 3.3) \ [= 46]$ | B1 | or $AO = 1.3(20) + \frac{1}{2}(0.1) \times 20^2$ |
| Distance $OB = 0.008 \times 20^3 \div 3 + 0.1 \times 20 \ [= 70/3 = 23.3]$ | B1 | |
| Distance $AB$ is $69.3$ m | B1 | **[Total: 3]** |
---4 Particles $P$ and $Q$ move on a straight line $A O B$. The particles leave $O$ simultaneously, with $P$ moving towards $A$ and with $Q$ moving towards $B$. The initial speed of $P$ is $1.3 \mathrm {~ms} ^ { - 1 }$ and its acceleration in the direction $O A$ is $0.1 \mathrm {~m} \mathrm {~s} ^ { - 2 } . Q$ moves with acceleration in the direction $O B$ of $0.016 t \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $t$ seconds is the time elapsed since the instant that $P$ and $Q$ started to move from $O$. When $t = 20$, particle $P$ passes through $A$ and particle $Q$ passes through $B$.\\
(i) Given that the speed of $Q$ at $B$ is the same as the speed of $P$ at $A$, find the speed of $Q$ at time $t = 0$.\\
(ii) Find the distance $A B$.
\hfill \mbox{\textit{CAIE M1 2014 Q4 [7]}}