| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| 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 with straightforward application of kinematic equations. Part (i) requires basic use of s=ut+½at² and v=u+at with given values. Part (ii) involves setting up a simple equation where distances are equal, requiring students to handle two phases of motion but with no conceptual challenges beyond routine mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_B = \frac{1}{2} \times 1.2 \times 5^2\); Distance travelled is 15 m | B1 | |
| \(v_B = 1.2 \times 5\); Speed is 6 ms\(^{-1}\) | B1 | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([4T = 15 + 6(T-10)]\) or \([4(T+5) = 15 + 6(T-5)]\) or \([4(T+10) = 15 + 6T]\) | M1 | For using \(s_A = s_B\) after \(T\) seconds or after \(T+5\) seconds or after \(T+10\) seconds |
| \(T = 22.5\) or \(T = 17.5\) or \(T = 12.5\) | A1 | |
| Distance OP \(= 4 \times 22.5 = 90\) m | B1 | [3 marks] |
## Question 2:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_B = \frac{1}{2} \times 1.2 \times 5^2$; Distance travelled is 15 m | B1 | |
| $v_B = 1.2 \times 5$; Speed is 6 ms$^{-1}$ | B1 | **[2 marks]** |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[4T = 15 + 6(T-10)]$ or $[4(T+5) = 15 + 6(T-5)]$ or $[4(T+10) = 15 + 6T]$ | M1 | For using $s_A = s_B$ after $T$ seconds or after $T+5$ seconds or after $T+10$ seconds |
| $T = 22.5$ or $T = 17.5$ or $T = 12.5$ | A1 | |
| Distance OP $= 4 \times 22.5 = 90$ m | B1 | **[3 marks]** |
---2 Alan starts walking from a point $O$, at a constant speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, along a horizontal path. Ben walks along the same path, also starting from $O$. Ben starts from rest 5 s after Alan and accelerates at $1.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 5 s . Ben then continues to walk at a constant speed until he is at the same point, $P$, as Alan.\\
(i) Find how far Ben has travelled when he has been walking for 5 s and find his speed at this instant.\\
(ii) Find the distance $O P$.
\hfill \mbox{\textit{CAIE M1 2016 Q2 [5]}}