| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Standard +0.3 This is a straightforward SUVAT and energy problem with clear structure. Part (i) is trivial calculation (a = v/t). Part (ii) requires calculating trapezoid areas from a velocity-time graph and solving a linear equation. Part (iii) applies work-energy principle with given values. All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 0.5\text{ ms}^{-2}\) | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(= 25 + 100 + 5(5+V) + 30V + 10V\) | M1 | For attempting to find the distance travelled |
| \(150 + 45V\) | A1 AG | |
| \(150 + 45V = 465 \rightarrow V = 7\text{ ms}^{-1}\) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 80 \times 7^2 - \frac{1}{2} \times 80 \times 5^2\ [= 960]\) | M1 | For change in KE |
| \(20 \times (5+7)/2 \times 10\ [=1200]\) | M1 | For work done against friction using \(F \times d\) |
| \([80gh = 960 + 1200]\) | M1 | For using PE loss \(=\) KE gain \(+\) WD against Res. |
| \(h = 2.7\text{ m}\) | A1 | [4] |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 0.5\text{ ms}^{-2}$ | B1 | [1] |
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= 25 + 100 + 5(5+V) + 30V + 10V$ | M1 | For attempting to find the distance travelled |
| $150 + 45V$ | A1 AG | |
| $150 + 45V = 465 \rightarrow V = 7\text{ ms}^{-1}$ | B1 | [3] |
## Question 5(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 80 \times 7^2 - \frac{1}{2} \times 80 \times 5^2\ [= 960]$ | M1 | For change in KE |
| $20 \times (5+7)/2 \times 10\ [=1200]$ | M1 | For work done against friction using $F \times d$ |
| $[80gh = 960 + 1200]$ | M1 | For using PE loss $=$ KE gain $+$ WD against Res. |
| $h = 2.7\text{ m}$ | A1 | [4] |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{a92f97e2-343f-4cac-ae38-f18a4ad49055-3_574_1205_260_470}
The diagram shows a velocity-time graph which models the motion of a cyclist. The graph consists of five straight line segments. The cyclist accelerates from rest to a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a period of 10 s , and then travels at this speed for a further 20 s . The cyclist then descends a hill, accelerating to speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a period of 10 s . This speed is maintained for a further 30 s . The cyclist then decelerates to rest over a period of 20 s .\\
(i) Find the acceleration of the cyclist during the first 10 seconds.\\
(ii) Show that the total distance travelled by the cyclist in the 90 seconds of motion may be expressed as $( 45 V + 150 ) \mathrm { m }$. Hence find $V$, given that the total distance travelled by the cyclist is 465 m .\\
(iii) The combined mass of the cyclist and the bicycle is 80 kg . The cyclist experiences a constant resistance to motion of 20 N . Use an energy method to find the vertical distance which the cyclist descends during the downhill section from $t = 30$ to $t = 40$, assuming that the cyclist does no work during this time.
\hfill \mbox{\textit{CAIE M1 2016 Q5 [8]}}