| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.8 This is a multi-part mechanics problem requiring integration of non-constant acceleration in two phases (upward and downward motion with different accelerations due to air resistance), application of kinematic equations, and work-energy principles. While the individual techniques are standard M1 content, the problem requires careful tracking of two distinct motion phases and synthesizing multiple concepts (kinematics, dynamics, and energy), making it moderately challenging but still within typical A-level scope. |
| Spec | 3.02h Motion under gravity: vector form6.02a Work done: concept and definition6.02c Work by variable force: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 5.2^2 - 2\times10.4s_1\) or \(s_1 = 5.2\times0.5 - \frac{1}{2}(10.4)(0.5)^2\) | M1 | For using \(0 = u^2 + 2as\), or \(0 = u + at\) and \(s = ut + \frac{1}{2}at^2\), or \(0 = u + at\) and \(s = \frac{(u+0)t}{2}\) |
| or \(s_1 = (5.2 + 0)\times0.5/2\) | A1 | |
| Greatest height is 7.5 m | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 2\times9.6\times7.5,\ v = 9.6\times1.25,\ v = 2\times7.5/1.25]\) | M1 | For using \(v^2 = 0 + 2as\), or \(s = \frac{1}{2}at^2\) and \(v = at\), or \(s = \frac{1}{2}at^2\) and \(0 + v = 2s/t\) |
| Speed is \(12 \text{ ms}^{-1}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE loss \(= 0.6g \times 6.2\ (= 37.2)\) or Initial total energy \(= 0.6g\times6.2 + \frac{1}{2}(0.6)(5.2)^2\ (= 45.312)\) or Energy loss upward \(= \frac{1}{2}(0.6)(5.2)^2 - 0.6g\times1.3\ (= 0.312)\); KE gain \(= \frac{1}{2}(0.6)(12^2 - 5.2^2)\ (= 35.088)\) or Final total energy \(= \frac{1}{2}(0.6)(12)^2\ (= 43.2)\); Energy loss downward \(= -\frac{1}{2}(0.6)(12)^2 + 0.6g\times7.5\ (= 1.8)\) | B1 | |
| B1ft | ft ans (ii); For using \(WD = \text{PE loss from start} - \text{KE gain from start}\) or \(WD = \text{Initial total energy} - \text{final total energy}\) | |
| \([WD = 37.2 - 35.088\ \text{or}\ 45.312 - 43.2\ \text{or}\ 0.312 + 1.8]\) | M1 | \(WD = \text{energy loss upward} + \text{energy loss downward}\) |
| Work done is 2.11 (2) J | A1 [4] | Accept exact or 3 sf |
| Alternatively: \([0.6g + R_{up} = 0.6\times10.4\ \text{or}\ 0.6g - R_{down} = 0.6\times9.6]\) | M1 | For applying Newton's second law to upward/downward motion and attempting to find \(R_{up}\) or \(R_{down}\) |
| \(R_{up} = 0.24\) or \(R_{down} = 0.24\) | A1, M1 | For using \(WD(\text{upward}) = 1.3R_{up}\) or \(WD(\text{downward}) = \text{ans(i)}R_{down}\) |
| Work done is 2.11 (2) J | A1ft [4] | ft ans (i) |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 5.2^2 - 2\times10.4s_1$ **or** $s_1 = 5.2\times0.5 - \frac{1}{2}(10.4)(0.5)^2$ | M1 | For using $0 = u^2 + 2as$, **or** $0 = u + at$ and $s = ut + \frac{1}{2}at^2$, **or** $0 = u + at$ and $s = \frac{(u+0)t}{2}$ |
| **or** $s_1 = (5.2 + 0)\times0.5/2$ | A1 | |
| Greatest height is 7.5 m | A1 [3] | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 2\times9.6\times7.5,\ v = 9.6\times1.25,\ v = 2\times7.5/1.25]$ | M1 | For using $v^2 = 0 + 2as$, **or** $s = \frac{1}{2}at^2$ and $v = at$, **or** $s = \frac{1}{2}at^2$ and $0 + v = 2s/t$ |
| Speed is $12 \text{ ms}^{-1}$ | A1 [2] | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE loss $= 0.6g \times 6.2\ (= 37.2)$ **or** Initial total energy $= 0.6g\times6.2 + \frac{1}{2}(0.6)(5.2)^2\ (= 45.312)$ **or** Energy loss upward $= \frac{1}{2}(0.6)(5.2)^2 - 0.6g\times1.3\ (= 0.312)$; KE gain $= \frac{1}{2}(0.6)(12^2 - 5.2^2)\ (= 35.088)$ **or** Final total energy $= \frac{1}{2}(0.6)(12)^2\ (= 43.2)$; Energy loss downward $= -\frac{1}{2}(0.6)(12)^2 + 0.6g\times7.5\ (= 1.8)$ | B1 | |
| | B1ft | ft ans **(ii)**; For using $WD = \text{PE loss from start} - \text{KE gain from start}$ **or** $WD = \text{Initial total energy} - \text{final total energy}$ |
| $[WD = 37.2 - 35.088\ \text{or}\ 45.312 - 43.2\ \text{or}\ 0.312 + 1.8]$ | M1 | $WD = \text{energy loss upward} + \text{energy loss downward}$ |
| Work done is 2.11 (2) J | A1 [4] | Accept exact or 3 sf |
| **Alternatively:** $[0.6g + R_{up} = 0.6\times10.4\ \text{or}\ 0.6g - R_{down} = 0.6\times9.6]$ | M1 | For applying Newton's second law to upward/downward motion and attempting to find $R_{up}$ or $R_{down}$ |
| $R_{up} = 0.24$ or $R_{down} = 0.24$ | A1, M1 | For using $WD(\text{upward}) = 1.3R_{up}$ or $WD(\text{downward}) = \text{ans(i)}R_{down}$ |
| Work done is 2.11 (2) J | A1ft [4] | ft ans **(i)** |
---6 A particle $P$ of mass 0.6 kg is projected vertically upwards with speed $5.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ which is 6.2 m above the ground. Air resistance acts on $P$ so that its deceleration is $10.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ when $P$ is moving upwards, and its acceleration is $9.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ when $P$ is moving downwards. Find\\
(i) the greatest height above the ground reached by $P$,\\
(ii) the speed with which $P$ reaches the ground,\\
(iii) the total work done on $P$ by the air resistance.
\hfill \mbox{\textit{CAIE M1 2008 Q6 [9]}}