| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.8 This is a standard M2 variable force problem requiring separation of variables and integration, but the algebraic manipulation to reach the given form in part (i) and the subsequent integration in part (ii) involve non-trivial steps. The setup is routine (F=ma with resistance), but executing the algebra and integration correctly places it moderately above average difficulty. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mv\frac{dv}{dx} = -mg - 0.02mv\) | M1 | Newton's 2nd law, 2 resistive forces |
| \(\frac{v + 50g - 50g}{50g + v}\frac{dv}{dx} = -0.02\) | DM1 | \(-mg - 0.02mv = -0.02(50g + v)\) used |
| \(\left(\frac{500}{500+v}-1\right)\frac{dv}{dx} = 0.02\) | A1 | [3] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\frac{500}{500+v}-1\,dv = \int0.02\,dx\) | M1 | Attempts integration |
| \(\left[500\ln(500+v)-v\right]_{14}^{0} = 0.02H\) | DM1 | Appropriate use of \(v=14\), \(0\) |
| \(H = 9.62\text{ m}\) | A1 | [3] |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mv\frac{dv}{dx} = -mg - 0.02mv$ | M1 | Newton's 2nd law, 2 resistive forces |
| $\frac{v + 50g - 50g}{50g + v}\frac{dv}{dx} = -0.02$ | DM1 | $-mg - 0.02mv = -0.02(50g + v)$ used |
| $\left(\frac{500}{500+v}-1\right)\frac{dv}{dx} = 0.02$ | A1 | [3] AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\frac{500}{500+v}-1\,dv = \int0.02\,dx$ | M1 | Attempts integration |
| $\left[500\ln(500+v)-v\right]_{14}^{0} = 0.02H$ | DM1 | Appropriate use of $v=14$, $0$ |
| $H = 9.62\text{ m}$ | A1 | [3] |
---3 A small ball of mass $m \mathrm {~kg}$ is projected vertically upwards with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The ball has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ upwards when it is $x \mathrm {~m}$ above the point of projection. A resisting force of magnitude $0.02 m v \mathrm {~N}$ acts on the ball during its upward motion.\\
(i) Show that, while the ball is moving upwards, $\left( \frac { 500 } { v + 500 } - 1 \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 0.02$.\\
(ii) Find the greatest height of the ball above its point of projection.
\hfill \mbox{\textit{CAIE M2 2014 Q3 [6]}}