| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question on non-constant acceleration with straightforward calculus. Part (a) applies Newton's second law directly, (b) involves separable differential equations with a given answer to verify, (c) is simple substitution, and (d) requires integrating v with respect to t. All steps follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Using \(F = ma\): \(-0.2mv^{\frac{1}{2}} = m\frac{dv}{dt}\) | ||
| \(\therefore \frac{dv}{dt} = -0.2v^{\frac{1}{2}}\) | B1 | AG Must see equation containing \(m\). Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int \frac{dv}{v^{\frac{1}{2}}} = -\int 0.2\, dt\) | M1 | |
| \(2v^{\frac{1}{2}} = -0.2t + c\) | A1m1 | m1 for \(+c\) |
| When \(t=0\), \(v=16\) \(\therefore C = 8\) | A1 | |
| \(2v^{\frac{1}{2}} = -0.2t + 8\) | ||
| \(v = (4 - 0.1t)^2\) | A1 | AG. Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| When \(v=1\): \(1 = (4-0.1t)^2\) | M1 | |
| \(4 - 0.1t = \pm 1\) | ||
| \(t = 30\) or \(50\) | A1 | If use \(2v^{\frac{1}{2}} = 8 - 0.2t\) no need to see 50 |
| \(t = 30\) | A1 | \(t \neq 50\) as ball stops when \(t=40\). Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Integrating \(v = (4-0.1t)^2\): \(v = 16 - 0.8t + 0.01t^2\) | ||
| \(x = 16t - 0.4t^2 + \frac{0.01}{3}t^3 + d\) | M1 | M1 for first 3 terms or \(-\frac{10}{3}(4-0.1t)^3\) |
| When \(t=0\), \(x=0 \Rightarrow d=0\) | A1 | |
| \(x = 16t - 0.4t^2 + \frac{0.01}{3}t^3\) | ||
| When speed is \(1\text{ ms}^{-1}\), \(t=30\): \(x = 480 - 360 + 90\) | m1 | dep on M1 above |
| \(= 210\) | A1 | [No '\(d\)', 3 marks only]. Total: 4 |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Using $F = ma$: $-0.2mv^{\frac{1}{2}} = m\frac{dv}{dt}$ | | |
| $\therefore \frac{dv}{dt} = -0.2v^{\frac{1}{2}}$ | B1 | AG Must see equation containing $m$. **Total: 1** |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\int \frac{dv}{v^{\frac{1}{2}}} = -\int 0.2\, dt$ | M1 | |
| $2v^{\frac{1}{2}} = -0.2t + c$ | A1m1 | m1 for $+c$ |
| When $t=0$, $v=16$ $\therefore C = 8$ | A1 | |
| $2v^{\frac{1}{2}} = -0.2t + 8$ | | |
| $v = (4 - 0.1t)^2$ | A1 | AG. **Total: 5** |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| When $v=1$: $1 = (4-0.1t)^2$ | M1 | |
| $4 - 0.1t = \pm 1$ | | |
| $t = 30$ or $50$ | A1 | If use $2v^{\frac{1}{2}} = 8 - 0.2t$ no need to see 50 |
| $t = 30$ | A1 | $t \neq 50$ as ball stops when $t=40$. **Total: 3** |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| Integrating $v = (4-0.1t)^2$: $v = 16 - 0.8t + 0.01t^2$ | | |
| $x = 16t - 0.4t^2 + \frac{0.01}{3}t^3 + d$ | M1 | M1 for first 3 terms or $-\frac{10}{3}(4-0.1t)^3$ |
| When $t=0$, $x=0 \Rightarrow d=0$ | A1 | |
| $x = 16t - 0.4t^2 + \frac{0.01}{3}t^3$ | | |
| When speed is $1\text{ ms}^{-1}$, $t=30$: $x = 480 - 360 + 90$ | m1 | dep on M1 above |
| $= 210$ | A1 | [No '$d$', 3 marks only]. **Total: 4** |5 A golf ball, of mass $m \mathrm {~kg}$, is moving in a straight line across smooth horizontal ground. At time $t$ seconds, the golf ball has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. As the golf ball moves, it experiences a resistance force of magnitude $0.2 m v ^ { \frac { 1 } { 2 } }$ newtons until it comes to rest. No other horizontal force acts on the golf ball.
Model the golf ball as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.2 v ^ { \frac { 1 } { 2 } }$$
\item When $t = 0$, the speed of the golf ball is $16 \mathrm {~ms} ^ { - 1 }$.
Show that $v = ( 4 - 0.1 t ) ^ { 2 }$.
\item Find the value of $t$ when $v = 1$.
\item Find the distance travelled by the golf ball as its speed decreases from $16 \mathrm {~ms} ^ { - 1 }$ to $1 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q5 [13]}}