| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of displacement (v dv/dx method) |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question on variable acceleration with air resistance. Part (a) requires applying F=ma with given forces (routine), part (b) involves separable variables integration (standard technique), part (c) is straightforward calculation, and part (d) requires basic energy reasoning. While it's a multi-step problem requiring calculus and mechanics understanding, all techniques are standard for Further Maths students with no novel insights needed. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.02d Mechanical energy: KE and PE concepts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| N2L on ball, upwards positive: \(-0.01v^2 - 0.4g = 0.4a\) | M1, A1 | dim correct; correct equation |
| \(0.4v\frac{dv}{dx} = -3.92 - 0.01v^2\) | ||
| \(40v\frac{dv}{dx} = -(392 + v^2)\) | A1 | convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(40\int \frac{v}{392+v^2}\,dv = -\int dx\) | M1 | separate variables |
| \(20\ln(392 + v^2) = -x + C\) | A1, A1 | \(\ln(392+v^2)\); everything correct |
| When \(t=0\), \(v=17\), \(x=0\): \(C = 20\ln(681)\) | m1, A1 | use of initial conditions |
| \(x = 20\ln\!\left(\frac{681}{392+v^2}\right)\) | ||
| \(\left(\frac{681}{392+v^2}\right) = e^{0.05x}\) | m1 | |
| \(v^2 = 681e^{-0.05x} - 392\) | ||
| \(v = \sqrt{681e^{-0.05x} - 392}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At greatest height \(v = 0\) | M1 | |
| \(x = 20\ln\!\left(\frac{681}{392}\right) = 11.05\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed of ball when it returns to \(O\) is less than \(17\ \text{ms}^{-1}\) | B1 | |
| This is because energy is lost in overcoming air resistance | E1 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L on ball, upwards positive: $-0.01v^2 - 0.4g = 0.4a$ | M1, A1 | dim correct; correct equation |
| $0.4v\frac{dv}{dx} = -3.92 - 0.01v^2$ | | |
| $40v\frac{dv}{dx} = -(392 + v^2)$ | A1 | convincing |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $40\int \frac{v}{392+v^2}\,dv = -\int dx$ | M1 | separate variables |
| $20\ln(392 + v^2) = -x + C$ | A1, A1 | $\ln(392+v^2)$; everything correct |
| When $t=0$, $v=17$, $x=0$: $C = 20\ln(681)$ | m1, A1 | use of initial conditions |
| $x = 20\ln\!\left(\frac{681}{392+v^2}\right)$ | | |
| $\left(\frac{681}{392+v^2}\right) = e^{0.05x}$ | m1 | |
| $v^2 = 681e^{-0.05x} - 392$ | | |
| $v = \sqrt{681e^{-0.05x} - 392}$ | A1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At greatest height $v = 0$ | M1 | |
| $x = 20\ln\!\left(\frac{681}{392}\right) = 11.05$ | A1 | cao |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed of ball when it returns to $O$ is less than $17\ \text{ms}^{-1}$ | B1 | |
| This is because energy is lost in overcoming air resistance | E1 | |
---\begin{enumerate}
\item A ball of mass 0.4 kg is thrown vertically upwards from a point $O$ with initial speed $17 \mathrm {~ms} ^ { - 1 }$. When the ball is at a height of $x \mathrm {~m}$ above $O$ and its speed is $v \mathrm {~ms} ^ { - 1 }$, the air resistance acting on the ball has magnitude $0.01 v ^ { 2 } \mathrm {~N}$.\\
(a) Show that, as the ball is ascending, $v$ satisfies the differential equation
\end{enumerate}
$$40 v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \left( 392 + v ^ { 2 } \right)$$
(b) Find an expression for $v$ in terms of $x$.\\
(c) Calculate, correct to two decimal places, the greatest height of the ball.\\
(d) State, with a reason, whether the speed of the ball when it returns to $O$ is greater than $17 \mathrm {~ms} ^ { - 1 }$, less than $17 \mathrm {~ms} ^ { - 1 }$ or equal to $17 \mathrm {~ms} ^ { - 1 }$.\\
\hfill \mbox{\textit{WJEC Further Unit 6 Q1 [14]}}