| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Power from force and derived speed (non-equilibrium) |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring standard differentiation of vectors, application of F=ma, calculation of dot product, and verification of the work-energy principle. All techniques are routine for Further Maths students, though the exponential component adds minor complexity. The verification in part (d) is guided rather than requiring independent insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a = \frac{dv}{dt}\) | M1 | Correct differentiation of at least one term |
| \(a = 6ti - 8j - 2e^{-t}k\) | A1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(F = ma = 0 \cdot 5(6ti - 8j - 2e^{-t}k)\) | B1 | FT \(a\) from part (a) |
| \(F \cdot v = (3t \times 3t^2) + (-4 \times -8t) + (-e^{-t} \times 2e^{-t})\) | M1 | Correct method for dot product |
| \(F \cdot v = 9t^3 + 32t - 2e^{-2t}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \(v \cdot v = (3t^2)^2 + (-8t)^2 + (2e^{-t})^2\) | M1 | |
| \(KE = \frac{1}{2}m \, v \cdot v\) | m1 | |
| \(KE = \frac{1}{2 \times 2}(9t^4 + 64t^2 + 4e^{-2t})\) | A1 | cao |
| \((KE = \frac{1}{4}t^4 + 16t^2 + e^{-2t})\) |
| Answer | Marks | Guidance |
|---|---|---|
| (d) rate of work (power) \(= \frac{d}{dt}(KE)\) | B1 | Any equivalent statement, mathematical or otherwise |
| \(\frac{d}{dt}(KE) = \frac{d}{dt}\frac{1}{4}(9t^4 + 64t^2 + 4e^{-2t})\) | B1 | Convincing |
| \(= 9t^3 + 32t - 2e^{-2t} = F \cdot v\) |
**(a)** $a = \frac{dv}{dt}$ | M1 | Correct differentiation of at least one term
$a = 6ti - 8j - 2e^{-t}k$ | A1 | All correct
**Subtotal: [2]**
**(b)** $F = ma = 0 \cdot 5(6ti - 8j - 2e^{-t}k)$ | B1 | FT $a$ from part (a)
$F \cdot v = (3t \times 3t^2) + (-4 \times -8t) + (-e^{-t} \times 2e^{-t})$ | M1 | Correct method for dot product
$F \cdot v = 9t^3 + 32t - 2e^{-2t}$ | A1 | cao
**Subtotal: [3]**
**(c)** $v \cdot v = (3t^2)^2 + (-8t)^2 + (2e^{-t})^2$ | M1 |
$KE = \frac{1}{2}m \, v \cdot v$ | m1 |
$KE = \frac{1}{2 \times 2}(9t^4 + 64t^2 + 4e^{-2t})$ | A1 | cao
$(KE = \frac{1}{4}t^4 + 16t^2 + e^{-2t})$ | |
**Subtotal: [3]**
**(d)** rate of work (power) $= \frac{d}{dt}(KE)$ | B1 | Any equivalent statement, mathematical or otherwise
$\frac{d}{dt}(KE) = \frac{d}{dt}\frac{1}{4}(9t^4 + 64t^2 + 4e^{-2t})$ | B1 | Convincing
$= 9t^3 + 32t - 2e^{-2t} = F \cdot v$ | |
**Subtotal: [2]**
**Total for Question 2: 10**
---2. A particle of mass 0.5 kg is moving under the action of a single force $\mathbf { F N }$ so that its velocity $\mathrm { v } \mathrm { ms } ^ { - 1 }$ at time $t$ seconds is given by
$$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } - 8 t \mathbf { j } + 2 \mathrm { e } ^ { - t } \mathbf { k }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the acceleration of the particle at time $t \mathrm {~s}$.
\item Determine an expression for F.v at time $t \mathrm {~s}$.
\item Find the kinetic energy of the particle at time $t \mathrm {~s}$.
\item Describe the relationship between the kinetic energy of a particle and the rate of working of the force acting on the particle. Verify this relationship using your answers to part (b) and part (c).
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q2 [10]}}