A2 June 2021 Paper 1 Q8
8 A particle of mass 4 kg moves horizontally in a straight line.
At time \(t\) seconds the velocity of the particle is \(v\) m s−1
The following horizontal forces act on the particle:
- a constant driving force of magnitude 1.8 newtons
- another driving force of magnitude \(30\sqrt{t}\) newtons
- a resistive force of magnitude \(0.08v^2\) newtons
When \(t = 70\), \(v = 54\)
Use Euler’s method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds.
Give your answer to four significant figures. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses Newton’s second law to form a four term differential equation. Must have correct terms, condone wrong signs PI | M1 | 3.3 |
| Obtains correct expression for \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\) or \(4\dfrac{\mathrm{d}v}{\mathrm{d}t}\) PI | A1 | 1.1b |
| Substitutes correct values into their first Euler equation | M1 | 1.1a |
| Obtains value for \(v_{70.5}\) which rounds to 56.4 | A1 | 1.1b |
| Uses Euler’s method exactly twice | M1 | 3.1a |
| Obtains correct answer to required degree of accuracy Condone lack of units | A1 | 3.2a |
| (6 marks) |
Typical solution
\[4\frac{\mathrm{d}v}{\mathrm{d}t} = 1.8 + 30t^{\frac{1}{2}} - 0.08v^2\]\[\frac{\mathrm{d}v}{\mathrm{d}t} = 0.45 + 7.5t^{\frac{1}{2}} - 0.02v^2\]\[v_{70.5} \cong 54 + 0.5(\dot{v}_{70})\]\[= 54 + 0.5\left(0.45 + 7.5 \times 70^{\frac{1}{2}} - 0.02 \times 54^2\right)\]\[= 54 + 0.5(4.8795\ldots)\]\[= 56.439751\]\[v_{71} \cong 56.439751 + 0.5(\dot{v}_{70.5})\]\[v_{71} \cong 56.439751 + 0.5(-0.2857\ldots)\]\[= 56.2969\]Velocity after 71 seconds = 56.30 m s−1