Further Math A2s 3.5.1 Variable-force motion

[Maximum number: 7]

A ball of mass 2 kg is projected vertically downwards with speed $5 \mathrm{~ms}^{-1}$ through a liquid. At time $t \mathrm{~s}$ after projection, the velocity of the ball is $v \mathrm{~ms}^{-1}$ and its displacement from its starting point is $x \mathrm{~m}$ . The forces acting on the ball are its weight and a resistive force of magnitude $0.2 v^{2} \mathrm{~N}$ .

(a)

Find an expression for v in terms of t.

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$2 \frac{\mathrm{~d} v}{\mathrm{~d} t}=2 g-0.2 v^{2}$
Separate variables and attempt to integrate $\frac{\mathrm{d} v}{0.1\left(100-v^{2}\right)}=\mathrm{d} t$
Integrate to a ln term of the correct form.

$\frac{1}{20} \ln \left(\frac{10+v}{10-v}\right)=0.1 t+c$ $t=0, \quad v=5, \quad c=\frac{1}{20} \ln 3$
Use initial condition.

$2 t=\ln \frac{10+v}{3(10-v)}, \mathrm{e}^{2 t}=\frac{10+v}{3(10-v)}$
Rearrange, removing ln.

$v=\frac{30-10 \mathrm{e}^{-2 t}}{3+\mathrm{e}^{-2 t}}$
AEF

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(b)

Deduce what happens to v for large values of t.

[ 1 ]

$v \rightarrow 10$

B1FT

FT from expression of correct form.

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CAIE A-Level Further Math A2 3.5.1 Variable Force Motion Questions

Exam points

CAIE A-Level Further Math A2 3.5.1 Variable Force Motion Questions question 1

Question (a)

Question (b)