CAIE A-Level Further Mathematics 3.5.1 Variable-Force Motion

Practise forming and solving separable differential equations for linear motion under forces that vary with time, speed or position.

[Maximum number: 6]

A particle P of mass 0.5 kg moves in a straight line. At time $t \mathrm{~s}$ the velocity of P is $v \mathrm{~ms}^{-1}$ and its displacement from a fixed point O on the line is $x \mathrm{~m}$ . The only forces acting on P are a force of magnitude $\frac{150}{(x+1)^{2}} \mathrm{~N}$ in the direction of increasing displacement and a resistive force of magnitude $\frac{450}{(x+1)^{3}} \mathrm{~N}$ . When t=0, x=0 and v=20.

Find v in terms of x, giving your answer in the form $v=\frac{A x+B}{(x+1)}$ , where A and B are constants to be
determined. determined.

$0.5 v \frac{\mathrm{~d} v}{\mathrm{~d} x}=\frac{150}{(x+1)^{2}}-\frac{450}{(x+1)^{3}}$

M1

Allow sign errors.

Integrate: $0.5 v^{2}=-\frac{300}{x+1}+\frac{450}{(x+1)^{2}}+A$

M1A1

Correct powers, allow sign errors.

x=0, v=20; A=50

M1

Use initial condition.

Rearrange: $v^{2}=\frac{100\left(x^{2}-4 x+4\right)}{(x+1)^{2}}$

A1

AEF

$v^{2}=\frac{100(x-2)^{2}}{(x+1)^{2}} \text { so } v= \pm \frac{10(x-2)}{(x+1)}$

From initial condition, sign must be negative, $v=\frac{20-10 x}{x+1}$

A1

Signs dealt with convincingly.

6

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3.5.1—Variable force motion question 1