CAIE A-Level Further Math A2 3.5 Linear Motion Under a Variable Force Questions

Practise selecting the acceleration form that matches the variables in a force law, separating and integrating it, then applying initial or terminal conditions.

Exam points

write Newton's second law and choose dv/dt or v times dv/dx to match the given force
separate variables and integrate before using the initial speed or position to fix the constant
use the velocity expression to find displacement, stopping distance or a terminal-speed limit

Covered syllabus points

3.5.1—Variable force motion

[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.

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

Integrate: $0.5 v^{2}=-\frac{300}{x+1}+\frac{450}{(x+1)^{2}}+A$
Correct powers, allow sign errors.

x=0, v=20;
A=50
Use initial condition.

Rearrange: $v^{2}=\frac{100\left(x^{2}-4 x+4\right)}{(x+1)^{2}}$
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, $v=\frac{20-10 x}{x+1}$
Signs dealt with convincingly.

6

[Maximum number: 9]

A particle P moving in a straight line has displacement $x \mathrm{~m}$ from a fixed point O on the line and velocity $v \mathrm{~ms}^{-1}$ at time $t \mathrm{~s}$ . The acceleration of P, in $\mathrm{ms}^{-2}$ , is given by $6 v \sqrt{v+9}$ . When t=0, x=2 and v=72.

(a)

Find an expression for v in terms of x.

[ 4 ]

$v \frac{d v}{d x}=6 v \sqrt{v+9}$ and attempt to separate variables and integrate
$2 \sqrt{v+9}=6 x+A$ $x=2, v=72 \quad A=6$

М1

Use initial condition to find constant.

$v=9(x+1)^{2}-9$
Correct, AEF .

4

(b)

Find an expression for x in terms of t.

[ 5 ]

$\left[\frac{d x}{d t}=9\left(x^{2}+2 x\right), \frac{d x}{x(x+2)}=9 d t\right] \frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+2}\right) d x=9 d t$
Separate variables and write in the form
$\left(\frac{a}{x}-\frac{b}{x+c}\right) d x=d t$ $\frac{1}{2} \ln \left(\frac{x}{x+2}\right)=9 t+B$
Integrate, any correct form.

$t=0, x=2 \quad B=\frac{1}{2} \ln \frac{1}{2}$
Use initial condition to find constant.

$18 t=\ln \left(\frac{2 x}{x+2}\right) e^{18 t}=\frac{2 x}{x+2}$
Take logarithms

$x=\frac{2 e^{18 t}}{2-e^{18 t}}$ or $x=\frac{2}{2 e^{-18 t}-1}$
Any correct form

5

[Maximum number: 10]

A particle P moving in a straight line has displacement $x \mathrm{~m}$ from a fixed point O on the line at time $t \mathrm{~s}$ . The acceleration of P, in $\mathrm{ms}^{-2}$ , is given by $\frac{200}{x^{2}}-\frac{100}{x^{3}}$ for x>0. When t=0, x=1 and P has velocity $10 \mathrm{~m} \mathrm{~s}^{-1}$ directed towards O.

(a)

Show that the velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ of P is given by $v=\frac{10(1-2 x)}{x}$ .

[ 5 ]

$\begin{aligned} v \frac{\mathrm{~d} v}{\mathrm{~d} x}=-\frac{100}{x^{3}}+\frac{200}{x^{2}} \frac{v^{2}}{2}=\frac{50}{x^{2}}-\frac{200}{x}+A \end{aligned}$
Correct equation and attempt to integrate
Correct

$x=1, v=-10: \quad A=200$
Use initial condition

$v^{2}=\frac{100(2 x-1)^{2}}{x^{2}}$
Rearrange to find $v^{2}$ $v= \pm \frac{10(2 x-1)}{x}$ and take negative sign to meet initial condition,
so $v=\frac{10(1-2 x)}{x}$
Convincingly shown (no mention of ± scores A0)

AG

5

PUBLISHED

(b)

Show that x and t are related by the equation $\mathrm{e}^{-40 t}=(2 x-1) \mathrm{e}^{2 x-2}$ and deduce what happens to x as t becomes large.

[ 5 ]

$\begin{aligned} \frac{x \mathrm{~d} x}{1-2 x}=10 \mathrm{~d} t \frac{1}{2}\left(\frac{1}{1-2 x}-1\right) \mathrm{d} x=10 \mathrm{~d} t -\frac{1}{4} \ln |1-2 x|-\frac{x}{2}=10 t+B \end{aligned}$
Rearrange and attempt to integrate

$t=0, x=1: \quad B=-\frac{1}{2}$
Use initial condition

$2 x-2=-40 t-\ln (|1-2 x|)$ so $e^{-40 t}=(2 x-1) e^{2 x-2}$
Convincingly shown, working required
AG

For large values of $t, x \rightarrow \frac{1}{2}$
CAO

5

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)