QCE Specialist Maths: Modelling motion Q255

Modelling motion — Question 255

Original QCE Vault practice · 12 marks

Q255 · Practice questionTechnology-freeVery complex unfamiliar12 marks

QUESTION 255 (12 marks)

In a magnetic-launch simulation, a particle moves on a straight track with acceleration

a(x)=-20/(1+x)^2

m/s

^2

for

x\ge0

. It starts at

x=0

with positive speed

v_0

. The acceleration depends only on position and applies on both outward and return motion, whenever a return occurs.

Derive an expression for

v^2

in terms of

x

and

v_0

[3 marks]

Find the smallest

v_0

for which no finite turning point exists. Then find the turning position for

v_0=\sqrt{20}

m/s.

[3 marks]

For

v_0=\sqrt{20}

, determine the exact time to reach the turning point. You may use

x=\cos\theta

to evaluate the required integral.

[4 marks]

For the launch in part (c), find the exact mean speed over the complete out-and-back journey and explain why the return time equals the outward time.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

v^2=v_0^2-40+\frac{40}{1+x}

. (b) Critical speed

\sqrt{40}

m/s; turning position

x=1

m for the specified launch. (c)

T=\frac{1+\pi/2}{\sqrt{20}}

s. (d) Mean speed

\frac{\sqrt{20}}{1+\pi/2}

m/s.

Worked solution

(a) Use

a=v\,dv/dx

. Integrating

v\,dv=-20(1+x)^{-2}dx

from launch to position

x

gives

(v^2-v_0^2)/2=20/(1+x)-20

, yielding the stated relation.

(b) As

x\to\infty

v^2\to v_0^2-40

. If

v_0^2<40

, the turning point is

x_*=40/(40-v_0^2)-1

. If

v_0^2\ge40

v^2

is positive at every finite

x\ge0

, so there is no finite turn. The smallest such speed is

\sqrt{40}

. With

v_0^2=20

x_*=1

v=\sqrt{20(1-x)/(1+x)}

. Thus

T=\frac1{\sqrt{20}}\int_0^1\sqrt{(1+x)/(1-x)}\,dx

. With

x=\cos\theta

, the limits become

\pi/2

and 0; the product of the square-root factor and

\sin\theta

1+\cos\theta

. Therefore

T=\frac1{\sqrt{20}}\int_0^{\pi/2}(1+\cos\theta)\,d\theta=(1+\pi/2)/\sqrt{20}

. This is a convergent improper integral at the turning point.

(d) The position-only acceleration gives the same

v^2(x)

on both trips; the return velocity has opposite sign but equal magnitude. Hence the return time is also

T

. Total distance is 2 m and total time is

2T

, giving mean speed

1/T=\sqrt{20}/(1+\pi/2)

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Use the position-dependent acceleration form.

[1 mark]

Part a: Integrate and apply the launch condition.

[1 mark]

Part a: Obtain the correct expression for

v^2

[1 mark]

Part b: Use the sign of the limiting squared speed to distinguish the regimes.

[1 mark]

Part b: Identify the critical speed, including the equality case.

[1 mark]

Part b: Obtain the specified finite turning position.

[1 mark]

Part c: Choose the positive outward velocity and form the time integral.

[1 mark]

Part c: Use the substitution with correct limits.

[1 mark]

Part c: Simplify the integrand to

1+\cos\theta

[1 mark]

Part c: Evaluate the finite exact time.

[1 mark]

Part d: Justify equal trip times from the equal speed at each position.

[1 mark]

Part d: Use total distance and total time to obtain the mean speed.

[1 mark]

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Modelling motion — Question 255

QUESTION 255 (12 marks)

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