Edexcel A-Level Physics A2 Unit 4 Fields and Particles

Practise Edexcel A-Level Physics A2 Unit 4 Further Mechanics, Fields and Particles with field, particle, energy and relativity tasks.

Covered syllabus points

4.3 - Further Mechanics 4.4 - Electric and Magnetic Fields 4.5 - Nuclear and Particle Physics

[Maximum number: 8]

Some asteroids pass very close to the Earth. Scientists are planning methods to deflect asteroids, to prevent them hitting the Earth.

One method would involve colliding a spacecraft into the surface of the asteroid, to change the path and speed of the asteroid. The spacecraft would remain joined to the asteroid after the collision.

(a)

This collision method is modelled for a spacecraft travelling in a direction at $90^{\circ}$ to the path of the asteroid.

Sketch a labelled vector diagram to show the momenta of the bodies before and after the collision.

[ 2 ]

(b)

Show that this collision method causes the asteroid to change its direction through an angle of about $10^{-7}$ radian.

Momentum of asteroid = $7.6\times10^{13}\,\mathrm{N\,s}$

[ 2 ]

Use trigonometry with the mutually perpendicular momenta.

$\tan\theta=\frac{1.1\times10^7}{7.6\times10^{13}}=1.45\times10^{-7}$

Hence $\theta=1.45\times10^{-7}\,\mathrm{rad}$ , approximately $1.5\times10^{-7}\,\mathrm{rad}$ .

(c)

After the collision, the asteroid and spacecraft remain joined and move together.

Calculate the component of their velocity at 90° to the original path of the asteroid after the collision.
mass of asteroid = 2.8 × 10⁹ kg

Component of velocity =

[ 2 ]

(d)

Another method would involve attaching a rocket motor to the asteroid and using the motor to apply a force to the asteroid.

In this method the force is applied at $90^\circ$ to the path of the asteroid. Deduce whether this would produce a change in momentum as great as the change produced by the collision method.

Force exerted by rocket motor = $5.1\times10^6\,\mathrm{N}$

Time for which rocket motor applies force = $6\,\mathrm{minutes}$

[ 2 ]

Use impulse $\Delta p=F\Delta t$ .

$\Delta p=(5.1\times10^6\,\mathrm{N})(6\times60\,\mathrm{s})=1.8\times10^9\,\mathrm{N\,s}$

This is greater than the $1.1\times10^7\,\mathrm{N\,s}$ change from the collision method.

[Maximum number: 9]

Some spiders may be seen suddenly moving up into the air, even when there is no wind. This happens when they let out strands of web.

The spiders are pulled upwards as the electrically charged strands of web interact with the Earth's electric field.

The average strength of the Earth's electric field is $120 \mathrm{~V} \mathrm{~m}^{-1}$ downwards.

(a)

State, with a reason, the polarity of the charge on the strands of web.

[ 1 ]

(b)

A spider moves upwards after letting out a strand of web. Determine the initial upwards acceleration of the spider.

mass of spider $=3.00\times10^{-6}\,\mathrm{kg}$

total magnitude of charge on a strand of web $=3.00\times10^{-7}\,\mathrm{C}$

[ 5 ]

Use of F=E Q

- Use of W=m g

- F-W to determine resultant force

- Use of F=m a

- $a=2.2 \mathrm{~m} \mathrm{~s}^{-2}$

Example of calculation
$\begin{aligned} & F=120 \mathrm{~V} \mathrm{~m}^{-1} \times 3.00 \times 10^{-7} \mathrm{C}=3.60 \times 10^{-5} \mathrm{~N} & W=3.00 \times 10^{-6} \mathrm{~kg} \times 9.81 \mathrm{Nkg}^{-1}=2.94 \times 10^{-5} \mathrm{~N} & \text { Resultant force }=3.60 \times 10^{-5} \mathrm{~N}-2.94 \times 10^{-5} \mathrm{~N}=6.57 \times 10^{-6} \mathrm{~N} & a=6.57 \times 10^{-6} \mathrm{~N} \div 3.00 \times 10^{-6} \mathrm{~kg} & =2.19 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

(5)

(c)

The Earth has an electric field because charge is distributed over its surface.

Determine the quantity of charge on $1 \mathrm{~m}^{2}$ of the Earth's surface which would cause an electric field of $120 \mathrm{~V} \mathrm{~m}^{-1}$ .

You should assume that the charge is distributed evenly.
radius of Earth $=6400 \mathrm{~km}$

[ 3 ]

Use $E=Q/(4\pi\varepsilon_0r^2)$ or $E=kQ/r^2$ , and $A=4\pi r^2$ .

$120=(8.99\times10^9)Q/(6.4\times10^6)^2$ , so $Q=5.47\times10^5\,\mathrm{C}$ . The surface area is $5.15\times10^{14}\,\mathrm{m^2}$ , giving $Q/A=1.1\times10^{-9}\,\mathrm{C\,m^{-2}}$ .

[Maximum number: 1]

Which of the following is the number of neutrons in a nucleus of ${ }_{51}^{121} \mathrm{Sb}$ ?

121

172

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Covered syllabus points

Unit 4: Further Mechanics, Fields and Particles question 1

Question (a)

Question (b)

Question (c)

Question (d)

Unit 4: Further Mechanics, Fields and Particles question 2

Question (a)

Question (b)

Question (c)

Unit 4: Further Mechanics, Fields and Particles question 3