Edexcel A-Level Physics A2: Further Mechanics

[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: 12]

Hockey is a sport played with a stick and a ball. The player tries to hit the ball with part of the stick called the 'heel', as shown.

Figure for Question 2 — Edexcel A-Level Physics A2

(a)

The player swings her stick so that the heel moves horizontally in a circle of radius 0.80 m across the ground, as shown below.

Figure for Question (a) — Edexcel A-Level Physics A2

It takes a time of 0.22 s for the heel to move through an angle of 1.3 radians.
Calculate the speed of the heel.

[ 3 ]

The angular speed is $\omega=\theta/t=1.3/0.22=5.9\,\mathrm{rad\,s^{-1}}$ . Hence $v=r\omega=(0.80)(5.9)=4.73\,\mathrm{m\,s^{-1}}.$

(b)

A ball has a speed of $13 \mathrm{~ms}^{-1}$ . The heel has a momentum $p_{\text {heel }}$ of 3.0 Ns , in the direction of the dashed line, as shown.

The heel collides with the ball and stops.
The speed of the ball after being hit by the heel is $16 \mathrm{~ms}^{-1}$ .

Figure for Question (b) — Edexcel A-Level Physics A2

[ 9 ]

(i)

Deduce whether momentum is conserved for this collision by completing the vector diagram below. A scaled line representing $p_{\text {heel }}$ is shown.
mass of ball $=160 \mathrm{~g}$

[ 5 ]

The ball momenta are $p_i=(0.160)(13)=2.08\,\mathrm{N\,s}$ and $p_f=(0.160)(16)=2.56\,\mathrm{N\,s}$ . Using the stated scale, draw these vectors with the given heel momentum $3.0\,\mathrm{N\,s}$ . Momentum is conserved if the three vectors form a closed triangle.

Figure for Answer to Question (i) — Edexcel A-Level Physics A2

(ii)

Deduce whether the collision is elastic.
speed of heel before collision = 5.0 m s^-1

[ 4 ]

Before collision, $E_k=\tfrac12p_{\rm heel}v_{\rm heel}+\tfrac12m v_i^2=7.5+13.5=21.0\,\mathrm J$ . After collision, $E_k=\tfrac12(0.160)(16)^2=20.5\,\mathrm J$ . The values are not equal, so the collision is not elastic.

[Maximum number: 1]

A non-relativistic particle of mass m has momentum p and kinetic energy E_k.
A second non-relativistic particle of mass m/2 has momentum 2p.
Which of the following expressions gives the kinetic energy of the second particle?

A

E_k/2

B

E_k

C

2E_k

D

8E_k

[Maximum number: 2]

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.

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 = $6\,\mathrm{minutes}$

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: 4]

Hockey is a sport played with a stick and a ball. The player tries to hit the ball with part of the stick called the 'heel', as shown.

Figure for Question 5 — Edexcel A-Level Physics A2

A ball has a speed of $13 \mathrm{~ms}^{-1}$ . The heel has a momentum $p_{\text {heel }}$ of 3.0 Ns , in the direction of the dashed line, as shown.

The heel collides with the ball and stops.
The speed of the ball after being hit by the heel is $16 \mathrm{~ms}^{-1}$ .

Deduce whether the collision is elastic.
speed of heel before collision = 5.0 m s^-1

Before collision, $E_k=\tfrac12p_{\rm heel}v_{\rm heel}+\tfrac12m v_i^2=7.5+13.5=21.0\,\mathrm J$ . After collision, $E_k=\tfrac12(0.160)(16)^2=20.5\,\mathrm J$ . The values are not equal, so the collision is not elastic.

[Maximum number: 3]

Hockey is a sport played with a stick and a ball. The player tries to hit the ball with part of the stick called the 'heel', as shown.

Figure for Question 6 — Edexcel A-Level Physics A2

The player swings her stick so that the heel moves horizontally in a circle of radius 0.80 m across the ground, as shown below.

It takes a time of 0.22 s for the heel to move through an angle of 1.3 radians.
Calculate the speed of the heel.

The angular speed is $\omega=\theta/t=1.3/0.22=5.9\,\mathrm{rad\,s^{-1}}$ . Hence $v=r\omega=(0.80)(5.9)=4.73\,\mathrm{m\,s^{-1}}.$

[Maximum number: 5]

An aeroplane flies in a horizontal circular path whilst waiting to land at an airport.

The aeroplane flies at a constant speed v around a horizontal circular path of radius r. The diagram shows two positions A and B of the aeroplane, on its circular path.

Figure for Question 7 — Edexcel A-Level Physics A2

The acceleration of the aeroplane is a.
Derive the expression $a=\frac{v^{2}}{r}$
You should include a vector diagram.

For a small angular displacement, the velocity-vector triangle gives $\Delta v/v\approx\theta$ . Also $v=r\theta/t$ , so $\theta/t=v/r$ . Therefore $a=\frac{\Delta v}{t}=v\frac{\theta}{t}=\frac{v^2}{r}.$

Figure for Answer to Question 7 — Edexcel A-Level Physics A2

[Maximum number: 4]

An aeroplane flies in a horizontal circular path whilst waiting to land at an airport.

The aeroplane flies in a horizontal circle by tilting to one side in a movement called 'banking'.

The aeroplane creates an upwards lift force, which acts in a direction perpendicular to its wings.

The diagram shows this lift force when the aeroplane is banking.

Figure for Question 8 — Edexcel A-Level Physics A2

Explain how banking allows the aeroplane to fly in a horizontal circular path.

[Maximum number: 4]

An aeroplane flies in a horizontal circular path whilst waiting to land at an airport.

The aeroplane flies in a horizontal circle by tilting to one side in a movement called 'banking'.

The aeroplane creates an upwards lift force, which acts in a direction perpendicular to its wings.

During banking, the angle between the wings and the horizontal is 5.2°.

Calculate the radius of the circular path when the aeroplane flies at a constant speed of 530 m s^-1.
mass of aeroplane = 4.1 × 10^5 kg

Radius =

Vertical equilibrium gives $L\cos5.2^\circ=mg$ , while the horizontal component supplies centripetal force: $L\sin5.2^\circ=mv^2/r$ . Thus $r=\frac{v^2}{g\tan5.2^\circ}=3.15\times10^5\,\mathrm m,$ about $3.2\times10^5\,\mathrm m$ .

Edexcel A-Level Physics A2 4.3 Further Mechanics Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9