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# ReVIEW - PowerPoint PPT Presentation

ReVIEW. Physics 2.3 Waves & 2.4 Mechanics. Review. For Calculation Questions : Go through every formula and make sure you know all the symbols, what they stand for, what the units are for each; and what the formula is used for and how to rearrange it.

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### ReVIEW

Physics 2.3 Waves & 2.4 Mechanics

• For Calculation Questions:

• Go through every formula and make sure you know all the symbols, what they stand for, what the units are for each; and what the formula is used for and how to rearrange it.

• Remember to give your answer to the same number of significant figures as the least number of significant figures given in the question.

• For Explanation Questions:

• Go through glossary list and make sure you know what every term means.

• Remember to make your answer as detailed as possible and refer to a relevant equation if you can.

• Read what you have written to check it makes sense and answers the question.

• Reflection & refraction

• Critical angle & total internal reflection

• Superposition of waves

• Mirrors – how to explain images

• Drawing ray diagrams & the 4 rays for mirrors

• Lenses – 3 rays

• Convex mirror & concave lens – virtual focus so the focal length is negative

Review – Mechanics 2.4

• Kinematic equations – used when a is constant

• Vectors – draw vector diagrams & use trigonometry

• Circular motion (atowards centre of circle, vtangential)

• Know Newton’s laws

• Draw free-body force diagrams

• Balance torques

• Use conservation of momentum

• Energy transfers

• Power & work

• Projectiles – consider vertical & horizontal components separately

Review – Mechanics 2.4

• Louise is playing soccer for her 1st XI soccer team. Louise has a mass of 65 kg.

• (a)  Louise is running towards the goal at 8.0 m s–1. She slows down to 6.0 m s–1in 3 s.

• Calculate the distance she travels over the 3 s. Write your answer to the correct number of significant figures.

• (b)  State the main energy change when she is decelerating.

d = v1 + v2

× t

= 21 m

= 7 × 3

2

Kinetic energy  heat (+ sound)

Review – Mechanics 2.4

• Louise is playing soccer for her 1st XI soccer team. Louise has a mass of 65 kg.

• (c) Calculate her kinetic energy while she is running at 8.0 m s–1.

• (d)  Calculate the size of Louise’s momentum while she is running at 8.0 m s–1. Write your answer with the correct units.

• (e)  While she is running at 6.0 m s–1 to the right, the ball rolls at 4.0 m s–1 to the left. Calculate the speed of the ball relative to Louise.

Ek = ½ mv2

= 2080 J

= ½ (65)(8)2

ρ = mv

= 520 kgms-1

= (65)(8)

6 + 4 = 10 ms-1

Review – Mechanics 2.4

• Louise is playing soccer for her 1st XI soccer team. Louise has a mass of 65 kg.

• (f)  Determine the size of the average net force acting on her when she is running at constant speed. Explain your answer.

For Louise to travel at a constant speed, all forces acting on her must be balanced, hence the net force acting on her is zero.

OR

If Louise is travelling at a constant speed then her acceleration must be zero, hence according to Newton’s second law,F = ma, her net force must be zero.

Review – Mechanics 2.4

• Louise kicks the ball vertically up. It rises and then falls. You may assume air resistance is negligible.

• (g) Describe and explain what happens to:

• (i)  the force on the ball

• (ii)  the ball’s acceleration

• (iii)  the ball’s velocity

• after the ball is kicked and as it rises and falls.

The only FORCE on the ball is gravity. This is down and constant the whole time.

ACCELERATION is constant and downwards

The VELOCITY is upwards, decreasing, then zero at the top, then downwards increasing.

Review – Mechanics 2.4

The collision is inelastic.

State what this means.

Inelastic means kinetic energy is not conserved / is lost.

ρbefore = ρafter

m1v1= (m1 + m2)vf

Eptop= mgh

= 55 × 9.8 × 4.0

= 2156 J

Ek bottom = ½ mv2 = 2156 J

v = √2Ek/m

= 8.85 ms-1

(55)(8.85) = (120)vf

Ek bottom = 2156 J

Vf = 4.06 ms-1

Review – Mechanics 2.4

Some painters are working at the airport. They have a uniform plank resting on two supports. The plank is 4.0 m long. It has a mass of 22 kg. The two legs that support the plank are 0.50 m from either end, as shown in the figure below.

(a) The plank is in equilibrium.

Draw labelled arrows of appropriate sizesin the correct position showing the forces acting on the plank on the diagram above.

Fs = 110 N

Fs = 110 N

W = mg

= 220 N

Review – Mechanics 2.4

(b) Calculate the support force on the plank at A if a painter of mass 60 kg sits 0.75 m from A, and another painter of mass 75 kg sits at a distance of 0.80 m from B.

Use g = 10 m s–2

Taking the moments about pivot point B

3 FA = (0.80 750) + (1.5  220) + (2.25 600)

3 FA= 600 + 330 + 1350

3 FA= 2280

FA = 760 N

?

Fb

Wp= 220 N

Review – Mechanics 2.4

• At a duty-free shop at the airport, a toy teddy bear is hanging at the end of a spring. The spring is 51.0 cm long when hanging vertically. When the teddy bear of mass 400 g is hung from the end of the spring, the length of spring becomes 72.0 cm.

F = mg

x = 72.0 – 51.0

(b) Calculate the energy stored in the spring when a second toy of mass 300 g is also hung along with the teddy bear on the spring.

(c) The 400 g teddy bear is now hung on a stiffer spring, which has double the spring constant. Discuss how this affects the extension and the elastic energy stored in the spring.

= 21.0 cm

F = kx

.4 x 9.81 = k (0.21)

k = 19 Nm-1

Ep = ½ kx2

F = mg

F = kx

m = 400 g + 300 g

=1.27 J

= 6.9 N

= 0.7 kg

x = F/k

= 0.369 m

Review – Mechanics 2.4

• At a duty-free shop at the airport, a toy teddy bear is hanging at the end of a spring. The spring is 51.0 cm long when hanging vertically. When the teddy bear of mass 400 g is hung from the end of the spring, the length of spring becomes 72.0 cm.

x = 21.0 cm

F = mg

(b) Calculate the energy stored in the spring when a second toy of mass 300 g is also hung along with the teddy bear on the spring.

(c) The 400 g teddy bear is now hung on a stiffer spring, which has double the spring constant. Discuss how this affects the extension and the elastic energy stored in the spring.

F = kx

k = 19 Nm-1

Ep = ½ kx2

F = 7.0 N

x = F/k

m = 0.7 kg

= 0.368 m

=1.29 J

A spring with double the spring constant for the same weight force would mean half the extension as F = kx.

As EP k, EP doubles when k doubles for the same extension. As EP x2it decreases by four when x is halved for the same spring constant. Overall a spring with double the spring constant for the same weight force would mean half the extension and hence half the Ep.

Review – Mechanics 2.4

The baggage at the airport is delivered on a horizontal circular conveyor belt that is moving at constant speed. The radius of the circular belt is 7.0 m.

Draw an arrow in the diagram below to show the direction of the velocity of the suitcase that is on the moving circular belt.

(b) Explainwhy the motion of the suitcase on the belt that is moving in a circle at constant speed is accelerated motion.

Object is changing direction

– velocity is a vector, rate of change of velocity is a

(c) Calculate the time it takes for the belt to complete one rotation if the unbalanced force on the suitcase is 5.5 N.

The mass of the suitcase is 18 kg.

v = √(F x r / m)

Vs

= 1.463 ms-1

= 30.07 s

d = 2πr = 2π×7.0 = 43.982 m

Review – Mechanics 2.4

Marama is a long-jumper. She runs down a track, and jumps as far as she can horizontally. Her take-offvelocity is shown in the diagram below.

You can assume there is no air resistance.Acceleration due to gravity = 9.8 m s–2.

Show that the horizontal component of her initialvelocity is 6.0 m s–1.

(b) Show that the vertical component of her initialvelocity is 2.2 m s–1.

(c) Calculate the distance she jumps horizontally.

Vx = 6.4 cos20°

= 6.01 ms-1

Vx = 6.4 sin20°

= 2.19 ms-1

(find the time to reach max height)

vf = vi + at

Vy

2 x 0.22

Time for entire flight =

0 = 2.2 – 9.81t

= 0.44 s

t = 0.22 s

d = vhx t

Vx

= 2.64 m

Review – Waves 2.3

Roy and Sally spent time on a beach watching the incoming waves. The diagram below shows wavefronts as they approach shallow water. The waves travel slower in shallow water.

(a) On thediagram, draw an arrow showing the wave direction and the refracted wavefronts in the shallow region.

(b) Clearly explain why the waves behave as you have drawn them in the diagram above.

The right-hand end of the waves slow down before the left-hand end. This causes the wave front to swing towards the normal.

Statewhat happens to the frequency and wavelength of the waves as they pass from deep to shallow water.

Frequency

Wavelength

stays the same.

Wavelength decreases as speed decreases as v= f..

Review – Waves 2.3

(d) Roy counted 8 complete waves passing a fixed point in 20 seconds. Calculate the frequency of the waves.

Statetwo possible units for frequency.

(f) The speed of the waves in shallow water is 2.8 m s–1. Calculate the wavelength of these waves.

F=

F = 0.40 Hz

Hz, s–1

V = f

 =

• =

•  = 7.0 m

Review – Waves 2.3

(g) Give a physics reason why Roy and Sally could hear the sound of the waves when they were sitting behind a building at the beach, even though they could not seethe waves.

StateTWO important physical differences between sound waves and light waves.

(1)

(2)

Sound waves diffract around barrier.

Sound waves have a long wavelength.

Light waves’ short wavelength means they will not be diffracted.

Light waves are electromagnetic waves, transverse

Travel at 3.0  108ms–1/ they do not need a medium.

Sound waves are mechanical waves, need a material medium,

longitudinal waves.

Review – Waves 2.3

Roy and Sally later shone a red laser beam through two narrow slits. They saw a pattern formed on the wall as shown in the diagram below.

(a) Statea name given to the bright fringe.

(b) Explainclearly why there are dark fringes on either side of the central bright fringe on the wall.

Antinode / constructive interference

The dark fringes on either side of the zero-order fringe are caused by destructive interference.  Light from one slit travels a distance that is ½ wavelength longer than the distance travelled by light from the other slit.  Crests meet troughs at these locations.

Review – Waves 2.3

Roy and Sally took skipping ropes to the beach. One rope was thicker than the other. They tied the two ropes together. Roy held the thin rope and gave it a flick so that a pulse travelled along the thin rope towards the thick rope as shown in Box One below.

Draw a diagram in Box Two below, to show the reflected and refracted pulses after the pulse hits the boundary. (The pulse travels faster in the thin rope.)

Review – Waves 2.3

Roy and Sally noticed a puddle of water with oil floating on top. The diagram below shows a ray of light travelling from air as it meets the air-oil interface.

Complete the path of the ray of light in the above diagram to show what happens to the ray as it enters the oil, and then the water.

The ray of light meets the air-oil interface at an angle of incidence of 40

Calculate the angle of refraction when the ray goes into the water.

n1sin1 = n2sin2

2 = sin–1(1.00 sin40 1.50) = 25.4

n2sin2 = n3sin3

3 = sin–1(1.50 sin 25.4 1.33) = 29

Review – Waves 2.3

Sally knelt down to take a closer look at the fish. While looking up at the surface of the water in the fish tank, she noticed that the surface of the water looked like a mirror and she could see the reflection of the fish in it.

(e) Explainconcisely, using physics principles, the conditions that are required for the rays of light from the fish to reflectoff the water / air boundary (interface).

Rays should travel from an optically dense to an optically less dense medium.

The angle of incidence should be greater than critical angle.

(f) Calculate the critical angle of the water / air interface. Express your answer to the correct number of significant figures.

The refractive index of water is 1.33

The refractive index of air is 1.0

n1sin 1 = n2sin2

Critical angle of water =

c = sin–1( )

c = 48.75 = 49

Refraction from more dense to less dense

Light bends awayfrom the normal