Auckland Sports College

"Moving Forward Together"

Physics - Level 3 Mechanics

Immediately below this paragraph is a table with links to the exam questions and answers for all the externals way back to 2013. Focus on the exams for the last three years. Recognise the type of question e.g This is a question about..... Know how you are going to set out your working when answering each type of question.

Scipad L3 Physics externals We use the L3 Physics externals Scipad.

If you click on the resource booklet image a new window will open that shows you the formula sheet you will have in the exams.

We also the free "College Physics" textbook from openstax

Mechanics

Mr Whibley has a helpful youtube video for this Level 3 Mechanics standard

Mr Whibley's notes are here

Translational Motion

Centre of Mass Q2a from the 2022 exam

Calculating the Centre of Mass as the distance from the heavier object

x_COM = (m₁x₁ + m₂x₂)/(m₁ + m₂)

= (100*4,000 + 300*0)/400 = 1,000
The Centre of Mass is 1,000m from the heavier object along the line joining the two objects

Q2b from the 2022 exam

F = G*M*m/r² = 2 x 10^-8 N

acceleration of the 100kg rock = 2 x 10^-10 m s ^-2

Need a justified comment about the validity of the assumption for merit.

Momentum, Impulse, Conservation of Momentum
Q2c from the 2022 exam

Total momentum is conserved so:
Since the initial total momentum is horizontal, the vertical components of momentum of both rocks must be equal and opposite / add to zero and since the angle to the horizontal is the same, the size of the momentum is the same

OR Since the mass of both rocks are the same and the angle to the horizontal is the same, they must have the same velocity therefore same momentum.

Q2d from the 2022 exam

initial momentum = 100*500 = 50,000 kg m s^-1 in the x-direction

The total of the two fragments’ final momentum in the x direction must equal 50,000 kg m s^-2

2*200*v*cosθ = 50,000

v = 133 m s^-1

Circular Motion

At level 2 you were expected to understand that the velocity of an object travelling at constant speed in a horizontal circle was constantly changing i.e there was an unbalanced external force acting that was directed toward the centre of the circle. When the object was a car moving around a track, the friction force between the tires and the track kept the car moving in a circle. If that friction force reduced (e.g. wet road or oil on the truck) the car might slide and stop moving in a circle. At level 3 you get to study circular motion in two dimensions - when there is a vertical component as well as a horizontal component to the external force - and circular motion around a vertical loop e.g a rollercoaster.

Mr Whibley starts his talk on circular motion at 19 minutes and 20 seconds into his video.

Banked Corner
Mr Whibley discusses this at 21m 25s into his video
Banked corners are discussed at p127 of the SciPad
The curve is banked at an angle Θ
The support force is perpendicular (normal, at 90 degrees) to the road. As gravity is acting vertically downward, the support force has a vertical component, equal to the force due to gravity, and a horizontal component, the centripetal force.

This is question 1b from the 2023 exam. It is a 'merit' question.

Tane has a toy car track set. Part of the track is a horizontal banked curve.
The mass of the toy car is 0.120kg
The banked curve of the car track has a radius of 0.750 m.
Calculate the angle of banking when there is no sideways friction on the wheels of the car as it goes around the banked curve at 1.55 m s–1.

Θ is the angle at the top of this right angled triangle. The centripetal force is the opposite side, the force due to gravity is the adjacent side, so
tanΘ = O/A = centripetal/gravity
tanΘ = 0.3844/1.1772 = 0.3265
Θ = tan^-1(0.3265) = 18.1^o

Q1b from the 2020 exam is another 'banked curve' question. It is an 'excellence' question.

Vertical Circular Motion
Mr Whibley discusses this at 24m 20s into his video
Vertical Circular motion is discussed at p137 of the SciPad

If you swing a bucket of water in a vertical circle fast enough, the water stays in the bucket even when the bucket is upside down at the top of the circle.

For this to happen, at the top of the circle the centripetal force must be at least equal to the force of gravity.

There is a helpful youtube video here

This is Q1d from the 2023 exam. It is an excellence question.

This is a conservation of energy question. At the top of the loop the car has kinetic energy and it has gravitational potential energy. By the bottom of the loop, the potential energy has been converted to kinetic energy.

Step 1. The speed at the top of the loop
F_c = F_g
mv²/r = mg
v² = gr = 2.45
v = 1.565ms^-1

Step 2.
Total energy, E = mgh + 0.5mv² = 0.588 + 0.147 = 0.735J

Step 3.
Speed at the bottom of the loop
0.735= 0.5*0.12*v²
v² = 0.735/0.06 = 12.25
v = 3.5ms^-1

Rotational Motion

Q2a from the 2023 exam

At the top of the ramp the barrel has gravitational potential energy
At the bottom of the ramp all the gravitational potential energy has been converted to linear (translational) kinetic energy and rotational kinetic energy.

Q2b from the 2023 exam

GPE = mgh = 5.5*9.81*1.34 = 72.3J
E_k(lin) gained = 0.5*5.5*3.4² = 31.8J

So, E_k(rot) gained = 72.3 - 31.8 = 40.5J

E_k(rot) = 0.5Iω²

and ω = v/r = 3.4/0.28 = 12.1rad s^-1

So I = (2*40.5)/12.1² = 0.553 kg m³

Q2c from the 2023 exam

If there is no net external torque, angular momentum is conserved.
Holding his arms out increases rotational inertia, since mass is further away from the axis of rotation.
Since L = Iω, when rotational inertia increases, angular velocity decreases, causing him to spin slower when his arms are outstretched.

Q2d from the 2023 exam

ω_f = ω_i + αt

7.00 = 3.00 + 4.50α

α = 0.889 rad s^-2

ω_f² = ω_i² + 2αθ

49 = 9 + 2*0.889θ

θ = 22.5rad

number of revolutions = 22.5/2π = 3.58 rev

Simple Harmonic Motion

Mr Whibley starts his talk on simple harmonic motion at 33 minutes and 32 seconds into his video.

We are going to start with this video from Khan Academy

The SciPad section on SHM (Simple Harmonic Motion) pages 188-248 is pretty ordinary.

I want you to imagine the 'Khan Academy' diagram as a block attached to a spring that is lying flat on a frictionless table in a vacumn i.e there are no unbalanced external forces and energy is conserved. When the spring is stretched to point 'A' in the Khan Academy video, and then held there, the block has potential energy (equal to kx²) but no kinetic energy. When the block is released, a Force acts on the block (it is called the restorative force as it acts to restore the spring to its natural length), the block accelerates, and potential energy begins to be converted to kinetic energy. When the block is at the neutral (original, before the spring was stretched) position the block is at its maximum velocity and has kinetic energy but no potential energy. Then, as the block compresses the spring its kinetic energy (velocity) decreases and its potential energy begins to build. By the time the block is at '-A' it has zero kinetic energy but the full amount of potential energy, and the block begins its journey back to 'A'.

In SHM acceleration is not constant. As F = -kx, ma = -kx. The acceleration is greatest at maximum displacement and is zero at the equilibrium position.

In SHM velocity is greatest at the equilibrium position and zero at maximum displacement

SHM can be modelled as circular motion. Watch this video The point is that the displacement of the weight on a spring is the same as the displacement in one dimension of an object moving in a circle at constant velocity. We call this circle, which has radius 'A' the reference circle.

Q3a from the 2023 exam