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.
Calculating the Centre of Mass as the distance from the heavier object
xCOM = (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) = 18.1o
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.
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
Fc = Fg
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.
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
ωf = ωi + αt
7.00 = 3.00 + 4.50α
α = 0.889 rad s-2
ωf2 = ωi2 + 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.
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
From the rotational motion formulas
ω = 2πf
so ω2 = 4π2f2
This can be rearranged to make 'f' the subject.
Q3b from the 2023 exam
ω = 2πf = 2π/T
gradient = 1,000 = ω2, so ω = 31.2 rad s-1 = 2π/T so T = 0.199 s.
Now, use the formula for the period of a mass on a spring
Q3c from the 2023 exam
Q3d from the 2023 exam
For excellence, all of these points
For achieved, one of these. For merit, 3 of them.
