Explore Science Elite Camp
Explore Science Elite Camp –
A2 Physics
Homework – Week 15
教学资料属于@探索国际教育教学部，不得转发或做其它⽤途。
1.
The angle subtended by an arc of a circle that has a length equal to the circle radius
a = w^2r
F = ma
a = F/m
F/m = w^2r
w^2 = F/mr
W = mg
m = W/g
w = sqrt(F/mr) = sqrt((18-3)/((3/9.81)*0.85)) = 7.6
7.6
2.
Repeated back and forth movements on either side of an equilibrium position
Ideal condition where the oscillation movements are not affected by any
external force
Oscillations when acceleration is proportional to displacement
from a fixed point and in the opposite direction, which is towards a fixed point
Not simple harmonic as acceleration of ball is constant due to gravity
3.
Loss of energy from a system due to resistive forces
Light damping as the mass has many oscillations
period = 0.6/2 = 0.3
1/0.3 = 3.3
3.3
KE = 1/2 mv^2
v = wr
KE = (1/2) *(65*10^-3)((2[pi]/0.3) *1.5*10^-2)^2 = 0.0032076... = 0.0032
= 3.2 mJ
1.5 -1.1 = 0.4, 1.1-0.4 = 0.7 so it would suggest linear decrease, but amplitude
does not decrease linearly so after 8 further complete oscillations the amplitude would
not be 0.7cm
4.
Newton's third law is that that an action force has a equal and opposite reaction force.
Therefore the weight should create a equal reaction force opposite to it if the pair
an example of Newton's third law. However a part of the normal contact force
would originate from the centripetal force, which means the force from weight
is not equal in magnitude to the reaction force from it.
v = wr
w = 2[pi]/T = 2[pi]f = 2[pi](1600/60)
v = 2[pi](1600/60) *(0.5/2) = 41.888 = 42 ms^-1
100/1600 * 41.888 = 2.618 = 3
42
F = ma
F = mv^2/r
ma = mv^2/r
a = v^2/r
gravity = centripetal acceleration for zero force
9.81 = v^2/0.25
sqrt(9.81*0.25) = 1.6
1.6
3
5.