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Notes p.24
Falling Objects & Terminal Velocity
Although the acceleration due to gravity is 9.8 ms-2,
downwards, objects do not maintain this acceleration in air
because ________
________________
acts upwards.
air
resistance
IMPORTANT EXPLANATION of TERMINAL VELOCITY
air
resistance
As a falling object speeds up, the _____
_________
increases . This causes the unbalanced
acting upwards ________
decrease . Eventually the
force and acceleration to ________
object will reach a speed where the upwards force due to
air resistance
weight of the object.
___
________ balances the _______
constant
From this point, the object will maintain a ________
velocity .
velocity
______ . We call this ________
terminal ________
Velocity (ms-1)
Example
Sketch a velocity – time graph to illustrate the
motion of a skydiver from the moment she jumps
from the plane. Your graph should clearly indicate
the point at which she opens her parachute.
‘a’ = 0 … balanced forces
Terminal Velocity.
‘a’ decreases as air
resistance increases with
speed.
0
initially ‘a’ = 9.8 ms-2
Parachute opens
Velocity decreases as air
resistance bigger than weight.
But air resistance decreases
with speed, so rate of
deceleration decreases.
New “terminal velocity”
when forces balance again.
Time (s)
Notes p.25
Conservation of Energy
From Nat 5 you should recall the following key
energy equations:
E p = mgh
Ek = ½ mv2
Ep = potential energy (in J)
m = mass (in kg)
g = gravitational field strength
(in Nkg-1)
h = height (in m)
Ek = kinetic energy (in J)
v = velocity (in ms-1)
EW = Fd
E=Pt
Ew = work done (in J)
F = force (in N)
d = distance (in m)
E = energy transferred (in J)
P = power (in W)
t = time (in s)
Revision questions for Higher Physics, page 26,
Q. 1 – 14.
Worked Example
A 20 g dart is travelling at 6 ms-1 when it strikes
a dart board. The dartboard exerts an average
frictional force of 30 N on the dart.
Determine the dartboard thickness required to
bring the dart to rest.
1st Ek = ?
m = 0.02 kg
v = 6 ms-1
2nd
Ek = ½ mv2
= 0.5 x 0.02 x 62
= 0.36 J
Kinetic energy changes to work done against friction!
Ew = 0.36 J
F = 30 N
d= ?
d =
=
=
=
EW / F
0.36 / 30
0.012 m
1.2 cm
Revision questions for Higher Physics, page 26,
Q. 1 – 14.