Transcript 2) i - Animated Science

Graphs of Motion
Physics – Equations of Uniform Motion 11.1.5
1) i
g = a = 10ms-2
80m
1) i
g = a = 10ms-2
80m
Simulation results
1) ii
u = 20ms-1
a = 10ms-2
80m
20
15
10
5
0
Note that since we have neglected air
resistance there are no decelerating
forces acting on VH as it is horizontal
to gravity.
Hence, velocity to constant in this
direction until it stops falling after 4
seconds
1
2
3
4
v
t=0 , v = 0
t=1 , v = 10
t=2 , v = 20
t=3 , v = 30
t=3 , v = 40
40
30
20
We can use the formulae v = u+at. This
simplifies to v = at (u=0). As gravity takes hold
on the ball its velocity increases.
10
If a = 10ms-2 we can simply say that v = 10t
0
1
2
3
4
h/m
We can use the formulae
s=ut+0.5at2.
80
This simplifies = s=0.5at2
But if a = 10ms-2 we can simply say that
s = 5t2
(NB s is actually s taken away from 80)
60
40
t=0 , s = 0
t=1 , s = 5
t=2 , s = 20
t=3 , s = 45
t=4 , s = 80
20
0
1
2
3
4
2) i
(ii) To work out the gradient or
velocity you take a tangential line
which fits the slope at that point i.e.
0.70s
(iii) We can explain the upwards
curve as each second the ball falls b
y more each second as the ball is
accelerating under the force of
gravity.
(iv) Grad is velocity of ball. To
estimate “g” from this experience we
can assume that ball started from
rest. Hence we can use our formulae
for uniformly accelerated motion;
Distance fallen (m)
s/ t = 5m / 0.71s = 7.04ms-1
v = u+at
(v – u)/t = a
(7.04ms-1 – 0 ) / 0.7s = 10.057ms-2
a = 10.1 ms-2
time in seconds (s)
2) ii
Newton's 3 Laws of Motion
Answer
I. Every object in a state of uniform
motion tends to remain in that state of
motion unless an external force is
applied to it.
The second law can be said to apply here as the
falling body is falling under constant acceleration
of 9.81Nkg-1 or 9.81ms-2 . This means that the
force applied by this acceleration is;
II. The relationship between an
object's mass m, its acceleration a,
and the applied force F is F = ma.
F = ma = constant of 9.81ms-1. This leads us to
the commonly used formulae of W = mg.
III. For every action there is an equal
and opposite reaction.
Hence, the force of weight acting on the body
does not change throughout the journey.
The third law also comes into play as there is an
opposite force of drag which is the reaction to the
acceleration which quickly increases as the body
falls until it is equal and opposite to W. At this
point the object must fall at a constant or terminal
speed.
3) a
5.0ms-1
20
2.5m
g = a = 10ms-2
Resolving horizontally
requires;
b) Resolving in a vertical direction can be
achieved simply using;
tan=opp/adj or adj x tan=opp
Hence;
s = opp = tan(20) x 2.5m
s = 0.909m
s = 0.91m
R = 5.0ms-1
adj = Rcos
= 5.0ms-1 x cos 
= 4.67ms-1
s/v = t = 2.5m / 4.67ms-1
= 0.54s