Physics
Revision
Tutorial
1
Kinematics
Solutions
Rectilinear Motion:
1
A
Equal areas under the graph above and below axes since distance travelled upward
equ
als distance travelled downwards.
2
A
Since
y
t
2
, distance fallen after 2 s will be (4
㈩2㴠㠠st潲敹献B慬lwill扥8
th
storey
below the 10
th
storey. Hence ball will be at 2
nd
storey.
Projectile Motion:
7
Let
t
be time to fall through vert
ical distance of 180 m.
2
2
1
2
9.81
180 0
2
6.058 s
y
y u t gt
t
t
Horizontal distance travelled =
v
x
t
= 500
6.〵8㴠㌰〰m(to㈠献f.)
DISCUSSION QUESTIONS
Rectilinear Motion
9
(a)
Thinking distance
s
t
=
u t
r
since t he car will be t ravelling at a const ant spe
ed
u
during
the reaction time
t
r
. Taking reaction time to be constant, thinking distance is therefore
proportional to speed.
When the brakes are applied, the car experiences a negative acceleration
a
. Using
2 2
2
2,
0 2( )
b
v u as
u a s
Braking distance
2
2
b
u
s
a
B
r
aking distance
is proportional to the square of the
speed.
v
t
Constant accln
upward
Accln of free fall
JC2
Raffles Institution (Junior College)
2009
2
(b)
2 2
2 2
b
b
u u
s a
a s
(b)
(c)
Speed /
km h
1
Speed /
m s
1
Thinking distance* /
m
Braking distance** /
m
a /
m s
2
t
r
/ s
40
11.1
6.7
9.4
6
.57
0.603
50
13.9
8.3
14.6
6.61
0.597
60
16.7
10.0
21.0
6.61
0.600
70
19.4
11.7
28.6
6.61
0.603
80
22.2
13.3
37.4
6.59
0.600
Average =
6.6 (2 s.f.)
0.60 (2 s.f.)
(c)
To find reaction time, use
t
r
s
t
u
for the data given and f
ind average (see table above).
For
u =
1
90
25.0 m s
3.6
,
Overall distance
2 2
25.0
25.0 0.60 62.3 m
2 2 6.6
t b r
u
s s ut
a
(d)
(i)
When the road is wet,
a
decreases in magnitude
and hence
s
b
will
be
larger
.
Thinking distance remains unchanged.
(ii)
When the dri
ver is not fully alert, reaction time is longer. Thinking distance is
longer whilst the braking distance is unchanged.
(e)
Assuming that d
uring
the
reaction time, the car
moves with constant velocity down the
slope,
10.0 m
t
s
During b
raking,
acceleration down the slope
2
6.6 sin 6.6 9.81sin10 4.897 m s
o
g
2
2
60/3.6
28.36 m
2 2 4.897
b
u
s
a
O
verall stopping distance
= 10.
0
+ 28.36 = 38.
4
m
JC2
Raffles Institution (Junior College)
2009
3
10
TYS Pg 43 Q12 (N97/II/1)
JC2
Raffles Institution (Junior College)
2009
4
TYS Pg 45 Q17 (
N03/III/1)
11
(c)
(i)
Intercept on the
t

axis = 1.76
s
(ii)
Acceleration can be found from the gradient of the tangent to the curve
.
(iii)
The ball was projected
with an upward velocity. Hence, just after the ball leaves
the thrower’s hand, it experiences a downward drag force. This drag force is
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灲潪散t敤. T桥 tot慬 摯w湷慲搠for捥 i猠湯
w t桥 s畭uof t桥 w敩g桴 of t桥 b慬l 慮搠
t桥 摲d朠for捥c潮 t桥 扡ll. e敮捥ct桥 扡ll 數灥ri敮捥猠愠摯w湷慲搠慣捥l敲慴ion
l慲aer t桡渠t桥 a捣敬敲慴i潮 摵攠t漠杲慶ity.
(iv)
The acceleration is equal to
g
only when the ball is momentarily at rest at its
max
imum height. At that instant
when
t =
1.76 s
, the drag force is zero and the
only force acting on the ball is the gravitational force of the Earth on the ball
(v)
(d)
On its way up, the drag force and the weight of the ball both act to slow
down the ball.
Hence the ball is brought to rest in a shorter time
than
if there is no drag force. On the
way down, the drag force acts opposite in direction to the weight. The net accelerating
force is smaller than if no drag force is acting. The average
velocity of the ball on the
way down is smaller than that on its way up. Hence the ball will take a longer time to
cover the same distance downward.
(e)
(i)
2
2
1
2
1
54 (26)
2
0.16 kg
k
E mv
m
m
(ii)
On its way up, kinetic energy of the ball is converted to gr
avitational P.E. with
some kinetic energy being used to overcome drag force (dissipated as heat). On
the way down, the gravitational P.E. is now converted to kinetic energy of the
ball with some of the gravitational P.E. also being used to overcome the dra
g
force. The ball will return to its starting point with less kinetic energy than when it
leaves the thrower’s hand.
1.
76
9.8
20
t /
s
a
/
m s
2
JC2
Raffles Institution (Junior College)
2009
5
Projectile Motion
12
1
100cos60 50 m s
o
x
u
1
100sin60 86.6 m s
o
y
u
Let
t
be time to fall through vertical distance of
500 m.
2
2
1
2
9.81
500 86.6
2
4.58 s
y
y u t gt
t t
t
Horizontal distance travelled =
v
x
t
= 50
4.㔸㴠㈲㤠9(3献f.)
13
2 2
2
1
Using 2
0 2( 9.81)(4.2)
2 9.81 4.2) 9.078 m s
y y
y
y
v u as
u
u
Using
0 9.078 9.81
0.925 s
y y
v u at
t
t
Horizontal speed
1
5.5
5.94 m s
0.9253
x
x
v
t
14
TYS Pg 46 Q19
(N2000/II/2)
(i)
1.
( cos )
cos
d v t
d vt
2.
Time taken to reach max height is
½
t
. At max height,
v
y
= 0.
Using
v
y
=
u
y
+
a
y
t
0 =
v
sin
g
(
½
t
)
2 sin
v
t
g
(ii)
2
2
o
sin2
32 sin2
94
9.81
32 or 58 (2 s.f.)
o
v
d
g
Note to Teachers: TYS gives answer as 32
o
but Cambr
idge Examiners accep
t
both.
60
o
100 m s
1
5.5
m
4.2
m
JC2
Raffles Institution (Junior College)
2009
6
(iii)
In practice, due to air resistance, the horizontal
range would become smaller
.
Based on
2
sin2
v
d
g
, the horizontal range increases as
increases up to a
maximum when
=
45
o
, i.e. sin 2
= 1. Beyond
=
45
o
, the horizontal range
decreases as
increases.
If
is originally 32
o
, the angle of projection must be increased
.
If
is originally 58
o
, the angle o
f projection must be decreased
.
Note to Teachers: TYS gives the wrong answer. Cambridge exami
ners only
mentioned greater than 32
o
.
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