KINEMATICS
Displacement
Distance
length
i.
track
Ii
iii.
points
A
:
&
B
is
,
you
the
The
sign
,
>
<
!
D
6m
I
>
4in
6in
•
B
E
.
the
from
object
lmag
+
Point
-
East
origin
.
direction /
B
of
is
A
.
40m
.
Distance
Displ
A
10m
of displacement DOES NOT tell
motion
direction
of
tells about the
position of
points
+
•
of displacement
two
:
Im
top
I.
possible distance
Example
North
due
•
sign
Iii
50m
origin
the
Vector
Ii
distance between
-
→
between
lmagnitnde only )
Example
points
I. Shortest
arbitrary
between two
Scalar
,
of
any
!•
0A
10m
-110m
A
OB
16m
+4m
0C
20m
Om
OD
2dm
-8m
OE
34m
-2m
.
Speed
Rate
i.
Velocity
of change of
i.
Rate
distance
displacement
OR
012
travelled
distance
unit
ii.
IV.
SI Unit
>
s
→
<
-
s
-
|
:
per unit
time
.
.
d : distance
If
v
Displacement
per
Scalar
iii.
+V
-
time
of change of
Ms
"
iv
+s
Vector
iii.
v
iv.
%
+ s
ii.
-
-
I
t
s :
SI Unit :
Ms
displ
.
"
-
→
velocity
The
sign
the
direction
of
of
tells about
motion !
Average
speed
=
total
distance
total
time
✓
✓
+
•
⇐
-
+
•
Average
velocity
=
total
aispl
total
time
•
.
^
V
-1
y
An
object moving
instantaneous
velocity
is
in
velocities
zero
as
a
circle
but
total
•
7
the
displ
•
was
average
.
is zero
✓
.
-
I
f- Is
t
-
-
Is
>>
ta
→
U -4ms-1
-
9=10-1-4
=
-
V :
10ms "
+
6m52
u
a<
-
-
8m51
U=
a=v¥
Is
-
t
4m51
U= -10ms
'
h=
a- Lu
t
t -20ms
-
a=6-f#
20×10-3
>
+6ms
Is
-
10m51
a=v÷
-1
<
v.
-
<
v= -4ms
4m52
Acceleration
-
→
<
-
-5ms "
=
+ a
1
__
3- 8
3ms "
Deceleration
9=-8-1-4 )
a
v=
g-
t
t
dms
-1
a-
a
v=
-
a=v-U_
Acceleration
a
s
>
-
'
a -_
800.ms -2
a=
-4-1-10 )=+6ms"
1
Acceleration
Rate
a-
of change of
V
-
SI Unit
u
velocity
:
Ms
respect
Vector
-2
t
Acceleration
occurs
Changes
2.
Changes
3.
Changes
its
1.
when
speed
both
V= 8m51
V¥
a- 4m52
so
.
object moving in
An
circle
accelerating
direction
a=
8-4
T
( acceleration)
10ms
a- VI
V -6ms
"
.
so
t
a--4ms
-2
-
-
a=6( deceleration )
<
✓ = -8ms
"
a=v-U_
t
a-
.
its
Is
<
>
"
as
keeps changing
direction
t
>
U=
continuously
is
-
>
with
velocity
time
to
a
speed and
-
t -1s
>
a-
o
direction
its
in
Quantity
object
f- Is
U= 4m51
change
OR
-2ms-2
U=
so
-
6m51
a= -8-1-67
1
( acceleration )
.
t
-
l
<
✓ = -4ms
t
!
"
u
-
10ms
a=V-u_
t
so
a=
-
-
Outs
A=V-¥
"
>
U= -10ms
-
"
4- C- 10 )
a= -160ms
0.1
-2
( acceleration )
1
a= -16ms -21 deceleration )
9=-6-1+101
SO
F- +6ms
"
A=
t
=
6- ( -107
0.1
=
160ms
-2
Equations
1.
2.
conditions
utat
v=
3.
s=
4.
V2
at
-
⇐
-
-
-
-
=✓-u-
%
s=
Vavgxt
to
""
"
9
•u
2.
straight line
straight line
U2 = Las
+
1g
at
'
s=(u¥)t
s=(u+uz+at)t
s=µu;at)t
ut
+
Izatt
v2
•
-
U2
Las
=
v=u+at
¥
s
"
s
:( u¥)t
:( u¥N¥ )
5=1%11%1
s
=
v2
-
u
2A
'
so
motion
analysis
Constant Acceleration
-2
v=u+at
-
a
for equations
-11gal
s= Ut
5-
-
-
t
It
5=1
•
Motion
of
v2 uZ= Las
-
f- at +
g-
at
'
e-
s=(u+u+_at)t
2
s=µ;at_It
s=|¥+a¥ )
s=
ut
+
f- Utat
Izatt
✗
2- u2=2as
1m¥ )t
V
-
u
-_
s=fu¥/ t
at
v¥=t
5 =
V2
-
e-
1%41%-1
U2
( Ta
lvtullv-ul-v2-uv2-UZ-2.ae
latblla b)
-
"
"
" "
"" " """
from rest from
"
"
""
height
^
+
of air
20m
Assuming
resistance
g¥
I
time
ii.
final
to
s= 20m
a
-_
velocity
ground
9.8ms "
t
as
Ii
,
u
=
-0ms
V2
v2
-
-
-
U'
-
-
=
-1
.
resistance
air
I
to
iii. time
??
I
u=
15m51
a=
-
9.8ms
,
height reached
Max
,
reach
iii. total time
=
of
determine
"
it
-2
max
of fight
✓ =D
s= ? ?
v ?u2=2as
?⃝
02-1157=21 -9.81s
5=11.5 in
Ii
s -20m
l
no
Assuming
"
height
2.02s
-
,
¥-9.8
.
tzat
20=0-11219.81-12
t
u -15ms
'
s= ut +
,
,
fall to ground
ground
,
,
, ,
the
hits
v
U= 0m51
negligible
determine
m
,
I
15ms
as
velocity
a
v. 0m51
.
"
vertically
upwards
with
a
ball
"" " " "
"
"
a =
9.8ms"
V=
?
U
v
=
=
15m51
u +
a =
at
0=15+1-9.8 )t
Zas
02=219.81120 )
,
V
-
-
19.8ms l
-
t
=
1.5s
-9.8ms
-2
V=0
1-
=
??
Iii
,
t= 1.5×2
1-
=
5=0,4
-
15m51 ,a= -9.8ms
;t=
?
U=
v=
3.0s
s=ut
+
fat
15ms
-
"
1-
15m51
=
??
'
15m51
0=151-+121-9.811-2
-15--15+1-9.8)t
0=151--4.91-2
-30=-9.8 t
0=1-115-4.91-7
t=Os
-9.8ms -2
a=
15-4.91--0
1-
=
3.0s
v=u+
ill
111
v
,
at
t -3.06s
-
"
-15ms "
Example # 3
A
.
man
launched
was
room
from the top
high cliff with a
a
a
1¥
velocity
*
vertically
t
"
" "
1µm /
'
"
'
of
20ms
upwards
resistance
calculate
a)
b)
20ms -1
u=
,
v
-
,
,
time
Iv
clifftop again he
9.8ms
-2
,
v =
0ms
t= ?
-1
,
U=
20m51
✓ 2- WZ
0
-
al
,
=
a =
-
=
9.8ms
-2
,
v =
0ms
-
s
I
=
?
,
2 as
20.4m
-
the
the
b) 100+20.4=120.4 m
life
his
.
the
crosses
faces
/ death
ground
most
flight
(205--21-9.8) s
s
Y,
of
-1=2.045
,
when he
before
speed
1
certain
wits
0=20+1-9.81 t
Ii
from ground
maybe
-
height
from clifftop
iii.
,
at
u +
-
a =
mat
.
ground
I
reach
ii. max weight reached
-1
Assuming negligible
air
I time to
thrilling
i. e.
a
i. e.
.
seconds
time
of
iii.
Method
tt
2.04×2
U= 20m51
4.08s
v=
=
t=
Method
a- 20ms-1
5=0
# 2
,
v.
-
a=
-9.8ms
1-
?
,
-
=
-2
Iv
# 3
Method # 1
,
20m51
a=
,
f-
uttlzat
=
-
l
s = -100m
✓ 2-
v2
=
✓ =
,
v
=
'
-
Lot
ret
way
.
.
9.8ms -2
a-
0
+
5=120 -4m
µ
? ?
=
v2 u2=2as
-
1- (4.9-1-20)=0
v2 -0=219.811120-41
t=O
V
-1=20
-4.9
1-
=
4.08s
-
"
-
✓
.
.
u -0ms
-
=
??
48.6ms"
0=201--4.91-2
4.9T
-2
Zas
=
t
t -4.08s
0=201-+121-9.811-2
uh
Another
-
a= -9.8ms
-12012=21-9.871-1001
-20=20+1-9.8 )t
-40
,
?
utat
-9.8
'
1-
-9.8ms
20m51
U=
,
=
48.6ms
-1
#
•*
.
-
.
¥-4m
%
Y
,
Method
20m51
u=
1-
=
?
Method
# 1
,
,
a= -9.8ms
s=
"
-100m
trise
=
tfall
=
# 2
u -0
-
¥20.4m
2.04s
?
Either
f-
uttlzat
'
s
uttlzat
-
'
OR
-100=201-+121-9.811-2
100=201--4.91-2
120.4×2=1-2
-1=4.965
fall
t= 4.96s
fall
-1=27.05
t-tn.se + 1- fall
=
E-
=
V=u + at
*
48.6=0 -19.8T
9.8
'
t ,= -2.19s
*
2
4.9T -201--100=0
height
.
120.4=0+119.811-2
-
Max
•
2.04+4.96
7.00s
I
Overtaking
during
Goods Train
>
Express
motion
10m51
v=
vjloms
-
_
"
At
÷
-
both bodies
must have
'
train
U=
0ms
-
i
=
"
-
-
stations
.
Goods Train
v=1
t
10=1
t
Express
f-
Train
ut +
Eat
SESE
'
5=0+110.5 )t
lot
0.25T
"
-25=0.251-2
=
0.25T
'
-
lot =D
t (0.251--10)=0
t -0s
-
same
displ
0.25T -10=0
'
,
travelled the
=
a
overtake
f. 40s
.
DISPLACEMENT
VELOCITY
GRAPHS
TIME
-
sM^
✓
-
TIME
1min
•
I
!
•
GRAPHS
>
>
•
Hs
Hs
✓
"
-4ms
-4m
GRADIENT : VELOCITY
m=D1=Ds_=v
DX
AREA :
A
"
:{
At
MEANINGLESS
bh
so
A-
-
tzltxs )
GRADIENT
:
ACCELERATION
m=D§n=¥÷=
AREA
A-
-
:
tbh
a
DISPLACEMENT
so
A :
Izltxvl
↳
S
"
VARIABLE
GRADIENT
CONSTANT
LINE
"
"
GRADIENT
CURVE
"
>
^
T
m=+inc
M=+dEC
M= -1
M=O
m=
-
m
,
m=N
•
If graph
gradient
•
is
If graph goes
gradient
upwards
positive
goes
is
downwards
negative
.
its
its
m=
-
incr
=
-
dec
>
If curve gets
vertical
→
gradient
increases
If
curve
gets
horizontal →
gradient
decreases
DISPLACEMENT
-
TIME
GRAPHS
?⃝
②
^
a
!
4m
4m
0
>
Hs
Hs
-4m
-4m
at the
object
origin
.
:
Zero
is
at
GRADIENT / VELOCITY
rest
MOTION :
rest
origin
s
Omfs
"
"
MOTION :
Omls
>
-
GRADIENT / VELOCITY
+5
;
or
\
zero
object is at
ahead
.
:
i.
behind
DISPLACEMENT
③
-
TIME
GRAPHS
①
"
"
①
①
4m
4m
0
>
>
Hs
Hs
②
-4m
"
"
GRADIENT / VELOCITY
① constant l l
GRADIENT / VELOCITY
① constant 1+1
:
②
constant
MOTION :
at
②
-4m
-
-
②
l
Object is moving
constant
away
l
velocity
from origin
:
.
constant 1+1
MOTION :
at
Object
constant
towards
origin
is
moving
velocity
.
DISPLACEMENT
⑤
TIME
GRAPHS
⑥
①
a
-
a
①
4m
4m
ACCELERATION
-4m
GRADIENT / VELOCITY
①
>
>
Hs
Hs
-4m
②
4
GRADIENT / VELOCITY
① decreasing 1+1
:
t
MOTION :
away
:
② decreasing l l
l
-
accelerating
the
Object
from
-
②
y
increasing ( )
② increasing l
DECELERATION
is
origin
.
MOTION :
away
Object
is
decelerating
from the origin
.
DISPLACEMENT
TIME
GRAPHS
?⃝
⑦
-
1'
①
4m
?⃝
am
①
>
>
tls
tls
②
-4m
-4m
"
"
GRADIENT / VELOCITY
:
①
C- I
decreasing Ltl
② decreasing
Object is moving
decreasing velocity
GRADIENT / VELOCITY
① increasing c- I
② increasing 1+1
MOTION :
MOTION :
with
with
towards
origin
.
:
object
is
moving
increasing velocity
towards
the
origin
.
Example # I
5=0
✓
4m
released
rest
"
4m
"
←
from
if
It
-
>
+
Hs
1111
-4mi
,
is
s
>
-
Hs
-4m
"
=
←
if
Iv -1
"
"
+
-
⇐o
-
Vt
Example # 2
tls
SIM
A
-
←
>
grad / veto
fall
be
←
¥
9rad / veto
stall
←
Ttt
grad / veto
-
-
1- It
v
5=0
O
>
y
1st collision
t
'
tis
2nd collision
-51mV
"
Example # 3
ii.
I
1^-1
1
i
,
'
,
←
-
"
0
am
Crosses origin
>
•
;
←
is
-
-
-
-
-
-
-
-
-
-
-
-
-
-
.
speeding up
below
v
ground
-
Hs
iii.
it
grad / Vel
!
u
5=0^1
s1m^
a
-51mV
origin
O
'
tis
if
ground
as
origin
was
taken
VELOCITY
"
quoi
TIME
GRAPHS
1-
"
yms
0
>
Its
tis
"
am
-
"
mm
"
GRADIENT / ACCELERATION
:
"
GRADIENT / ACCELERATION
ZERO
ZERO
MOTION :
MOTION :
REST
OBJECT
IS
CONSTANT
I
MOVING
AT
VELOCITY
.
:
VELOCITY
"
qwi
①
•
-
TIME
sign of
velocity
GRAPHS
"
"
tells
4ms
①
about the
direction
Its
motion
of
Its
.
②
②
"
Ams
"
Nms
"
GRADIENT / ACCELERATION
①
CONSTANT
②
CONSTANT l
acceleration
GRADIENT / ACCELERATION
:
(t)
-
,
19K ve
not deceleration )
l
-
①
CONSTANT
l
②
CONSTANT
(t)
IS
CONSTANT
Onys
-
:
l
MOTION :
MOTION :
OBJECT
"
MOVING
OBJECT
WITH
ACCELERATION
2m15
Amps
IS
CONSTANT
MOVING
DECELERATION
.
6mA
8m15
8m15
→
tonys
WITH
→
Amp
→
.
2mA → Omfs
VELOCITY
"
quoi
①
Ams
①
"
①
>
>
Hs
Hs
"
Ams
GRADIENT / ACCELERATION
④ INCREASING
GRAPHS
yms
"
INCREASING
TIME
"
④
"
-
.
"
GRADIENT / ACCELERATION
:
( t)
①
DECREASING
1+7
l
②
DECREASING
I -7
-
l
MOTION :
MOTION :
MOVING WITH
INCREASING ACCELERATION
OBJECT
OBJECT
Omls
②
MOVING WITH
DECREASING ACCELERATION
IS
1- m/s -73m / s
6m15
:
.
lomls
Omfs
IS
5m15 -49mA
12mA
.
14mA
VELOCITY
"
quoi
②
GRAPHS
"
yms
①
①
>
>
Hs
Hs
②
②
"
mm
"
GRADIENT / ACCELERATION
①
TIME
"
"
am
-
GRADIENT / ACCELERATION
① DECREASING l l
:
INCREASING l l
INCREASING I -11
②
OBJECT
IS
MOVING WITH
INCREASING DECELERATION
9ms
DECREASING ( t )
MOTION :
MOTION :
longs
:
-
-
OBJECT
"
7-
m/s
.
4m15
IS
DECREASING
15mA
MOVING
WITH
DECELERATION
10m11 -46m /s
3m15
.
"
Hms
A
case # I
-
tv
.
NO AIR RESISTANCE
-
-
ii.
+
: :
T
:
✓ T
:
←
t
g
t
dist
m
-
-
a
-
la g I
g
-
-
-
.
rise
O_0
o
r
i
dist
t
'
TIS
i
.
fall !
-
-
-
v
!
-
Vlmsj
i
4M$
Case # 2
NO AIR RESISTANCE
@
const
'
"
5mls
i :
i
.
-
it
1- T
7-
i
-
i
-
-
max
.
'
deist
fall
i
i
i
,
r
cliff top
t
hits
d
i
i
.
i
,
ground
'
tis
,
i
in distance
i
:
i
i
ifmi
-
acceleration
height
k
rise
O
gravitational
-
at
when
dist
,
vi.
ace
from cliff
to ground
←
gradient
tell
ace
.
Hms
•
Case # 3
-
+
.
rest
rest
t
µ
t
t
hit the
ground
v
-
AIR RESISTANCE
No
w
i'
'
'
rims
"
l
l
- _
Max
V
t
height
→
•
'
with
n
t
'
& rebounds
no
energy loss
collisions
.
End
start
-
"
Hs
7
'
'
'
n
v
-
-
V
ym5
+
5
-4ms
9
'
'
leaves The
^
-
"
o
-
v
'
Ins
ground
of
i
4
-
i
i
°
2
,
:
-
-4
-
iris
i
z
't
-4ms
i
t
i
'
-
ok
:
-
4ms l
a
'
q
Mat
height
'
1
th
" "
T
t
.
5ms
-
2
3
.
.
'
I
ill
i
TI
energy
loss
ex collision
during collision
time
tf
.
PROJECTILE
A
motion
2D
MOTION
the
under
effect of gravity
"
→
-
Assumptions
No
1.
air
.
resistance
No
external
b)
No
loss
'
Vy
force
'
.
a
of energy
Acceleration
vertical
"
^
•
a
V' "
'•
>
Vx
•
>
.
^
and
0
v×|
ace
>
.
.
Vx
✓ Vx will
stay constant
throughout motion as
no
is
there
force along
acceleration
.
direction
^
grav
y
Y•=0 vx
t
vy
✓
heat
as
constant
is
gravitational
to
the
in
.
gravitation acceleration
No
'
a)
equal
motion
.
•
2.
projectile
called
is
X-axis to
>
>
-
•
>
vyv
vx
Hmm,
✓
-
Vy
falls
Vx
,
v
•
Vx
>
Vx
-
Vy
✓
decreases
and
rises
>
•
Vy
I ✓
change
.
as
it
.
object
increases
as
it
^
T
vy
^^
h
✓
V
Vx
Vy n
•
.
Vy
70
>
b
.
.
.
+
Equations
:
•
!
We
'
•
-
VYZ + ✓
2
✗
applied on
They can only work
breakdown
x-axis
=
velocity
&
constant acceleration
I
t
-
sin 0=1
by
h
sin D=
Vy
T
T
v×=vcosO
vy=VsinO
straight line
analysis along
on
the
y-axis
-
!!
Y-axis
axis
constant
Vx
WSO
motion cannot
be
of the
curved track !
4=07
✗
'
Tv
"
•
+b
tan D= Vy
.
•
=
'
cos 0=4
Vx
!•
.
v2
=p
P
g-
÷:*
.
hi
•
Time
links
°
Time
x-axis
found from and
vice
used
Vu =
,
Uyt
sy= ut
Y
}
v
-
u
+
at
Eat
:-. Zasy
'
both
x-axis
&
y-axis
can
in
y-axis
be
versa
.
Type #
Type # 1
Type # 3
2
•
>
Uy= 0
>
V
=
Vx
>
Vy
•
projected horizontally
U
,
a
> Vx
0
"
'
•
Uy
=
Vx
.
>
Vy
>
"
P Ema,
•
✓
T
:O
Type # 4
>
✗
•
"
✓×
•
i
•
.
=
MGH
>
Vx
height
Max
vy=0
K
.
Emin
-_
Iz MVT
Vx
vy
Ivy
V1
^
•
✓
>
•
Vx
•
•
K E Max
-
P .E
=
0
=
Lzmv
'
•
•
K E Max
-
P .E
=
0
=
Lzmv
'
7 of 65
4
Answer all the questions in the spaces provided.
1
For
Examiner’s
Use
(a) Distinguish between scalar quantities and vector quantities.
Scalar have
..........................................................................................................................................
magnitude only
and direction
Vectors have both
..........................................................................................................................................
magnitude
.
..................................................................................................................................... [2]
(b) In the following list, underline all the scalar quantities.
acceleration
force
kinetic energy
mass
power
weight
[1]
(c) A stone is thrown with a horizontal velocity of 20 m s–1 from the top of a cliff 15 m high.
The path of the stone is shown in Fig. 1.1.
ii.i
20 m s–1
find
horizontal
cliff
15 m
>ms
-
t
l
20 =
ground
1.75
✓
>
speed
Fig. 1.1
S×= 35m
Air resistance is negligible.
For this stone,
calculate the time to fall 15 m,
'
utttgat ?
15=0+1219.811
5- 15m
s=
-
Uy
=
a
-_
1-
=
0
9.8ms -2
?
1-
=
1.75s
1. 75
time = ..............................................
s [2]
(ii)
calculate the magnitude of the resultant velocity after falling 15 m,
4--0
a-
-
9.8ms
V2
-
✓
=
?
17 / mfs
.
-
s -15m
20m$
✓ 2- u2= 2 as
-2
0=219.81115)
✓ =
17.1ms
-
'
u
>
p2-ibh-flh.li#- 26.3M5
h'
=
'
26.3
resultant velocity = ........................................ m s–1 [3]
© UCLES 2011
9702/22/M/J/11
displ
V×=£
20
•
(i)
the
.
20 of 65
6
2
(a) Explain what is meant by a scalar quantity and by a vector quantity.
they
have
magnitude
only
scalar: .......................................................................................................................................
.
...................................................................................................................................................
They
have both
magnitude and direction
vector: .......................................................................................................................................
.
...................................................................................................................................................
[2]
(b) A ball leaves point P at the top of a cliff with a horizontal velocity of 15 m s–1, as shown in
Fig. 2.1.
Uy=O
ball
P
25 m
=
15 m s–1
path of ball
tg-9.cm/s2cliff
Q
ground
Fig. 2.1
try
The height of the cliff is 25 m. The ball hits the ground at point Q.
Air resistance is negligible.
(i)
Calculate the vertical velocity of the ball just before it makes impact with the ground at Q.
y-axis
v2
-
Uy
=
0
a- +9.8ms
V2
-2
-
-
S
Vy
=
=
-
25m
??
uZ=2as
02--219.811251
V=
22.1
22.1
vertical velocity = ................................................. m s–1 [2]
(ii)
Show that the time taken for the ball to fall to the ground is 2.3 s.
V=u + at
22.1
=
0
+
( 9. 8) t
t -2.3
-
[1]
© UCLES 2014
9702/23/M/J/14
21 of 65
7
(iii)
Calculate the magnitude of the displacement of the ball at point Q from point P.
P
Iasi
V =L
h
t
25m
=p
2. 255
'
h
"
-1
Sx 33.8m
-
-
"
b.
=p 2+62
hZ= (255+133.8)
h
15=-4
'
"
33.8m
=
42m
Q
n
42
displacement = ...................................................... m [4]
(iv)
Explain why the distance travelled by the ball is different from the magnitude of the
displacement of the ball.
a
curried
the length
was
Distance travelled
...........................................................................................................................................
of
Q
...........................................................................................................................................
straight line from P
Displ
-
track
.
.
was
a
to
.......................................................................................................................................[2]
© UCLES 2014
9702/23/M/J/14
[Turn over
'
31 of 65
5
A ball is thrown from a point P with an initial velocity u of 12 m s–1 at 50° to the horizontal, as
illustrated in Fig. 2.1.
b) Max height
"£
v=U+at
reached
'
path of ball
t
5.
Q 4=0
2
°
•
•
☐
utttgat
V2 U'
-
1-
=
→
c)
^
Las
X =12 m s–1
Sirin
128in 500
"
P
50°
dis.pt
.
Pto Q
horizontal
12005>500
Fig. 2.1
The ball reaches maximum height at Q.
Air resistance is negligible.
(a) Calculate
(i)
the horizontal component of u,
12m50
7.7
horizontal component = .................................................
m s–1 [1]
(ii)
the vertical component of u.
lLsin50= 9. Zmfs
vertical component = .................................................
m s–1 [1]
(b) Show that the maximum height reached by the ball is 4.3 m.
Uy= 9.2ms
0ms
Vy
=
"
"
v2 n' =2as
-
02-19.211=21-9.8/5
a= -9.8ms"
s -4.3m
-
s= ?
[2]
(c) Determine the magnitude of the displacement PQ.
Q
h
P
7.2m
4.3m
v=¥
in :p ' -15
77=50%4
Ñ=( 4.314-17.25
h
-
iv-u-atsx-7.IM
8. 4m
0=9.2+1 -9.81T
t -0.94s
-
displacement = ...................................................... m [4]
© UCLES 2016
9702/21/M/J/16
[Total: 8]
[Turn over
45 of 65
7
3
A ball is thrown against a vertical wall. The path of the ball is shown in Fig. 3.1.
Vy
kg
:O
P
fit
For
Examiner’s
Use
>
-1
15.0 m s–1
wall
60.0°
S
F
6.15 m
9.95 m
Fig. 3.1 (not to scale)
The ball is thrown from S with an initial velocity of 15.0 m s–1 at 60.0° to the horizontal.
Assume that air resistance is negligible.
(a) For the ball at S, calculate
(i)
its horizontal component of velocity,
15hrs60
7.5mA
horizontal component of velocity = ........................................ m s–1 [1]
(ii)
-_
its vertical component of velocity.
158in 60--12.9=13 mfs
vertical component of velocity = ........................................ m s–1 [1]
(b) The horizontal distance from S to the wall is 9.95 m. The ball hits the wall at P with a
velocity that is at right angles to the wall. The ball rebounds to a point F that is 6.15 m
from the wall.
Using your answers in (a),
(i)
calculate the vertical height gained by the ball when it travels from S to P,
Uy
-
-
Vy=
13m15
v2 uh
0
02-132=21-9.815
a= -9.8Mt
s =
?
© UCLES 2011
-
-
-
Las
S -8.6m
height = ............................................. m [1]
-
9702/21/O/N/11
[Turn over
46 of 65
8
(ii)
show that the time taken for the ball to travel from S to P is 1.33 s,
v
-
¥
7. 5-
-
909¥
[1]
1.33
show that the velocity of the ball immediately after rebounding from the wall is about
4.6 m s–1.
t
(iii)
For
Examiner’s
Use
=
✓ =
✓
=
¥
✓
=
4. tends
same time to rise
60¥
,
& fan
[1]
(c) The mass of the ball is 60 × 10–3 kg.
(i)
Calculate the change in momentum of the ball as it rebounds from the wall.
change in momentum = ........................................... N s [2]
(ii)
State and explain whether the collision is elastic or inelastic.
..................................................................................................................................
..................................................................................................................................
............................................................................................................................. [1]
© UCLES 2011
9702/21/O/N/11
U
→
u
40M
i
O
X
Y
-
-
s
*
40M
-
-
ILS
:
??
a :
? ?
U
Z
t
-
les
XZ
←
.
-
t
I
O
125
KY
=
40in
I
8
t
r
s=utt{
40
-_
UHH
at
'
°
s= 80in
t
+
Iza
( 121
40=12 U -172A
185
-
'
U
-
-
a=
-
5-
80
??
??
①
Uttlzaf
-_
cello
'
)tIza( 1872
80=18 u -1162A
"
U=
1.1ms
a -
0.37ms
-2
-
②
5- Ut +
My
5-
n
a-
uttlzat
-
fat
MI 's
-_
nth
U
2
h
,
-_0t{
-
O
'
,
2
uttlzat
-
=
ath
fat
,
'
2
fat
-
-
th
-
-
,
-
-
z
fate
tati tati
tatti til
-
-
2h_
( ti til
at ,
fat :
-
-
h
-
xth
'
'
Ottat
-
{
at
←
a
U
→
a
v
-
O
v
@
•
y
⇐
O
.
=
Las
U'
'
-
Clout
at
20in
=
'
'=xffI1s
l
U
Las
Method # 2
as
u
-
a
U2
-
V' U2 -_2as
R
.
✓ 2-
'
'
:# ÷
s
-
O
-
-
U2
=
2am
wa
-
2x
u
-10.24
f
1.4462=-10×5
i
.
mu
=
%
i
.
n
new
a
1.44
=
I
n
5=1.44 n
U2 x s
s
-
.
!
( 1. 2h)
=
I
-44121k
ut
S=
-
1.44N
-
n
O
2. 5ms
u=o
-
Z
→
10ms
10ms"
"
do
I
•
Om
4+8
100m
v
Ut
-
-
at
-
4S
-
t
10
?
=
I
✓ 2-
u2=2as
'
(O
-
S
-
-
const
speed
10
80
02--2 ( 2. 5) S
20M
125
(
S
-
10=0+2.5 t
-
=
race
O
v
-
t
for
total time
→
=
-
=
t
AECEELERATED
t =8s
I
•
dist
-
-
80mi