Projectile motion
1) The horizontal and vertical components of motion are independent of
each other.
2) Air resistance is negligible.
3) The only force acting on the projectile is gravity.
ux
t=0
uy
ux
t=1
uy
t=3
ux
ux=u cos
uy=u sin
g=9.81
ux
uy
R
ux
t=4
u= ux
ux
ux
t=2
uy
ux
u
ux
t=5
hmax
can be used because a=g,
a is constant
Ep=mgh
Ek=1/2mv2
uy
Projectile motion
27
(c)
The experiment is repeated with the clay block placed at the edge of the table so that it
is fired away from the table. The initial speed of the clay block is 4.3 m s1 horizontally.
The table surface is 0.85 m above the ground.
SPEC/4/PHYSI/HP2/ENG/TZ0/XX
A5. This question is about throwing a stone from a cliff.
(Question B4, part 1 continued)
10
N10/4/PHYSI/HP2/ENG/TZ0/XX
table
clay block
Antonia stands at the edge of a vertical cliff and throws a stone vertically upwards.
v 8.0 ms
1
path
0.85 m
ground
(i)
(not to scale)
Ignoring air resistance, calculate the horizontal distance travelled by the clay block
before it strikes the ground.
[4]
sea
The stone leaves Antonias hand with a speed v 8.0 m s1. Ignore air resistance,
the acceleration of free fall g is 10 m s2 and all distance measurements are taken from
the point where the stone leaves Antonias hand.
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(ii) The diagram in (c) shows the path of the clay block neglecting air resistance.
On the diagram, draw the approximate shape of the path that the clay block will
take assuming that air resistance acts on the clay block.
[3]
(This question continues on the following page)
(a)
Determine,
(i)
(ii)
the maximum height reached by the stone.
the time taken by the stone to reach its maximum height.
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(b)
The time between the stone leaving Antonias hand and hitting the sea is 3.0 s.
Determine the height of the cliff.
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[3]
21
Projectile motion
20
M10/4/PHYSI/HP2/ENG/TZ1/XX+
(Question B2, part 2 continued)
(Question B2 continued)
Part 2
Projectile motion
A sphere is projected horizontally. The sphere is photographed at intervals of 0.10 s. The images
of the sphere are shown against a grid on the diagram. Air resistance is negligible.
horizontal distance / m
0
0.5
1.0
1.5
2.0
2.5
0.4
0.6
0.8
vertical distance / m
1.0
[3]
(a) Use data from the diagram to determine the acceleration of free fall.
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0.2
M10/4/PHYSI/HP2/ENG/TZ1/XX+
[5]
(b) Determine the speed of the sphere 1.2 s after release.
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1.2
(c) On the grid, draw the path of the sphere assuming air resistance is not negligible.
[2]
1.4
1.6
1.8
(This question continues on the following page)
2210-6508
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Projectile motion
15
B4. Part 1
(a)
12
N10/4/PHYSI/HP2/ENG/TZ0/XX/M
Part 2
Collisions
the total momentum of a system is constant;
provided external force does not act;
[2]
(a)
or
the momentum of an isolated/closed system;
is constant;
4
SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M
Award [1] for momentum before collision equals collision afterwards.
(b)
A4. (a)
(b)
(c)
3
initial1momentum
1 2.0 10 140 ;
3
P or V 2.0or 10
pressure
inversely proportional to volume etc.;
140
P
final Vspeed
;
2
5.6 10
2.0 10 3
s1
(ii) V 4.8
Tmetc.;
Watch for incorrect mass values in equation.
P1 P2
(i)
= ;
(ii) initial
energy of pellet clay block 12 mv 2 ;
T1 Tkinetic
0.5 0.058 4.82 ( 0.67 J) ;
V1 V2 work done
(ii) force
= ;
;
T T2 distance travelled
0.24 N;
or T / = P2T1 ;
from (i)
P1
V1T2
use of/ appropriate
kinematic equation with consistent sign usage e.g. a
from (ii) T =
;
2V2
4.8
a
; PV
PV
1 1
equate to get
= 2 2;
2 2.8
T1
T
0.058 4.82 2
FPV
;
2.8
so that
or PV = KT ;
= 2constant
T N;
0.24
A5. (a)
(i)
A4. (a)
(i)
(ii)
(ii)
(b)
(b)
[1]
[2]
[1]
(c)
[1]
[4]
u 2 v2
;
2s
[4]
4
SPEC/4/PHYSI/HP2/ENG/TZ0/XX/M
v2
;
2g
1
1
P
m ; or pressure inversely proportional to volume etc.;
[1]
to
give or
h =V3.2
[2]
V
P
0.80 s ;
V T etc.;
time to go from top of cliff to the sea = 3.0 1.6 = 1.4s ;
P1 P2
(i)
; s = ut + 1 at 2 with correct substitution, s = 8.0 1.4 + 5.0 (1.4) 2 ;
recognise=to use
2
T1 T
to give s = 21m ;
Candidates
find the speed with which the stone hits the sea from v = u + at ,
V1 Vmight
2
(ii)
; then use v 2 = u 2 + 2as .
(42 m sT1 )= and
T
P2T1
;
P1
VT
from (ii) T / = 1 2 ;
V2
PV
PV
equate to get 1 1 = 2 2 ;
from (i) T / =
Projectile motion
2s
;
t2
s 1.75 m, t 0.6 s ;
g 9.7 0.1 m s 2 ;
Award [2 max] if data not from last three data points.
g
[3]
horizontal speed 3.2 m s 1 ;
use of v gt or s 12 gt 2 ;
vertical speed 11.6 or 11.7 m s 1 ;
use of Pythagoras theorem;
speed 12 m s 1 ;
[5]
line always to left of spheres;
becoming more vertical;
[2]
[1]
h=
(c)
(b)
(i)
(i)
M10/4/PHYSI/HP2/ENG/TZ1/XX/M+
[1]
[1]
[1]
[3]
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