NAME______________________________
IB PHYSICS HL
REVIEW PACKET: KINEMATICS IN 1 & 2 DIMENSIONS
1.
This question is about projectile motion and the use of an energy argument to find the speed with which a thrown stone lands in
the sea.
Christina stands close to the edge of a vertical cliff and throws a stone. The diagram below (not drawn to scale) shows part of the
trajectory of the stone. Air resistance is negligible.
P
15 m s –1
O
Q
25 m
sea
Point P on the diagram is the highest point reached by the stone and point Q is at the same height above sea level as
point O.
(a)
At point P on the diagram above draw arrows to represent
(i) the acceleration of the stone (label this A).
(ii)
(1)
the velocity of the stone (label this V).
(1)
−1
The stone leaves Christina’s hand (point O) at a speed of 15 m s in the direction shown. Her hand is at a height of 25 m above
−2
sea level. The mass of the stone is 160 g. The acceleration due to gravity g = 10 m s .
(b)
(i)
Calculate the kinetic energy of the stone immediately after it leaves Christina’s hand.
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(1)
(ii)
State the value of the kinetic energy at point Q.
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(1)
(iii)
Calculate the loss in potential energy of the stone in falling from point Q to hitting the sea.
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(iv)
(1)
Determine the speed with which the stone hits the sea.
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(2)
1
2.
This question is about projectile motion.
A small steel ball is projected horizontally from the edge of a bench. Flash photographs of the ball are taken at 0.10 s intervals.
The resulting images are shown against a scale as in the diagram below.
0
0
20
distance / cm
40
60
80
100
20
40
60
distance / cm
80
100
120
140
(a)
Use the diagram to determine
(i)
the constant horizontal speed of the ball.
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...........................................................................................................................
(2)
(ii)
the acceleration of free fall.
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(2)
(b)
Mark on the diagram the position of the ball 0.50 s after projection.
In the space below, you should carry out any calculations so that you can accurately position the ball.
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(3)
(c)
A second ball is projected from the bench at the same speed as the original ball. The ball has small mass so that air
resistance cannot be neglected. Draw on the diagram the approximate shape of the path you would expect the ball to take.
(3)
(Total 10 marks)
2
3.
This question is about throwing a stone from a cliff.
Antonia stands at the edge of a vertical cliff and throws a stone vertically upwards.
v = 8.0ms –1
Sea
–1
The stone leaves Antonia’s hand with a speed v = 8.0ms .
–2
The acceleration of free fall g is 10 m s and all distance measurements are taken from the point where the stone leaves
Antonia’s hand.
(a)
Ignoring air resistance calculate
(i)
the maximum height reached by the stone.
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...........................................................................................................................
...........................................................................................................................
(2)
(ii)
the time taken by the stone to reach its maximum height.
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(1)
The time between the stone leaving Antonia’s hand and hitting the sea is 3.0 s.
(b)
Determine the height of the cliff.
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(3)
(Total 6 marks)
3
4.
This question is about projectile motion.
A marble is projected horizontally from the edge of a wall 1.8 m high with an initial speed V.
V
1.8 m
ground
A series of flash photographs are taken of the marble. The photographs are combined into a single photograph as shown below.
The images of the marble are superimposed on a grid that shows the horizontal distance x and vertical distance y travelled by the
marble.
The time interval between each image of the marble is 0.10 s.
0
0.50
x/m
1.0
1.5
2.0
0
–0.50
y/m
–1.0
–1.5
–2.0
(a)
On the images of the marble at x = 0.50 m and x = 1.0 m, draw arrows to represent the horizontal velocity VH and
vertical velocity VV.
(2)
(b)
On the photograph, draw a suitable line to determine the horizontal distance d from the base of the wall to the point
where the marble hits the ground. Explain your reasoning.
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(3)
(c)
Use data from the photograph to calculate a value of the acceleration of free fall.
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(3)
(Total 8 marks)
4
5.
This question is about linear motion.
A police car P is stationary by the side of a road. A car S, exceeding the speed limit, passes the police car P at a constant speed of
–1
–2
18 m s . The police car P sets off to catch car S just as car S passes the police car P. Car P accelerates at 4.5 m s for a time of
6.0 s and then continues at constant speed. Car P takes a time t seconds to draw level with car S.
(a)
(i)
State an expression, in terms of t, for the distance car S travels in t seconds.
...........................................................................................................................
(1)
(ii)
Calculate the distance travelled by the police car P during the first 6.0 seconds of its motion.
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...........................................................................................................................
(1)
(iii)
Calculate the speed of the police car P after it has completed its acceleration.
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(1)
(iv)
State an expression, in terms of t, for the distance travelled by the police car P during the time that it is travelling
at constant speed.
...........................................................................................................................
(1)
(b)
Using your answers to (a), determine the total time t taken for the police car P to draw level with car S.
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(2)
(Total 6 marks)
5
6.
This question is about circular motion.
A geo-stationary satellite is one that orbits the Earth in an equatorial plane in the same direction of rotation as that of the Earth
and with an orbital period of 24 hours. Since the period of rotation of the Earth is 24 hours, this means that the satellite is
stationary relative to a point on the Equator.
(a)
The diagram below shows a geostationary satellite in orbit about the Earth.
not to scale
satellite
Earth
On the diagram above, draw an arrow to show the direction of acceleration of the satellite.
(1)
(b)
State the name of the force causing the satellite’s acceleration.
...................................................................................................................................
(1)
(c)
7
The distance of the satellite from the centre of the Earth is 4.2  10 m. Calculate the acceleration of the satellite.
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(3)
(Total 5 marks)
6
7.
Linear motion
(a)
Define the term acceleration.
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...................................................................................................................................
(2)
(b)
An object has an initial speed u and an acceleration a. After time t, its speed is v and it has moved through a distance s.
The motion of the object may be summarized by the equations
v = u + at,
s=
(i)
1
2
v  u t.
State the assumption made in these equations about the acceleration a.
.........................................................................................................................
(1)
(ii)
Derive, using these equations, an expression for v in terms of u, s and a.
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(2)
(c)
The shutter speed of a camera is the time that the film is exposed to light. In order to determine the shutter speed of a
camera, a metal ball is held at rest at the zero mark of a vertical scale, as shown below. The ball is released. The shutter
of a camera is opened as the ball falls.
0 cm
scale
camera
196 cm
208 cm
The photograph of the ball shows that the shutter opened as the ball reached the 196 cm mark on the scale and closed as
–2
it reached the 208 cm mark. Air resistance is negligible and the acceleration of free fall is 9.81 m s .
(i)
Calculate the time for the ball to fall from rest to the 196 cm mark.
.........................................................................................................................
.........................................................................................................................
(2)
(ii)
Determine the time for which the shutter was open. That is, the time for the ball to fall from the 196 cm mark to
the 208 cm mark.
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(2)
7
8.
This question is about circular motion.
A stone is attached to an inextensible string. The stone is made to rotate at constant speed v in a horizontal circle. Diagram 1
below shows the stone in two positions A and B.
Diagram 1
Diagram 2
v
B
A
A
v
Diagram 2 above shows the velocity vector of the stone at point A.
(a)
On diagram 2, draw vectors to show the change in velocity v of the stone from point A to point B.
(3)
(b)
Use your completed diagram 2 to explain why a force, directed towards the centre of the circle, is necessary to cause
circular motion.
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(2)
(Total 5 marks)
8
9.
Linear motion
At a sports event, a skier descends a slope AB. At B there is a dip BC of width 12 m. The slope and dip are shown in the diagram
below. The vertical height of the slope is 41 m.
A
(not to scale)
slope
41m
C
B
D
1.8m
dip
12m
The graph below shows the variation with time t of the speed v down the slope of the skier.
25.0
20.0
15.0
v / ms
–1
10.0
5.0
0.0
0.0
1.0
2.0
3.0
4.0 5.0
t/s
6.0
7.0
8.0
The skier, of mass 72 kg, takes 8.0 s to ski, from rest, down the length AB of the slope.
(a)
Use the graph to
(i)
calculate the kinetic energy EK of the skier at point B.
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.........................................................................................................................
(2)
(ii)
determine the length of the slope.
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9
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(4)
(b)
(i)
Calculate the magnitude of the change EP in the gravitational potential energy of the skier between point A and
point B.
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.........................................................................................................................
(2)
(ii)
Use your anwers to (a)(i) and (b)(i) to determine the ratio
EP  EK  .
E P
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.........................................................................................................................
(2)
(iii)
Suggest what this ration represents.
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(1)
(c)
At point B of the slope, the skier leaves the ground. He “flies” across the dip and lands on the lower side at point D. The
lower side C of the dip is 1.8 m below the upper side B.
(i)
Calculate the time taken for an object to fall, from rest, through a vertical distance of 1.8 m. Assume negligible
air resistance.
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(2)
(ii)
The time calculated in (c)(i) is the time of flight of the skier across the dip. Determine the horizontal distance
travelled by the skier during this time, assuming that the skier has the constant speed at which he leaves the slope
at B.
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(2)
(Total 15 marks)
10
MARK SCHEME!
1.
(a)
V
P
A
(b)
(i)
acceleration;
(ii)
velocity;
(i)
KE =
1
2
1 max
1 max
2
mv = 0.08 × 225 = 18 J;
1 max
(ii)
18 J;
1 max
(iii)
loss in PE = mgh = 40 J;
1 max
(iv)
total KE = 58 J;
116
–1
= 27 m s ;
0.16
v=
2 max
[7]
2.
(a)
(i)
speed =
distance
72
or
;
time
0.40
–1;
= 180 cm s
2
Award [1 max] if time incorrect.
(ii)
s=
1
2
2
gt or 80 =
1
2
2
× g × 0.4 ;
–2
g = 10 m s ;
2
Award answer with no working [0].
If it is clear that same mistake as in (i) has been made for the timing, then award full
marks in (ii).
(b)
horizontal distance moved = 90 cm; (allow ecf from (a) (i))
vertical distance moved = 125 cm; (allow ecf from (a) (ii))
correct plot from candidate’s working;
3
Award full marks if the plot is correct but there is no working shown.
(c)
sketch: overall reasonable shape (smooth curve “below” given path);
horizontal distance moved always decreasing when compared to
given path; angle to vertical always greater than given path;
3
[10]
3.
(a)
(i)
(ii)
(b)
h=
v2
;
2g
to give h = 3.2 m;
2
0.80 s;
1
time to go from top of cliff to the sea = 3.0 – 1.6 = 1.4 s;
2
recognize to use s = ut + 12 at with correct substitution,
2
s = 8.0 × 1.4 + 5.0 × (1.4) ;
to give s = 21 m;
3
–
Answers might find the speed with which the stone hits the sea from v = u + at, (42 m s
1
2
2
) and then use v = u + 2as.
[6]
11
4.
(a)
0
0.50
x/m
1.0
1.5
2.0
0
–0.50
y/m
–1.0
–1.5
–2.0
Mark both together.
VH : horizontal arrows equal in length;
VV : two vertical arrows, the one at 1.0 m noticeably longer than the one at 0.5 m;
2
If arrows correct but wrong point(s) award [1].
(b)
(c)
curve that goes through all data points;
stops at y = 1.8m as this is the height of the wall;
from graph d = 1.5(± 0.1)m;
3
travels vertically 1.8m in 0.6s / 1.25m in 0.5s;
g=
2s
;
t2
–2
to give g = 10 (±1) m s ;
3
Award [2 max] for any time shorter than 0.5 s.
[8]
5.
(a)
(b)
(i)
18t;
1
(ii)
s=
(iii)
v = at = 6 × 4.5 = 27 m s
(iv)
27(t – 6);
1
2
2
× 4.5 × 6 = 81m;
–1
idea of (a) (i) = (a) (ii) + (a) (iv);
18t = 81 + 27(t – 6)
t = 9.0 s;
1
;
1
1
2
[6]
12
6.
(a)
towards the centre of Earth;
1
(b)
gravity / gravitational / centripetal;
1
(c)
4
T = 24  3600 = 8.64  10 s;
substitute into
a
4  2 R 4 2  4.2 10 7
;

T2
8.642 108
= 0.23ms
2
;
3
[5]
7.
Linear motion
(a)
(b)
change in velocity / rate of change of velocity;
per unit time / with time; (ratio idea essential to award this mark)
2
(i)
acceleration is constant / uniform;
1
(ii)
t
v  u  ;
2s
and t 
u  v 
a
clear working to obtain v2 = u2 + 2as;
(c)
(i)
1.96 =
1
2
2
 9.81  t ;
t = 0.632s;
(ii)
2
time to fall (1.96 + 0.12)m is 0.651s;
shutter open for 0.019s;
If the candidate gives a one significant digit answer treat it as an SD-1.
2
Award [0] if the candidate uses s = 12 at and s = 12cm.
2
2
[9]
18.
(a)
arrow drawn (from A) of about correct length;
arrow drawn (from A) at about correct angle;
vector v labelled clearly and in correct direction;
3
Award [1 max] if vectors are added and [1 max] if v is opposite to
correct direction.
(b)
v is directed towards the centre of the circle;
force necessary to cause change in velocity / v;
Response must clearly refer to diagram and be consistent with it.
2
[5]
13
9.
Linear motion
(a)
(i)
E K = 12 × 72 × 23 2 ;
4
= 1.9  10 J;
(ii)
(b)
(i)
2
uses area between the t-axis and the line;
correctly converts area  distance (one 1cm  1cm square  5.0m);
distance between 90 m and 105 m;
improved accuracy, distance between 95 m and 100 m;
Do not accept kinematic formulas. Distance can only be found
from area.
4
EP = 72  9.8  41;
4
= 2.9  10 J;
2
2
4
Accept 3.0  10 J for responses using g = 10ms .
(ii)
ratio =
(2.9 ×10 4 1.9 ×10 4 )
(2.9 ×10 4 ) ;
= 0.34;
2
Accept 0.37 for responses using g = 10ms
(iii)
(c)
(i)
(ii)
2
.
fraction of energy lost due to air resistance / friction between skis and
slope / work to push snow away from skis;
1
1.8 = 2 × 9.8 × t
t = 0.61s;
2
1
;
2
distance = 23  0.61;
= 14m;
2
[15]
14