School Science Lessons
14. Kinematics, vectors, describing motion path, displacement, distance,
velocity, acceleration, g
2011-12-26
Please send comments to: J.Elfick@uq.edu.au Table of contents 14.0.0 Kinematics, motion 14.1.0.1 Equations of uniformly accelerated motion,
gravity 14.2.0 Acceleration, falling 14.3.0 Measuring g 14.1.01 Scalars and vectors 14.1.0.0 Velocity 14.2.0 Acceleration, falling 14.2.12 Acceleration down an incline 14.3.0 Acceleration due to gravity, g, Measuring,
g 4.148 Acceleration of marbles down an incline 4.147 Ball bearings fall together 4.108 Ball projected upwards from a cart 14.2.6 Coin and feather, penny and feather, guinea
and feather 6.11 Coins on a slope (Primary) 14.2.5 Coins fall together 15.1.3 Coupled pendulums 14.2.16 Gun and tunnel 14.2.1 Drop strings with attached weights 37.35 Dust in the air 14.2.7 Falling heavy and light balls 16.2.0.1 Falling object 4.154 Falling washers on a string 36.108B Gravitational field of the
Earth, g 36.109 Gravitational potential energy 8.3.0 Gravity, gravitational field 4.152 Paths of projectiles, free fall 14.2.15 Parabolic trajectory of thrown chalk 14.2.13 Path of a projectile, water from a hose,
flight of a shuttlecock, monkey and hunter 14.2.18 Range of a gun 14.2.4 Simultaneous fall, ball bearings falling
together 14.3.10 Swing bucket in vertical circle, swinging
pail 4.153 Three holes can, 3-hole can, a vase with three
holes, spouting cylinder 3.22 Throw up and fall down (Primary) 4.151 Time a falling body with a stopwatch 14.2.3 Time a falling body with an electric "ticker
timer" (recording timer) 14.3.0 Measuring g
14.3.0 Measuring g 14.3.1 Free fall timer 14.3.2 Dropping balls 14.3.3 Ink jet marker 14.3.4 Falling drops 14.3.5 Catch a metre stick, reaction time of a person 14.3.6 Catch a coin 14.3.7 Rotating turntable 14.3.8 Pendulum-timed free fall 14.3.9 Many bounce method 14.1.01 Scalars and vectors
14.1.01 Scalars and vectors 14.1.02 Vector addition, parallelogram law, vector
diagram 14.1.03 Vector resolution 14.1.04 Addition of vectors when P and Q are at right angles 14.1.05 Polygon of vectors 14.1.0 Velocity
14.1.0.0 Velocity 14.1.0.1 Velocity, Equations of uniformly accelerated motion,
gravity 15.4.2 Velocity, position, and acceleration 4.16 Air resistance (Primary) 14.1.5.1 Examine a table of velocities ranging
from continental drift to the speed of light. 1.41 Falling parachutes (Primary) 14.1.4 Muzzle velocity, bullet timers 14.1.2 Photograph uniform motion 4.1.2 Qualitative analysis of graphs,
distance / time graph 14.1.3 Terminal velocity 14.1.5 Time of flight 14.1.1 Toy car running on moving paper 4.147 Ball bearings fall together See diagram 8.234: Ball bearings fall together
Use two clothes pegs, a pair of ball bearings and a wide rubber band.
Fix the band lengthways around one peg. Then open the peg and force a ball
against the tension of the rubber band between the prongs of the peg. Grip
the other ball with the second peg. Hold the pegs side by side, pointing
away horizontally above the floor. Squeeze both pegs at once. At the same
moment, one ball begins to fall vertically, and the other is shot forwards.
Note what happens by looking and listening very carefully. Repeat the experiment
from different heights and with a tighter rubber band. If the experiment
is done correctly, while the ball bearings land in different places they
strike the ground simultaneously. 4.148 Acceleration of marbles
down an incline
Use a 3 m plank of wood with a groove down the centre. Incline the plank
so that marbles can roll down the groove. Arrange small tin flags hung from
wires so that the marbles hit them and make "clinks" sounds. Put the flags
at regular intervals, e.g. 25, 50, 75, 100 cm, from the end of the plank.
Roll a marble down the groove and listen to the time intervals between "clink"
sounds. The time intervals between the "clinks" reduce as the ball rolls
down the incline.
Arrange the flags so that the clinks occur at equal intervals of time.
Measure the distance between the flags. The distance between the flags increases
down the incline in the ratio 1:3:5:7:9.
4.151
Time a falling body with a stopwatch See diagram 8.151.1: Throw up and fall down
| See diagram 8.151.2: Graph of throw up and fall
down
1. Time a falling body. Throw a ball as high as you can. Use a stopwatch
to measure
1.1 from when it leaves your hand to when the ball reaches the greatest
height and stops rising,
1.2 from when the ball starts to fall and reaches the height of your hand.
Time up equals time down. 4.152
Paths of projectiles, free fall See diagram 8.239: Path of a projectile | See 2.0.5: Conic sections | See
2.0.6: Parabola equation
1. The apparatus is used to show that the vertical and horizontal velocities
of a projectile are independent. The projectile is a metal ball, e.g. a
ball bearing. The target is a metal drink-can, suspended by an electromagnet.
The circuit includes two wires attached to a copper gate at the entrance
to the cardboard tube. When the circuit is closed, the drink-can is kept
in position by the electromagnet. Sight along path p1 then blows the ball
up the cardboard tube. The ball hits the copper gate to open the circuit
and let the drink-can fall. The ball travels through path p2 and hits the
drink-can target in mid-air.
2. The following experiment can be applied to different projectiles, e.g.
golf balls, cannon balls, darts, discus, shot put, slingshot, catapult and
long jumper. If a projectile has initial horizontal velocity before it starts
falling, its trajectory is a parabola, e.g. ball rolling off a table. The
only force on a projectile is gravity.
Throw a ball vertically as high as you can. Note the time between when
the ball leaves your hand and the ball stops rising. Also, note the time between
when the ball stops rising and the ball descends to the height of your hand.
The times are the same. Throw the ball up at different angles to the horizontal
and note the times taken for rising and descent.
3. Hold ball 1 just over the edge of the table so that it can hit the
floor when you drop it. Put an identical ball 2 in the middle of the table
and use a stick to push it steadily towards the edge of the table. When ball
2 passes the edge of the table immediately drop ball 1. Ball 2 has original
horizontal velocity but ball 1 has no horizontal velocity. However both balls
fall and land on the floor simultaneously because the acceleration because
of gravity is the same, whatever their state of motion. Both balls are projectiles.
4. A body in free fall descends height, h = 1/2 gt2, where
t = time and g = acceleration because of gravity, 9.8 m / sec2. The
time interval for both balls to reach the floor, t = √ (2h / g). If the
table is 1 m high and velocity of ball B is 5 m / sec, the time of fall,
t = √ (2 X 1) / 98 = 0.45 sec. The distance ball B travelled before reaching
the floor, d = vt = 5 X 0.45 = 4.25 m.
Find your "hang time", i.e. the time you are off the ground when you make
a vertical jump. Hold a piece of adhesive tape between your thumb and finger,
jump up next to a wall, and leave the sticky tape on the wall at the top
of your jump. Use the formula: t = √ (2h / g), where h is the distance
from the sticky tape to the floor. How high can you jump? An American basketball
player can do a 1.25 metres vertical jump!
5. Use a bow and arrow by pulling the bowstring back a certain distance
and pointing the arrow at a certain angle to the horizontal before releasing
the bowstring. Note the maximum height of the arrow and the distance it
travels before hitting the ground. Repeat the experiment by pulling the
bowstring back the same distance but change the angle to the horizontal.
Find the angle to the horizontal where the arrow attains the greatest height
and longest distance before hitting the ground. 4.153 Three holes can, 3-hole
can, a vase with three holes, spouting cylinder, Torricelli's tank
Direct relationship of pressure and depth. See diagram 8.153: Three hole can
1. Use a 2 litre plastic drink bottle or a vertical acrylic tube with
3 holes 6 cm vertically apart and covered with adhesive tape to seal the
three holes. Fill the bottle with water and close tightly. Make the holes
with a heated knitting needle. Or punch three smooth identical holes in
the side of a plastic drink bottle, or in a 1.5 m length and 12 cm diameter
water down pipe, at 1 / 4, 1 / 2 and 3 / 4 of the height, but offset so that
the streams of water do not interfere with each other. Plug the holes then
fill the bottle with water. Put the bottle on a table with a sink draining
top. Attach a tube to a tap to keep a constant head of water when you remove
the plugs. Remove the plugs. Note the speed the water through the three holes.
The water streams follow parabolic paths towards the sink. Feel the water
with your finger as it comes out of the hole. The fastest water stream is
through the lowest hole. Note how much water passes through each hole in the
same period. Note the path of the water streams. Draw a diagram of the three
water streams showing the distances travelled by each stream to the table
top. Diagram 5.1.1a is incorrect, although it occurs in some textbooks. Diagram
5.1.1b is correct. Water from the middle hole hits the table at the greatest
distance from the bottle, d2. Water from the bottom and top holes both hit
the table at the same lesser distance from the bottle, d1. The greater the
depth, the greater the pressure. Liquid pressure increases with depth.
2. Apply projectile motion to water streaming out of a hole in a 3-hole
can: For the time for a water drop to fall vertically, use s = ut + ½
gt2, so t = √ (s / 0.5 g), where g = the acceleration because
of gravity, 9.8 m / sec2, and s = vertical distance from the
hole to table.
For the top hole, s = 3, so t = √ (3 / 0.5 g), for the middle hole,
s = 2, so t = √ (2 / 0.5 g), and for the bottom hole, s = 1, so t = sqrt
(1 / 0.5 g).
For the horizontal velocity of the drop, v, use Torricelli's Theorem,
v = √ (2gh) where h is the height from the hole to the TOP of the can.
For the top hole, h =1, so √ (2gh) = √ (2g), for the middle hole,
h = 2, so √ (2gh) = √ (4g), and for the bottom hole, h =3, so √ (2gh)
= √ (6g)
To find the horizontal distance the drop travels, use distance = vt. For
the top hole, vt = √ (3 / 0.5) X √ (2g) = 3.464, for the middle hole,
vt = √ (2 / 0.5) X √ (4g) = 4, and for the bottom hole, vt = √ (1 / 0.5) X √ (6g)
= 3.464. So water from the middle hole hits the table at the greatest distance,
4, from the bottle, and water from the top and bottom holes both hit the
table at the same distance, 3.464, from the bottle.
3. If a bottle has five equidistant holes, water passing through hole
3 and hits the table top at distance 3, d3. Water passing through hole 2
and hole 4 and hits the table together at distance 2, d2. Water passing
through holes 1 and 5 and hit the table together at distance 1, d1. The
farthest distance from the bottle is d3, then d2, then d3. Also, (d3 - d2)
= (d2 - d1). The range of water hitting the table is greatest from the middle
hole and less the farther the hole is from the middle hole.
4. Torricelli's tank allows water to stream from holes at different heights
in a vertical glass tube. The flow from the holes changes as you move the
tube up and down. Determine the velocity of efflux of water by the comparing
the parabolic trajectory of each stream, or attach a manometer to the different
openings. Holes of different size at the same height show that the flow
is independent of the diameter. 4.154 Falling washers on a
string See diagram 8.154: Falling washers on a string
1. Tie an iron washer to one end of a 2 m string. Tie another six washers
every 30 cm along the string. Stand on a chair, hold the end of the string
with no washer attached and let the string hang down. Let the bottom washer
be 30 cm above the floor. Release the string and listen to the sound of
the washers hitting the floor. The sound intervals are not the same, they
get smaller as the washers hit the floor.
2. To repeat the experiment, tie an iron washer to one end of a 4 m string,
then tie on other washers 15 cm, 45 cm. 75 cm, 105 cm and 135 cm apart.
So the distances of the washers from the end of the string are 15 cm, 60
cm, 135 cm, 240 cm and 375 cm. Stand on a chair, hold the end of the string
with no washer attached and let the string hang down. Let the bottom washer
be 30 cm above the floor. Release the string and listen to the sound of
the washers hitting the floor. The sound intervals are the same. (0.175 seconds).
The acceleration because of gravity (acceleration of free fall) g = 9.8 m
/ sec2. For a free falling body, h = gt2 / 2, so as
time increases the speed of a free falling body is faster and faster. Using
the formula h = gt2 / 2, the falling distances of a free falling
body at 1, 2, 3, 4, 5 seconds are 0.5g, 2g, 4.5g, 8g, 14.5g. The falling
distances every second are (0.5g), (2g - 0.5 g = 1.5g), (4.5g - 2 g = 2.5g),
(8 g - 4.5 g = 3.5g), (14.5 g - 8 g = 4.5), i.e. the ratio is 0.5:1.5:4.5:3.5:4
= 1:3:5:7:9. For similar falling objects separated by the ratio of distances
1: 3: 5: 7:9, the time intervals of falling are equal. 14.1.0.0 Velocity
Velocity, motion in one dimension, linear uniform motion, kinematics,
motion in a straight line, velocity, acceleration, vectors, kinematics,
motion in two dimensions, displacement in one direction, motion in a straight
line, kinematics
If an object changes position from A to B along any path, you measure
its displacement by a straight line from starting point A to finishing point
B. This is a vector quantity, which has magnitude (size) and direction., linear
uniform motion, description of motion (kinematics), how objects move and
an analysis of motion (dynamics), scalar / vector quantities. distance /
displacement. speed / velocity, acceleration, construction and interpretation
of graphs of speed / velocity and acceleration with time, quantitative analyses
of these, problems involving equations for linear uniform motion, uniform
velocity:
v = s / t,
uniform acceleration: v = u + at, s = ½(u + v)t, s = ut + 1 / 2at2,
s = (v2 - u2) / 2a, free fall involving terminal velocity
Average velocity, uniform motion = displacement / time taken. Instantaneous
velocity is the rate of change of displacement with respect to time as measured
by a speedometer.
Average acceleration = change in velocity / time taken. Instantaneous
acceleration is the rate of change of velocity with respect to time as measured
by an accelerometer in an aircraft. You can read velocity and acceleration
from displacement-time graphs and velocity-time graphs.
Equations of motion, motion in one dimension, motion in a straight line
u = initial velocity, v = final velocity, a = acceleration, t = time,
s = distance (displacement), V = u + at
s = (u + v) / 2 X t
s = ut + ½ at2
v2 = u2 + 2as
Velocity and speed, velocity (vector) (steady speed), speed (scalar),
speedometer, velocity-time graph (V-t graph)
Other experiments: 1. Toy bulldozer on moving sheet, 2. Constant velocity
(air track), linear air track and glider, 3. Examine a table of velocities
ranging from continental drift to the speed of light, 4. Ballistics, maximum
height and range, 5. Stroboscope, hand stroboscope, 6. Dynamics trolley,
7. Skate board 14.1.0.1 Equations of uniformly accelerated motion,
gravity
1. v = u + at
2. s = (u + v) t / 2
3. s = ut + at2 / 2 4. v2 = u2 + 2as
where
u = initial velocity
v = velocity after time t
s = distance travelled in time t
a = constant acceleration
(v, s, and a are positive in the direction of u) Gravity
g = the acceleration due to gravity, 9.8 m sec.-2 14.1.01 Scalars and vectors
A scalar is a quantity with magnitude, size. but no direction, e.g. density,
energy, mass, work. Addition of scalars is algebraic addition, e.g. a 100
g body is added to a 200 g body, so its residual mass is 200 g. Scalar quantities
are like the rungs of a ladder.
A vector is a quantity with magnitude, size, and direction, e.g. force, momentum,
velocity. A vector can be represented by an arrow with its length proportional
to the magnitude of the vector and its direction being parallel to the the
direction of the vector and so showing its direction. More than one vector
acting at a point may be represented by a single resultant vector being
equivalent to all of them in magnitude and direction.
If all the forces on a body are acting in the same direction the net force
is the simple arithmetical sum of the forces. So 90 kg force is needed to
raise a 90 kg mass but if three people share the load each person needs
exert only 30 kg force.
Vector quantities have both size and direction and you can represent vectors
by lines, with length proportional to the size and direction indicated relative
to some reference direction, e.g. displacement, velocity, acceleration,
force and momentum. You can add vectors to find the resultant vector, by
placing drawn vectors head to tail. You can resolve vectors into two perpendicular
component vectors, rectangular components, so that the two components are
equivalent to the single vector. 14.1.02 Vector addition,
parallelogram law, vector diagram See diagram 14.1.1: Parallelogram of vectors,
Triangle of vectors
1. Addition of vectors using the parallelogram law
If two vectors, AB and AD, acting at a point, may be represented by in
magnitude and direction by two adjacent sides of a parallelogram, ABCD, the
resultant, R, is the diagonal AC.
R2 = AD2 + AC2 + 2 × AD × AC × cos BAD
The usual practice is to call the two vectors P and Q and call the angle
between them θ (theta)
So R2 = P2 + Q2 + 2PQ cos θ
P and Q are called the components of R.
Also, the direction of the resultant relative to AD is the angle α (alpha).
tan α =
Q sin θ
/ P + Q cos θ See diagram 14.1.2: Verify
the parallelogram law
1.1 Verify the parallelogram law by vertically supporting a 1 kg mass
by two spring balances as in the diagram. Place the apparatus next to a
vertical board with graph paper stuck to it. Trace the pathway of the three
strings onto the graph paper and read the forces measured by the two spring
balances.
Draw a parallelogram with side P and Q proportional to the values on the
spring balances using the scale 1 cm = 1 N (newton). The diagonal OC should
be 9.8 cm. 2. Addition of vectors using a vector diagram See diagram 14.1.1: Addition of two vectors
The magnitude and direction of the resultant may be found by drawing to
scale AB, AD and the angle θ, completing
the drawing of the parallelogram and carefully measuring the length of R and
the size of angle α
relative to AD.
In the diagram, to add vector B to vector A, move B parallel to itself
until the tail of B is at the end of A. In the new position of B it has its
original length and direction. The sum A + B is the vector R drawn from the
tail of A to the head of B. 2.1 See diagram 14.1.1:
Boat speed
A boat travelling due east at 4 km / hour is in a river flowing due south
at 2.0 km / hour. The velocity of the boat relative to a point on the river
bank is 4.47 km / hour in direction 26o33' south of east. 2.2 An aeroplane is travelling north with airspeed
240 km h-1 through a wind blowing 70.0 km h-1 from
the west. Its ground speed is √ (2402 + 702) = 250
km h-1 and its track is about N 16o E, (cos θ = 240 / 250 = 0.96, cos 16o =
0.9613). Ground speed is the speed of an aircraft relative to the ground. 3. Triangle of vectors See diagram 14.1.3: Triangle of vectors, boat
travel
If vectors P and Q may be represented by AB and AC as sides of a triangle,
the resultant is represented by the third side AC. So the resultant of vectors
acting on a point or body at different directions may be calculated or found
with a vector diagram.
A boat travels 4.5 km north then 6.5 km north east. The total displacement
of the boat is represented by the third side of the triangle of vectors. 14.1.03 Vector resolution See diagram 14.1.4: Resolution of vector into
components
Often it is convenient to break up a single vector, F, into two vectors,
i.e. resolve one force into two component forces
If P and Q are at right angles, P is called the resolved part of R in
the direction of P.
P = R cos α
Q= R sin α
A student uses a rope to pull a wagon along a road with force F. The angle
of the rope to the horizontal is θ. The force F on the rope can be resolved
into a horizontal component F cos θ that causes the wagon to move along the road.
and a vertical component F sin θ that tends to lift the wagon. So some of
the force needed to move the wagon is lost because the rope is not pulled
parallel to the road. 1. A big box is pulled along a bench by a cord
at 30o to the horizontal bench top. The tension in the cord measured
by spring balance is 80 N. What is (i) the effective part of the pull in
moving the box, and (ii) the force tending to lift the box off the bench?
(i) F = 80 cos 30o = 80 × 0.866 = 69.28 N, (ii) F = 80 sin 30o
= 80 × 0.5 = 40 N. 2. A canal boat is pulled along a canal by a tug
boat with a rope at angle 10o horizontally to the direction of
movement along the canal. The captain of the tug boat sets the rudder to
ensure that the canal boat moves directly along the middle of the canal.
If the tension of the rope is 150 kN (kilonewton), the resolved part of
the force on the canal boat in that direction is 150 cos 10o
= 150 × 0.98 = 147 kN
14.1.04 Addition of vectors when P and Q are at right angles See diagram 14.1.1: P and Q are at right angles 4.1 P is the resolved part of R in the direction OX. P = R cos α and Q = R sin α. 4.2 The resolved part of any vector F in a direction making an angle α with the direction of the vector is F cos θ.
The resolved part is the effective part of the given vector in the required
direction. For example if a force of 20 N be acting in a given direction
the fraction of the force in a direction at an angle of 60 o with the direction
of the available force = 20 cos 60o = 10 N. 4.3 The potential energy of a body mass m above the level AB = mgh. If the
mass m was rolled up the slope length l, the work done against the force
F down the slope, i.e. the resolved part of of the weight of m down the slope,
through distance l = mgh.
So Fl = mgh, F = mg × h / l = mg cos (90 - θ) = mg sin θ
14.1.05 Polygon of vectors See diagram 14.1.3: Polygon of vectors
Four vectors, acting on one point can be represented in size and direction
by the sides AB, BC, CD, DE of a polygon. The resultant of these four forces
is represented in size and direction by the resultant, R, i.e. the side AE
that completes the polygon. If the vectors are forces, a force equal and
opposite to R, the equilibrant, would produce equilibrium in the system.
If four coplanar forces 40, 40, 10 and 30 newton are acting at a point at angle 60o in succession to each other, the equilibriant is 50 newton parallel to the second force of 40 newton. 14.1.1 Toy car running on moving paper
Run a toy car at constant speed in both directions on moving paper to
show how velocities add and subtract. Time a toy bulldozer with a stop clock
as you pull it across the table at constant velocity in front of a metre
stick. 14.1.2 Photograph uniform motion
Take an open shutter photograph of a toy clockwork motor car 14.1.3 Terminal velocity
Let a marble roll down a tube of water at a slight incline to reach terminal
velocity and allow you to measure slow constant velocity. 14.1.4 Muzzle velocity, bullet timer
Fire a bullet to pass through two aluminium foil strips and show the signal
on an oscilloscope Fire an air gun through two rotating cardboard discs
separated by a known distance. 14.1.5 Time of flight:
Release a projectile fired from a pendulum apparatus by timing signals
from two microphones. 14.1.5.1 Examine a table of velocities
ranging from continental drift to the speed of light. 14.2.1 Drop strings with attached weights
Drop two strings with attached weights, one with equal distance intervals
and the other with equal time intervals (1, 4, 9, 16). Let strings hit a
hard floor and listen to the sounds of weights hitting the floor. Drop a
string with wood blocks or lead balls tied at unit intervals and equal time
intervals. 14.2.3 Time a falling body with an electric "ticker
timer" (recording timer) See diagram 8.238.1: Tape | See diagram 8.238: Ticker timer | See diagram 8.238.1: Ticker timer tape
1. Time a falling body with a ticker timer. Attach a weight to a strip
of paper tape. Pass the tape between the armature of an electric bell and
a pad of carbon paper. Release the paper tape so that the weight falls and
drags the paper after it. The end of the arm of the timer hits the carbon
paper against the tape and makes marks on it at equal time intervals. Measure
the distance between the marks.
2. Make a ticker timer from an electric bell mechanism. Remove the clapper
and extend the armature by soldering a strip of metal to it. At the end
of this extension, drill a hole to fit a small round head screw. Fix the
screw head downwards to act as a marking hammer. Fasten the mechanism to
a wooden base. Fix a 3 cm diameter disc of a carbon paper disc to the base
with a drawing pin. The drawing pin holds the disc loosely at the centre
so that the disc can rotate to expose a new surface as the tape passes under
it. Attach staples to the base to guide the path of the ticker tape. If the
extension to the armature strikes the paper too hard, the timing may be uneven.
3. Attach a weight to a strip of paper tape. Pass the tape between the
armature of an electric bell and a pad of carbon paper. Release the paper
tape so that the weight falls and drags the paper after it. The end of the
arm of the timer hits the carbon paper against the tape and makes marks on
it at equal time intervals. Measure the distance between the marks. A ticker
timer uses low volt AC power source to make a vibrating spring print a series
of points on paper tape using circular carbon paper. When the paper tape moves
with the object, the greater the space between points the further the object
has moved. The time between any two points is equal to the reciprocal of
the AC frequency, 50 HZ, so the time interval is 0.02 s.
4. To study the motion of a long window blind, tape the paper tape to
the bottom of the blind and let the paper tape moves up through the ticker
timer. Find a convenient starting point A on the paper tape and record it
as 0, even if it is not the first point. Then mark each point every 5 dots,
i.e. 5 x 0.02 = 0.1 second. Measure the distance between points AB, BC, CD,
DE and calculate the average speed in every space, e.g. from A to B, v =
distance / time = 5.0 cm / 0.1s.
5. Ticker timer using an electric bell mechanism See diagram 32.5.4.4: Electric bell circuit
Remove the clapper and extend the armature by soldering a strip of metal
to it. At the end of this extension drill a hole to fit a small round headed
screw. Fix the screw head downwards to act as a marking hammer. Fasten the
mechanism to a wooden base. Fix a 3 cm diameter disc of a carbon paper disc
to the base with a drawing pin. The drawing pin holds the disc loosely at
the centre so that the disc can rotate to expose a new surface as the tape
passes under it. Attach staples to the base to guide the path of the ticker
tape. If the extension to the armature strikes the paper too hard, the timing
may be uneven.
6. To draw the (V - t) graph, calculate the instantaneous speed equal
to the average speed. Note that the window blind had begun to move before
the selecting point A of t = 0, so there is a part of the graph before t
= 0. The graph shows that the curtain moves in constant acceleration before
0.35 seconds, it moves in constant speed of 2 m / s between 0.35 s and 0.65
s, and then it moves in constant deceleration. Calculate the acceleration
in AD stage from changes in the instantaneous speed in the centre of the
time intervals AB and CD: (1.5 - 0.5) / (0.25 - 0.05) = 5.0 (m / s2).
Calculate the acceleration in each stage by this method and draw an acceleration-time
graph of motion of the window blind.
Points
Distance cm
travelled in 0.1 sec
Average v, m / s
AB
05.0
0.5
BC
10.0
1.0
CD
15.0
1.5
DE
20.0
2.0
EF
20.0
2.0
FG
20.0
2.0
GH
20.0
2.0
HI
17.0
1.7
IJ
08.0
0.8
JK
04.0
0.4
Commercial
Recording timer
Recording timer, AC 5 - 12 V, ticker timer, heavy base with rubber feet, includes disc and small recording tape
Carbon paper disc, suits recording timer
Recording tape, large, suits recording timer 14.2.4 Simultaneous fall, ball bearings falling
together See diagram 8.234: Simultaneous fall
1. Use two clothes pegs, a pair of ball bearings A and B, a wide rubber
band about 8 cm long. Fix the rubber band lengthways around one clothes peg.
Then open the clothes peg and force ball bearing A against the tension of
the rubber band between the prongs of the clothes peg. Grip ball bearing B
with the second peg. Hold the two pegs side by side, pointing away horizontally
above a sheet of tin on the floor. Squeeze the ends of the arms of both
clothes pegs. At the same moment, ball bearing A begins to fall vertically,
and ball bearing B is shot forwards. Observe what happens by looking and
listening when the ball bearings hit the sheet of tin. Repeat the experiment
at heights 0.5 m, 1.0 m, 1.5 m, 2.0 m and with a tighter rubber band. If
the experiment is done correctly, ball bearings A and B land in different
places on the sheet of tin but they strike the ground simultaneously.
2. Two balls simultaneously dropped and projected horizontally hit the
floor together. Drop one billiard ball and shoot another out simultaneous.
One ball is projected horizontally as another is dropped simultaneously. Instructor
rolls a ball off the hand while walking at a constant velocity. 14.2.5 Coins falling at the same time See diagram 16.159: Flicking
Observe the falling object and horizontal projectile object start off
at the same time and fall down to the ground at the same time too. Fold
a paper card in half, then fold each side one third from the end outwards
to form a convex. Place coins on each side of the centre ridge of the card
and hold one end on the table edge. Flick the ridge of the card to the side
with your middle finger of right hand. One of the coins will be thrown several
metre s away, the other will fall straight to the floor at the exact same
moment. First observe if the two coins start motion at the same time. Then
repeat the steps above, note which coin hits the floor first. Flick the
card lightly, hear the click. If you hear only one click, that means the
direction of the falling object at first has not any influence on the falling
times. This is because the accelerations of the two falling coins are the
same if the air resistance is disregarded. 14.2.6 Coin and feather, penny and feather, guinea
and feather
1. Drop a coin and feather in a 150 cm sealed glass tube full of air.
Stand the glass tube on its end. Invert the glass tube. The coin falls much
faster because air resistance affects the feather more. Connect the glass
tube to an air pump. Turn on the pump to evacuate the glass tube. Stand
the glass tube on its end. Invert the glass tube. The coin and feather fall
at same the uniform acceleration in the absence of air resistance.
2. Repeat the experiment with same size cork and lead ball.
3. Practice dropping heavy and light objects simultaneously with one hand.
Use two identical sheets of paper. Crumple one sheet of paper into a tight
ball. Simultaneously drop the ball of paper and sheet of paper from the
same height.. 14.2.7 Falling heavy and light balls
Estimate the height a light ball must be dropped so it hits the floor
at the same time as the heavy ball. 14.2.12 Acceleration down an incline See diagram 8.235: Acceleration down an incline
1. Use a grooved three metre plank. Incline it so that marbles will roll
down the groove. Arrange small tin flags hung from wires so that the marbles
hit them and make "clinks" sounds. Put the flags at regular intervals, e.g.
25, 50, 75, 100 cm, from the beginning of the plank. Roll a marble down
the groove and listen to the intervals between "clink" sounds. The time
between the "clinks" will reduce as the ball rolls down the incline. Arrange
the flags so that the clinks occur at equal intervals of time. Measure the
distance between the flags. The distance between the flags increases down
the incline.
2. Use an inclined air track. For timing use a stop clock and metre stick.
Put lights that flash every second along an incline and horizontal track
such that they are flashing at the moment the ball passes.
3. Observe a car on an inclined wire. Stretch a long wire diagonally across
the chalkboard with chalk marks at every metre . Time a car as it accelerates
past the marks.
4. Observe a ball on an incline. Roll a ball bearing down the groove of
a plastic metre stick Use a slow roller solid wheel turning on a small axis
to roll down an incline.
5. Use a Duff's plane, chalk ball on incline. Use a ball that leaves a
trail while rolling down a chalk covered trough. 14.2.13 Path of a projectile
1. Water from a hose
Observe the path of water from a hose held at different angles and with
different water pressure. Note the angle that results in the greatest distance
reached by the water, 45o.
2. Flight of a shuttlecock
Hit a shuttlecock in front of a blackboard marked with chalked squares.
Observers can draw similar curves on graph paper to show the flight path
of the shuttlecock.
3. Monkey and hunter, monkey and bullet, mid-air target, path of a projectile See diagram 8.239: Monkey and hunter
3.1 Acceleration of a freely falling body is equal to the acceleration
of the vertical component of a projectile. This experiment shows that the
vertical and the horizontal velocities of a projectile are independent of
each other and that the acceleration of a freely falling body is the same
as the acceleration of the vertical component of a body in projectile motion.
A "gun" is aimed at a "monkey" suspended by an electromagnet. When the
"bullet" leaves the barrel of the "gun" it activates a light-sensitive switch
to release the "monkey". This experiment is called "The monkey and hunter"
but nowadays use a drink-can instead of a poor monkey!
. The projectile is a ball-bearing and the target "monkey" is a metal
drink-can hanging from an electromagnet. The circuit of the electromagnet
includes two bared wires fixed parallel to and each side of the axis of a
cardboard tube. They project about 2.5 cm beyond the end of the tube. Complete
the electrical circuit with a short length of copper wire resting on the
projecting wires as a switch. Fix the cardboard tube in a stand so that it
points towards the drink-can. Note the angle of the cardboard tube above
the horizontal. Blow the ball-bearing up the cardboard tube. When the ball-bearing
passes the end of the cardboard tube, it displaces the piece of copper wire,
opens the switch and no electric current flows through the electromagnet.
So the metal can is released to fall. The ball-bearing can hit the metal
drink can in mid-air if the angle of the cardboard tube above the horizontal
is correct. The ball bearing will hit the drink-can at different angles
of the cardboard tube (between 0o and 90o), different
distances, and different initial speeds of the ball-bearing up the cardboard
tube. 14.2.15 Parabolic trajectory of thrown chalk See 2.0.5: Conic sections, parabola | See 2.0.6: Parabola equation
Throw a piece of chalk so it follows a parabolic path drawn on the chalkboard.
Roll ink dipped balls down an incline onto a tilted stage on an overhead
projector. Roll a tennis ball covered with chalk dust across a tilted blackboard.
14.2.16 Gun and tunnel
A spring loaded gun on a cart shoots a ball vertically and after the cart
passes through a tunnel the ball lands in the barrel. A ball fired vertically
from cart moving horizontally falls back into the barrel. 14.2.17 Mid air target
A hunter shoots a compressed air at a target released when the gun is
fired The ball hits the target in mid air. 14.2.18 Range of a gun
Fire a spring loaded gun at various angles. Use a tennis ball serving
machine to find muzzle velocity and range of a gun 14.3.0 Measuring g
Acceleration due to gravity, falling objects, air resistance, electric
stop clock, Distance / time graph, g = 2S / t2 14.3.1 Free fall timer
Time a ball as it drops 0.5 m, 1.0 m, 1.5 m, or 2.0 m. Drop a magnet through
several equally spaced coils of wire and examine the induced voltage on
an oscilloscope. 14.3.2 Dropping balls
Use a latching relay system for turning a standard timer on and off so
that an electromagnet releases the ball and starts the clock and a catcher
stops the clock. Drop light and heavy balls through a multiple pass light
beam and show the output on an oscilloscope. 14.3.3 Ink jet marker
Let slab of wood be dropped by a ink squirter which leaves lines at equal
time intervals. Use a rotating ink jet to spray a paper sleeve on a falling
metre stick. 14.3.4 Falling drops
Time 10 drips of water dripping through a tap at uniform rate. 14.3.5 Catch a metre stick, reaction time of a person, drop the ruler See diagram 9.249: Reaction time
Drop a metre stick and use the distance it drops before catching it to
find the reaction time of the catcher. If a body falls from a height s, the
distance it falls after t seconds = gt2 / 2. So if you measure
s, you can obtain t, t = √2S / g. Hold metre ruler vertically with the
zero on the scale down and the 100 on the zero on the scale up. With your
arm stretched horizontally, hold the ruler vertically between the thumb and
first finger with the lower edge of the first finger at the zero on the scale.
Open your fingers then close them again as quickly as possible to catch the
ruler again. Record the distance to the downward edge of your first finger.
Repeat the experiment and calculate the average distance down the metre stick.
Use t = √2S / g to calculate the time of the falling ruler, i.e. your
reaction time. Repeat the experiment under the following pairs of conditions:
1. Hold the ruler first with your left hand. Hold the ruler first with
your right hand.
2. Talk to others while doing the experiment. Do not talk to others while
doing the experiment.
3. Allow loud background music. Do not allow loud background music.
4. Try other contrasting conditions to see whether your reaction time
is affected. 14.3.6 Catch a coin
Catch a coin starting with the fingers at the midpoint of the coin. 14.3.7 Rotating turntable
Drop a ball on a turning phonograph turntable. 14.3.8 Pendulum-timed free fall
Let a pendulum released from the side hit a ball dropped from the height
that gives a fall time equal to a quarter period of the pendulum. 14.3.9 Many bounce method
Time a bouncing ball for many bounces and find g using the coefficient
of restitution. 14.3.10 Swing bucket in vertical circle, swinging
pail
This experiment is best done outside the classroom on the grass.
1. Fill a bucket with water and attach a heavy rope or chain to the handle.
Hold on to the end of the rope or chain and swing the bucket to and fro
until you can swing it quickly in a vertical circle without the water spilling
out. Ask an observer to time the circular swings. Start swinging the bucket
slower and slower until water starts to fall out. Ask the observer to tell
you the slowest swing speed before the water started falling out.
2. Repeat the experiment with a bucket full of nails. Observe when the
nails start to drop down or hit the bucket making a noise at the top of the
swing.