Experiment 1 Virtual experiment - Imam U · - Prepare a sheet of graph paper for plotting t2 versus d. You should make t2 the vertical axis and d the horizontal axis. - Plot the measured

sicshyDepartment of p – PHY 119 & PHY 103 College of Science

1

Experiment 1

Virtual experiment

1- Objects of the experiment

- To determine the constant acceleration of a body moving in a straight line.

2- Principles

From the laws of motion for constant acceleration the distance travelled "d " is given by

the following equation:

20 2

1attvd (1)

Where: v0 is the initial velocity, t is the interval of time taken to travel the distance d and

a is the constant acceleration.

Since 00 v then

2

21

atd (2)

Rewriting the above equation as:

da

t22 (3)

3- Measurements:

An experiment was carried out on an object moving from rest at constant acceleration

"a". The time taken "t" was measured for different distances "d ". The experiment was

repeated five times for the same distance. The data were then entered in a table as

follows.


2

Distance Time taverage

d (m) t1 (s) t2 (s) t3 (s) t4 (s) t5 (s) tavg (s) 22avgtt (s

2)

1 2.09 2.34 2.18 2.14 2.24

2 3.16 3.31 3.19 3.26 3.01

3 3.87 3.95 4.02 3.77 3.72

4 4.44 4.32 4.62 4.47 4.57

5 5.00 5.10 5.15 4.93 4.85

6 5.63 5.38 5.48 5.41 5.33

7 5.92 6.02 5.94 5.82 5.77

8 6.47 6.07 6.32 6.22 6.17

9 6.71 6.81 6.50 6.61 6.56

10 6.92 7.17 7.22 7.00 7.07

- Prepare a sheet of graph paper for plotting t2 versus d. You should make t

2 the vertical

axis and d the horizontal axis.

- Plot the measured values on a t2-d graph.

- Find a relationship between the slope and the constant acceleration.

- Determine the value of the acceleration.


3

Experiment 2

Free fall: displacement-time law


- To determine the numerical value of the acceleration due to gravity labeled g.

2- Principles

If a ball falls, starting at rest ( 00 v ) at time t = 0, it covers the fall displacement 0yy

in the time t:

20 2

1gtyyy (1)

where g is acceleration due to gravity.

The fall distance, h, is written as:

2

21

gtyh (2)

By rearranging the Equation 2 to be in the form:

hg

t22 (3)

3- Setup

A steel ball is suspended from an electromagnet. The ball starts to fall when the magnet

current is interrupted. At the same time, a digital counter is started. The ball cuts a light

barrier, thus stopping the counter. The fall time is read off on the digital counter, and the

fall distance can be measured directly.


4

Figure 1. Experiment setup

for determining the

displacement-time law for

free fall.

4- Carrying out the experiment:

- Attach the ball to the holding magnet.

- Position the forked light barrier at a distance h=0.2m from the underside of the ball as

shown in Figure 1.

- Release the ball by pressing the START/STOP key of the digital counter. The ball

passes through the light barrier and then automatically the digital counter stops to count

the time.

- Read off the fall time t.

- Reset the counter to zero by pressing the RESET key.

- Attach the ball to the holding magnet again.

- Repeat the measurement up to 3 times, and calculate the mean value of the measured

times.

- Repeat the measurements as described above for h=0.3m; 0.4m; 0.5m; 0.6m and 0.7m.

5- Measurements:

0yyh

(m)

t1

(s)

t2

(s)

t3

(s) 3321 ttt

t

t2

0.2

0.3

0.4

0.5

netelectromag

ballsteelcounterdigital

yh

clamp

y

0y

barrierlightForked


5

0.6

0.7

- Prepare a sheet of graph paper for plotting t2

versus h. You should make h the

horizontal axis and t2 the vertical axis.

- Plot the measured values on the t2-h graph.

- Draw a best fit line to the points on your graph.

- Find a relationship between the slope and the acceleration due to gravity g (see

Equation 3)

- Determine the value of g.

- Find the percentage error if you know that the value of g in literature is equal to

9.81m/s2

(Hint: 1009.81

9.81-g error % )


6

Experiment 3

Vector Addition of Forces

1- Objectives of the experiment

- Determine the force which balance other forces,

- Determine the resultant force,

- Check the results by using the polygon and components methods for adding forces.

2- Principles

2-1- Polygon method:

Vectors may be added graphically by repositioning each one so that its tail coincides with

the head of the previous one (see Figure 1). The resultant force, RF

, (sum of the forces) is

the vector drawn from the tail of the first vector to the head of the last vector. The

magnitude (length), rF , and angle, r , of the resultant force are measured with a ruler

and a protractor, respectively.

Note: In order to measure the angle, a set of axes must first be defined.

Figure 1-a. Vector addition of two vectors

1F

and 2F

by the polygon method.

Figure 1-b. Vector addition of three

vectors 1F

, 2F

and 3F

by the polygon

method.

2-2- Components method

Vectors may be added by selecting two perpendicular directions called the x- and y-axes,

and projecting each vector on to these axes. This process is called the resolution of a

1F

2F

RF

x

y

3F

321 FFFFR

r

3F

2F

1F

2F

RF

x

y

21 FFFR

r

2F


7

vector into components in these directions. If the angle that the vector makes from the

positive x-axis in the anticlockwise direction is used (see figure 2), these components are

given by

sin and cos FFFF yx (1)

Figure 2. Finding the two perpendicular

components Fx and Fy of a vector F

.

The x component of the resultant is the sum of the x components of the vectors being

added, and similarly for the y component.

11

and i

yiyRi

xixR FFFF (2)

The angle that the resultant makes with the positive x-axis in the anticlockwise direction

is given by

2or 1 ,0 where180tan1 nnF

F

xR

yRr

(3)

The right value of n is given by the signs of xRF and yRF as follows:

Sign of xRF Sign of yRF Value of n

+ + 0

_ + 1

_ _ 1

+ _ 2

and the magnitude is given by

22yRxRR FFF (4)

x

y

F

xF

yF


8

2-3- Equilibrium Conditions

Newton's second law predicts that a body will not accelerate when the net force acting on

it is zero. So, for an object to be at rest, the resultant force acting on it must be zero. In

equation form, the above statement can be written

1

0i

iF

(5)

Thus, if four forces act on an object at rest, the following relationship has to be satisfied.

04321

FFFF (6)

An equivalent statement is

RFFFFF

3214 (7)

so that 4F

is equal in magnitude and opposite in direction to the resultant force, RF

, of

the other three forces ( 321 FFFFR

).

The force 4F

is called the equilibrant force, eqF

, since it is the force which establishes

equilibrium.

Req FFF

4 (8)

Figure 3. Finding the two directions r and eq , respectively, of rF

and eqF

.

180 eqr (9)

3- Equipments

- Force table.

- Set of slotted weight with hanger.

RF

x

y

eqFF

4

eqRF

x

y

eqFF

4

r1F

2F

RF

x

3F

y

eqFF

4

2F

3F


9

- Pulleys.

- Rings.

- String.

- 30-cm ruler

- Protractor

- Level

- Two graph papers.

4- Setup and carrying out the experiment.

4-1 Part 1 (use of three pulleys)

4-1-1 Measurement of equilibrant force experimentally

- assemble the force table and use three pulleys. Two pulleys for the forces that will be

added and one pulley for the force that will balance their sum.

- fix the center pin on the table.

- Hang two set of weights on pulleys at right angles to each other, one equal to the weight

of 150g and the other to the weight of 200g.

- Let the 200g weight be at 0o and the 150g weight be at 90

o.

Figure 4. Force table showing four pulleys

and weights on hangers.

Figure 5. Adjusting the forces on the ring.

- Balance these two forces with several hundred grams on the third string so that the ring

of the force table remains at rest at the center (see Figure 5).

- Carefully, push the ring to one side and see if it goes back to the center. If it does, then

it is in equilibrium (the net force on the ring is zero).

- Find the angle for the third pulley and the masses which must be suspended.

- Fill Table 1.


11

Table 1. Measurement of eqF

directly on force table. Take the acceleration due to gravity

g=10m/s2.

1F

2F

eqF

rF

Hanged mass 1m =200g 2m =150g eqm = XXXXXXXX

Magnitude 1F=2N 2F =1.5N eqF = rF =

Direction 1 =0° 2 =90° eq = r =

4-1-2 Measurement of resultant force by the polygon’s method

- Prepare a graph paper to draw 1F

, 2F

and rF

(take 0.2N = 1cm).

- Measure the magnitude and direction of rF

, and write down these values in Table 2.

Table 2. Measurement of resultant force by the polygon’s method.

1F

2F

rF

Hanged mass 1m =200g 2m =150g XXXXXXXX

Magnitude 1F=2N 2F =1.5N rF =

Direction 1 =0° 2 =90° r =

4-1-3 Determination of resultant force by the components’ method

- From the magnitudes and directions of 1F

and 2F

(columns 2 and 3 of Table 2), fill by

calculation Table 3.

Table 3. Determination of resultant force by the components’ method

1F

(N) 2F

(N) rF

(N)

xF1 (N) yF1 (N) xF2 (N) yF2 (N) xrF (N) yrF (N) rF (N) r

- Compare your results by calculating the % difference in magnitude and in direction

between the calculated resultant force (polygon’s method and components method) and

the measured resultant force (on force table) and write down the results in Table 4.


11

100magnitudein difference %

calculatedr

calculatedrmeasuredr

F

FF

100directionin difference %

calculatedr


Table 4. Comparison between the measured and calculated methods

Table force Polygon’s

method

Components method % difference1 % difference

2

Magnitude: rF (N)

Direction: r

% difference1: comparison between table force and polygon’s method

% difference2: comparison between table force and components method

4-2 Part 2 (use of four pulleys)

4-2-1 Measurement of equilibrant force experimentally

- assemble the force table and use four pulleys. Three pulleys for the forces that will be

added and one pulley for the force that will balance their sum.

- fix the center pin on the table.

- Hang three sets of weights (m1, m2 and m3) on pulleys as shown in Table 5.

- Balance these three forces with several hundred grams on the fourth string so that the

ring of the force table remains at rest at the center.

- Carefully, push the ring to one side and see if it goes back to the center. If it does, then

it is in equilibrium (the net force on the ring is zero).

- Find the angle for the fourth pulley and the masses which must be suspended.

- Fill Table 5.

Table 5. Measurement of eqF

directly on force table. Take the acceleration due to gravity

g=10m/s2.

1F

2F

3F

eqF

rF

Hanged mass 1m =200g 2m =150g 3m =100g eqm = XXXXXXXX

Magnitude 1F=2N 2F =1.5N 3F =1N eqF = rF =

Direction 1 =0° 2 =90° 3 =135° eq = r =


12

4-2-2 Measurement of resultant force by the polygon’s method

- Prepare a graph paper to draw 1F

, 2F

, 3F

and rF

(take 0.2N = 1cm).

- Measure the magnitude and direction of rF

, and write down these values in Table 6.

Table 6. Measurement of resultant force by the polygon’s method.

1F

2F

3F

rF

Hanged mass 1m =200g 2m =150g 3m =100g XXXXXXXX

Magnitude 1F=2N 2F =1.5N 3F =1N rF =

Direction 1 =0° 2 =90° 3 =135° r =

4-2-3 Determination of resultant force by the components’ method

- From the magnitudes and directions of 1F

, 2F

and 3F

(columns 2, 3 and 4 of Table 5),

fill by calculation Table 7.

Table 7. Determination of resultant force by the components’ method

1F

(N) 2F

(N) 3F

(N) rF

(N)

xF1 yF1 xF2 yF2 xF3 yF3 xrF yrF rF (N) r

- Compare your results by calculating the % difference in magnitude and in direction

between the calculated resultant force (polygon’s method and components method) and

the measured resultant force (on force table) and write down the results in Table 8.

100magnitudein difference %

calculatedr


F

FF

100directionin difference %

calculatedr



13

Table 8. Comparison between the measured and calculated methods

Table force Polygon’s

method

Components method % difference1 % difference

2

Magnitude: rF (N)

Direction: r

% difference1: comparison between table force and polygon’s method

% difference2: comparison between table force and components method

5- Conclusion

Discuss your results


14

Experiment 4

Simple Pendulum

1- The experiment goal:

- To measure the acceleration due to gravity.

2- Experiment theory

A simple pendulum consists of a body (ball) of small dimensions suspended by an

inextensible weightless string. When pulled to one side of its equilibrium position and

then released, the ball vibrates about this position.

The motion of this ball is periodic. Theory suggests that the time period T can be

calculated using the following equation

gL

2T (1)

where, L is the length of the pendulum and g is the acceleration due to gravity.

Squaring both sides

Lg

4T

22 (2)


- Start with a length L=0.3m and pull the ball to one side with a small displacement then

release it.

L


15

- Using the stopwatch, determine the time (t) of 10 full oscillations.

- Repeat the measurement up to 3 times, and calculate the mean value of the measured

times.

- Repeat the measurement for L=0.4m, 0.5m, 0.6m, and 0.7m.

4- Measurements:

Table 1:

L (m) t1 (s) t2 (s) t3 (s)

3ttt

t 321 (s)

22

10t

T

0.3

0.4

0.5

0.6

0.7

- Prepare a sheet of graph paper for plotting T2 versus L. You should make T

2 the vertical

axis and L the horizontal axis.

- Plot the measured values on the T2-L graph.


- Find a relationship between slope and g

- Determine the value of g.

- Find the percentage error if you know that the value of g in literature is equal to

9.81m/s2

(Hint: 1009.81

9.81-g error % )


16

Experiment 5

Expansion of a Helical Spring


- Determining the change of length "s " of two helical springs with different turn

diameters as a function of the gravitational force exerted by the suspended weights.

- Confirming Hooke’s law and determining the spring constant k of the two helical

springs.

- Determine the mass of an object.

2- Experiment theory

If the spring is stretched or compressed a small distance from its equilibrium position, the

spring will exert a force on the body given by Hooke’s law, namely

)( 0LLkskFs (1)

where Fs is known as the spring force. Here the constant of proportionality, k, is known

as the spring constant, and s=L-L0 is the displacement of the body.

The negative sign in Equation 1 indicates that the direction of Fs is always opposite the

direction of the displacement.

When a mass, m, is suspended from a spring and the system is allowed to reach

equilibrium, as shown in Figure 1, Newton’s second law tells us that the magnitude of the

spring force equals the weight of the body

mgksFs (2)


17

Figure 1. Schematic illustration of the expansion of a helical spring

3- Measurements:

a) Part 1: Spring 1

- Measure the length, L01, of the spring.

- Suspend a mass of 50g to the end of the spring and measure its length L.

- Repeat the procedure as above with 100g, 150g and 200g, and fill the following table.

Mass added (g) L (mm) s = L – L01 (mm) sF (N)

0 L01= 0 0

50

100

150

200

- Prepare a sheet of graph paper for plotting sF versus s. You should make sF the

vertical axis and s the horizontal axis.

- Plot the measured values on a sF -s graph.


- Find a relationship between the slope and the constant of the spring, k1.

- Determine the value of k1.

0L

L

m

gm

sF

m

0LLs


18

b) Part 2: Spring 2

Mass added (g) L (mm) s = L – L02 (mm) sF (N)

0 L02= 0 0

50

100

150

200

- Prepare a sheet of graph paper for plotting sF versus s. You should make sF the

vertical axis and s the horizontal axis.

- Plot the measured values on a sF -s graph.


- Find a relationship between the slope and the constant of the spring, k2.

- Determine the value of k2.

Part 3: Determination the mass of an unknown object.

- Suspend the object of unknown mass to the end of spring 1.

- Measure the displacement L- L01

- Determine its mass m1 by using Equation 2.

- Repeat the experiment with spring 2 and determine the object’s mass m2.

- Weigh the object with an electronic balance and write down its mass m0.

- Fill the table below.

Spring 1 Spring 2 Electronic balance

Object’s mass m1= m2= m0=

% error XXXXXXXXXXX

(Hint: 100error %0

0

m

mmi where i=1,2)

- Discuss the results obtained.


19

Experiment 6

Simple Harmonic Motion: Oscillating Mass on a Spring


Determine the spring constant by using the simple harmonic motion of masses attached to

a spring.

2- Principles

Periodic motion is motion of an object that regularly returns to a given position after a

fixed time interval. A special kind of periodic motion occurs in mechanical systems when

the force acting on an object is proportional to the position of the object relative to some

equilibrium position. If this force is always directed toward the equilibrium position, the

motion is called simple harmonic motion.

As a model for simple harmonic motion, consider a block of mass m attached to the end

of a spring, with the block free to move on a vertical plane. When the spring is neither

stretched nor compressed, the block is at the position called the equilibrium position of

the system, which we identify as x = 0. We know from experience that such a system

oscillates back and forth if disturbed from its equilibrium position.

Figure 1. Schematic illustration of the oscillation of a

mass on a helical spring.

A

0

A

x

m


21

We can understand the motion in Figure 1 qualitatively by first recalling that when the

block is displaced to a position x, the spring exerts on the block a force that is

proportional to the position and given by Hooke’s law

xkF (1)

We call this a restoring force because it is always directed toward the equilibrium

position and therefore opposite the displacement from equilibrium. Applying Newton’s

second law to the motion of the block, we obtain

xkma (2)

xmk

a (3)

Recall that, by definition, 2

2

dtxd

dtdv

a , so we can express Equation 3 as

xmk

dtxd

2

2

(4)

If we denote the ratio mk

with the symbol 2 , then Equation (4) can be written in the

form

xdt

xd 22

2

(5)

One may show that the solution x(t) is sinusoidal, with one form of the solution being

0 tcosAtx (6)

where

A: amplitude of the oscillation (maximum distance from equilibrium position).

0 : initial phase angle.

: angular frequency of the oscillation.

If we pull the mass m to stretch the spring by a distance A, and then release it from rest,

the initial conditions on the motion (at t = 0) are

0000 0 ;v;Ax (7)


21

Applying these initial conditions to Equation (6) give the equation of motion of the

spring-mass system as

tcosAtx (8)

The parameter that we used is the angular frequency of oscillation. Therefore, the

frequency of oscillation of the mass is

mk

f

21

2 (9)

And the period of oscillation is

km

fT 2

1 (10)

Squaring both sides,

km

T 22 4 (11)

This relation of period squared to mass can be written as a linear equation. So a graph of

T2 versus mass should be a straight line with the slope is written as

kSlope

24 (12)

3- Setup

Setup the experiment as shown in Figure 1.


1- Displace the mass a small distance and let it oscillate

2- Record the time for the mass to make 20 complete oscillations (20 T).

3- Repeat steps 1 and 2, for a total three recorded times (see Table 1)

4- Repeat the experiment with different masses as shown in Table 1.


22

Table 1. Determination of the period of oscillations as function of masses.

m (kg) 20 T1 (s) 20 T2 (s) 20 T3 (s)

60202020 321 TTT

T

(s) T

2 (s

2)

0.05

0.10

0.15

0.20

0.25

5- Prepare a sheet of graph paper for plotting T2 versus m. You should make T

2 the

vertical axis and m the horizontal axis.

6- Plot the measured values on a T2-m graph.

7- What is the slope from the linear regression fit?

8- Calculate the spring constant from this slope (use Equation 12)


23

Experiment 7

Linear motion and Newton’s second law

1-Objects of the experiment:

- Measuring the time required by a trolley of mass m1 to cover a certain path "d".

- Representing the relation between path and time in an d-t2 diagram.

- Calculating the acceleration "a" of the trolley of mass m1 with different masses of the

falling object of mass m2.

2-Principles

Figure 1.

If the acceleration is constant, we can use the following kinematics equation:

20 2

1attvd (1)

v0=vA= 0, then

2

21

atd (2)

Rearrangement of Equation 2 gives us:

da

t22 (3)

The Newton’s second law:

2m

a

a

AB xxd A B

1m


24

amF

1 (4)

where F

is the resultant force exerted on the mass m1 (or m2) and a

is its

acceleration.

Fig. 2: Free- body diagrams for the two masses:

By using (Equation 4), we can find the acceleration as:

gmm

ma

21

2

(5)

3- Carrying out the experiment

- Align the track horizontally.

- Adjust the voltage at the holding magnet so that the trolley with the additional weight is

just held.

- Define the starting point with the movable interrupter flag on the trolley, and read it

from the scale of the track.

- Position the light barrier at a distance of 20 cm from the starting point.

- Release the motion by pressing the START/STOP key at the stopclock.

- Wait until the interrupter flag passes the light barrier, and read the time from the

stopclock.

- Reset the stopclock to zero by pressing the RESET key.

- Repeat the measurement at distances 30 cm, 40 cm, 50 cm, and 60 cm from the starting

point.

4- Measurements

Two masses connected by a light cord.

n

T

1m

2m

motion

2m

1W

2W

T

1m


25

Table 1. Distance as a function of time with m1=0.486kg and m2=0.0252kg.

d (m) t1(s) t2 (s) t3 (s) Average t t2

0.2

0.3

0.4

0.5

0.6

- Graph distance d versus time squared t2 (d is the axis-x and t

2 is the axis-y)

- Draw the best line.

- Determine the acceleration "a" by finding a relation between the slope and the

acceleration (use Equation 3).

- Determine the acceleration "a" by repeating the measurement as above but with m2=

0.0452kg.

Table 2. Distance as a function of time with m1=0.486kg and m2=0.0452kg.

d (m) t1(s) t2 (s) t3 (s) Average t t2

0.2

0.3

0.4

0.5

0.6

- Discuss your results.


26

Experiment 8

Free fall-Conservation of mechanical energy


To observe the changes in potential energy, kinetic energy, and total mechanical energy

of a rolling body, and to ascertain graphically whether the total mechanical energy

remains constant.

2- Principles

Kinetic energy of a mass m moving with speed v, is defined as:

2

21

mvK (1)

The product of the magnitude of the gravitational force mg acting on an object and the

height h of the object is named the gravitational potential energy U, and so the defining

equation for gravitational potential energy is

hgmU (2)

An object held at some height h above the floor has no kinetic energy (v=0).

However, the gravitational potential energy of the object-Earth system is equal to mgh. If

the object is dropped, it falls to the floor; as it falls, its speed and thus its kinetic energy

increase, while the potential energy of the system decreases. In other words, the sum of

the kinetic K and potential energies U – the total mechanical energy E – remains

constant. This is an example of the principle of conservation of mechanical energy.

UKE (3)


27

Figure1. Experiment’s setup.

2

2

2121

BBB

AAA

mvmghE

mvmghE (4)

)21

()21

( 22AABBAB mvmghmvmghEEE

)(21

)( 22ABAB vvmhhmgE

vA is equal to zero because at point A, the trolley is at rest (vA=0), and hhh AB

It follows that:

2

21

BmvmghE (5)

where

2

21

BAB mvKKK (6)

mghUUU AB (7)

In Equation (5), 0E represents the condition for the mechanical energy to be

conserved.

4- Method and results

- Setup the experiment as shown in Figure 1.

netelectromag

ballsteelcounterdigital

clamp

cmy 5.3

0y

barrierslightForked2

A

B

Ah

Bh

BA hhh


28

- Position the two combination light barriers in such a way that they touch each other and

the half distance between them corresponds to the position hB.

- Make the distance h=hA-hB=0.3m

- The distance between the two light barriers is equal to Δy=0.035m and it is fixed. To

determine the speed at point B, you should follow this expression:

avrB t

yv

- Release the motion by pressing the START/STOP key at the counter S.

- Write down the time from the counter S.

- Reset the counter S to zero by pressing the RESET key.

- Position the two combination light barriers at other distances as shown in Table 1 and

repeat the measurement as mentioned above.

Table 1. Time ti as function of height h. Take the position hA always constant.

hA =

BA hhh (m) t1 (s) t2 (s) t3 (s) tavr (s) Bv (m/s) ΔK (J) ΔU (J) ΔE(J)

0.3

0.4

0.5

0.6

0.7

- Prepare a sheet of graph paper for plotting ΔK, ΔU and ΔE versus d. You should make

d the horizontal axis, and ΔK, ΔU and ΔE the vertical axis.

- Plot the measured values.

- Draw the three best fit lines to the points on your graph.

- Determine the slopes, SΔK, SΔU and SΔE of best fit lines.

5- Conclusions

Discuss your results


29

Experiment 9

Conservation of mechanical energy of a uniformly

accelerated mass


To observe the changes in potential energy, kinetic energy, and total mechanical energy

of a rolling body, and to ascertain graphically whether the total mechanical energy

remains constant.

2- Principles

Kinetic energy of a mass m moving with speed v, is defined as:

2

21

mvK (1)

The product of the magnitude of the gravitational force mg acting on an object and the

height h of the object is named the gravitational potential energy U, and so the defining

equation for gravitational potential energy is

hgmU (2)

An object held at some height h above the floor has no kinetic energy (v=0).

However, the gravitational potential energy of the object-Earth system is equal to mgh. If

the object is dropped, it falls to the floor; as it falls, its speed and thus its kinetic energy

increase, while the potential energy of the system decreases. In other words, the sum of

the kinetic K and potential energies U – the total mechanical energy E – remains

constant. This is an example of the principle of conservation of mechanical energy.

UKE (3)


31

Figure1. Setup of the experiment.

2

2

2121

BBB

AAA

mvmghE

mvmghE (4)

)21

()21

( 22AABBAB mvmghmvmghEEE

)(21

)( 22ABAB vvmhhmgE

vA is equal to zero because at point A, the trolley is at rest (vA=0), and hhh AB

It follows that:

2

21

BmvmghE

2

21

sin BmvmgdE (5)

where

2

21

BAB mvKKK (6)

sinmgdUUU AB (7)

In Equation (5), 0E represents the condition for the mechanical energy to be

conserved.

Ah

dxx AB

Bh

B

A

Track

Trolley

BA hhh

x

g

y

barrierslight n combinatio 2


31

3- Equipments

Apparatus Catalogue Number

Multi-core cable, 6-pole Track Trolley Counter S Steel tape measure Two combination light barriers

4- Method and results

- Make the track inclined with an angle 10 .

- Define the starting point with the movable interrupter flag on the trolley as xA.

- Position the two combination light barriers in such a way that they touch each other and

the half distance between them corresponds to the position xB. The distance between the

two light barriers is equal to Δx=0.047m and it is fixed. To determine the speed at point

B, you should follow this expression:

avrB t

xv

- Release the motion by pressing the START/STOP key at the counter S.

- Wait until the interrupter flag passes the light barriers, and write down the time from the

counter S.

- Reset the counter S to zero by pressing the RESET key.

- Position the two combination light barriers at other distances from the starting point as

shown in Table 1 and repeat the measurement as mentioned above.

Table 1. Time ti as function of distance d. Take the position xA always constant.

AB xxd (m) t1 (s) t2 (s) t3 (s) tavr (s) Bv (m/s) ΔK (J) ΔU (J) ΔE(J)

0.3

0.4

0.5

0.6

0.7


32

- Prepare a sheet of graph paper for plotting ΔK, ΔU and ΔE versus d. You should make

d the horizontal axis, and ΔK, ΔU and ΔE the vertical axis.

- Plot the measured values.

- Draw the best fit lines to the points on your graph.

- Determine the slopes, SΔK, SΔU and SΔE of best fit lines.

5- Conclusions

Discuss your results.


33

Experiment 10

Parallel and series connection of resistors


- Confirming Ohm's law.

- Determining the equivalent resistance of two resistors in parallel connection and

comparing with the resistances of the individual resistors.

- Determining the equivalent resistance of two resistors in series connection and

comparing with the resistances of the individual resistors.

2- Principles

The current flowing through a conductor is proportional to the potential difference

applied to its ends, it therefore fulfills the equation:

IV

RVR

I

1

(1)

In SI units, resistance is expressed in volts per ampere. A special name is given: Ohms

().

When two resistors with the resistances R1 and R2 are parallel connected, they

take the same voltage: V1=V2

The applied current is the sum of the individual currents I1 and I2: I=I1+I2.

The resistances of the parallel connection, therefore fulfill the equation:

21

1111RRRR peq

(2)

In a series connection, both resistors take the same current I=I1=I2. The applied

voltage is the sum of the individual voltages V= V1+V2. The resistances of the series

connection therefore fulfill the equation:


34

21 RRRR seq (3)

3- Equipments

- Plug-in board A4 (Do not connect DC voltages greater than 60V or AC voltages greater

than 25V)

- Set of resistors

- Power supply

- Ammeter

- Voltmeter

- Potentiometer 1k, 3W (variable resistor)

- Set of bridging plugs

- Connection wires


4-1- Confirming the law of connecting resistors in series

In this section we measure the current through a series connected resistors R1=100, and

R2=150.

Figure 1. Connection for measuring the resistance Rs of the series connected resistors

- Turn the power supply off, and then turn the voltage to its minimum setting.

1001R

V

AVV 121

Potentiometer

abc

Ammeter

Voltmeter

1502R


35

- Realize the electric circuit as shown in Figure 1.

- Turn the ammeter and the voltmeter on. Make sure they are set to “DC” and not “AC”.

- Turn the potentiometer to the maximum position.

- Switch on the power supply and turn it to its value (V=12V) by using a voltmeter.

- Turn the potentiometer slowly knob down until you get the desired current.

- Record the voltage and the current in Table 1. Make a series of five measurements.

- Use Ohm’s law to find the resistance Rs of the series connected resistors at each couple

of values I and V.

Table 1. Current-voltage characteristics of a resistance Rs of the series connected

resistors

I (A) V (V) )(

IV

Rs

- Prepare a sheet of graph paper for plotting I versus V (I is the y-axis and V

is the x-axis)

- Plot the measured values on a I - V graph.

- Determine the slope from the graph.

- Find a relationship between the slope and Rs.

- Determine the value of Rs.

- Compare the measured resistance with the resistance calculated from equation 3 by

using the following equation:

100error %,

,,

calculateds

calculatedsmeasureds

R

RR

4-2- Confirming the law of connecting resistors in parallel

In this section we measure the current through a parallel connected resistors R1=100,

and R2=150.


36

Figure 2. Connection for measuring the resistance Rp of the parallel connected resistors

- Turn the power supply off, and then turn the voltage to its minimum setting.

- Realize the electric circuit as shown in Figure 2.

- Turn the ammeter and the voltmeter on. Make sure they are set to “DC” and not “AC”.

- Turn the potentiometer to the maximum position.

- Switch on the power supply and turn it to its value (V=12V) by using a voltmeter.

- Turn the potentiometer slowly knob down until you get the desired current.

- Record the voltage and the current in a table like the one below. Make a series of five

measurements.

- Use Ohm’s law to find the resistance Rp of the parallel connected resistors at each

couple of values I and V.

Table 2. Current-voltage characteristics of a resistance Rp of the parallel connected

resistors

I (A) V (V) )(

IV

Rp

1001R

V

AVV 121

Potentiometer

abc

Ammeter

Voltmeter

1502R


37

- Prepare a sheet of graph paper for plotting I versus V (I is the y-axis and V

is the x-axis)

- Plot the measured values on a I – V graph.

- Determine the slope from the graph.

- Find a relationship between slope and Rp.

- Determine the value of Rp.

- Compare the measured resistance with the resistance calculated from equation 2 by

using the following equation:

100error %,

,,

calculatedp

calculatedpmeasuredp

R

RR

5- Conclusions



38

Experiment 11

Resistance and Resistivity of a Wire


- To find the resistance, xR , of a given wire using a meter bridge.

- Determine the resistivity, , of the material of the wire.

2- Principles

The meter bridge consists of a one meter long wire (wire 2) of uniform cross sectional

area, fixed on a wooden block. A scale is attached to the block. Two gaps DE and HI are

formed on it by using thick metal strips in order to make the Weatstone’s bridge. The

terminal F between the gaps is used to connect a galvanometer, G, and a jockey.

Figure1. Setup of the experiment.

A resistance wire (wire 1) is introduced in the gap DE and a known resistance is in the

gap HI. One end of the galvanometer is connected to terminal F and its other end is

connected to a jockey. As the jockey slides over the wire 2, it shows zero deflection in

the galvanometer at the balancing point B (null point). The length of the wire 2 is taken to

be equal to 100cm.

If the length AB is (in cm), then the length BC is (100- ) cm.

G

RxR

cm100

SV

A B

F

ED

C

H I

1 Wire

2 Wire

1I2I


39

The resistance of the portion and that of the portion (100- ) of the wire 2,

respectively, are

ARAB

(1)

ARBC

100 (2)

where : resistivity of the wire 2 (in Ω.cm)

A: cross sectional area of the wire 2 (in cm2)

21 IRIRVVVV ABxBAED (3)

21 IRIRVVVV BCCBIH (4)

BC

ABx

BC

ABx

RR

R

R

IRIR

RI

IR

2

2

1

1 (5)

100100BC

AB

BC

AB

RR

A

ARR

(6)

From (5) and (6), then

RRx

1100

(7)

The resistivity of the material of the wire 1 can be then calculated by using the relation:

L

Rr x2

(8)

Where L is the length of the wire 1 and r is the radius of its sectional area A.

3- Equipments

- Meter bridge

- Galvanometer with jokey (G)

- Switch (S)

- Set of resistors (100Ω, 150Ω, 330Ω, 470Ω)

- Power supply


41

- Precision micrometer (31183)

- Connection wires


1. Measure of the length, L, of the wire 1.

2. Measure the radius, r, of the cross sectional area of the wire 1 by using the precision

micrometer.

3. Connect the apparatus as shown in Figure 1 by choosing R = 100Ω

4. Adjust the value of the power supply to a value between 4 and 5V.

5. Move the jockey on the wire 2 until you reach the null point B for which no deflection

in the galvanometer (don’t slide the jockey continuously on the wire).

6. Take the reading of on B.

7. Repeat the measurement up to 3 times, and calculate the mean value of the measured

lengths.

8. Repeat the steps 2 to 7 for different values of resistors.

9. Record your measurements in the following table,

R (Ω) 1 (cm) 2 (cm) 3 (cm)

3321

(cm)

100

100

150

330

470

10. Prepare a sheet of graph paper for plotting

100 versus R

(

100

is the y-axis

and R is the x-axis)

11. Plot the measured values on a

100 – R

graph.

12. Determine the slope from the graph.

13. Find a relationship between the slope and xR .


41

14. Determine the value of xR .

15. Determine the resistivity, , of the material of the wire.

5- Conclusions


Documents

Experiment 1 Virtual experiment - Imam U · - Prepare a sheet of graph paper for plotting t2 versus d. You should make t2 the vertical axis and d the horizontal axis. - Plot the measured