Chapter 2
Motion and Newton’s 2nd Law
2.1
Purpose
In this lab, we will explore the relationship between position, velocity and acceleration. In
this experiment, friction will be neglected. Constant (uniform) acceleration will be investigated.
2.2
Introduction
Speed is just how fast something is moving. It is measured in how far something moves in
some time. Speed is given by the equation:
Speed =
∆x
∆t
(2.1)
The symbol ’∆’ indicates a change is some quantity such as position (’x’) in this case. The
unit of speed is always distance divided by time. We might measure speed in miles/hour,
meters/second etc. In this lab we will use meters/second (m/s).
Speed is how far something moves in some time. Obviously this depends on the time
period we use. Things rarely move at constant speed. We measure the speed of something
over a very short time, we can talk about the instantaneous speed. Instantaneous speed
is like taking a snap shot of the speed. The average speed is what we have in equation 2.1
for some time period that may not be very small.
Just as the position changing gives us velocity, when velocity changes we have an acceleration. Acceleration is:
Acceleration = a =
change in velocity
∆v
=
time interval
∆t
(2.2)
Acceleration has units of distance divided by time2 . For this lab, we will use meter/second2
(m/s2 )
When we specify a speed and the direction, we are giving a velocity. Velocity is a
vector. Like all vectors it has magnitude (how much) and direction (which way). Acceleration
being the change in velocity is also a vector. In this lab, we will restrict the motion to one
7
dimension so the vector nature of velocity and acceleration are no so important. However,
you should remember that velocity and acceleration are vectors.
If the acceleration is constant, there are relatively simple relationships between position,
velocity and acceleration in one dimension. If an object at time zero (t = 0) is at the origin
(x = 0), the the velocity and position are given by:
v(t) = v0 + a · t
(2.3)
and:
1
(2.4)
x(t) = v0 · t + a · t2
2
where a is the constant acceleration, t is the time and v0 is the initial velocity of the object
at time zero. For this experiment, we will have motion in one dimension with zero initial
velocity.
Next we should define force. Force is simply a push or a pull. It can be from gravity
i.e. gravity pulls on a mass, a magnetic ’pull’ between a north and south magnetic pole or
simply you pushing a book across a table.
Newton’s 2nd law relates mass, acceleration and force. In words:
• The acceleration of an object is directly proportional to the net force acting
on the object, is in the direction of the net force and is inversely proportional
to the mass of the object.
Force
We can can write this as:
mass
F~ = m~a
acceleration
(2.5)
The units of force in the metric system is the newton (see below).
There are several things we should notice about this statement. The equation for Newton’s 2nd law tells us force is a vector like velocity or acceleration. When you push something,
you push in a direction e.g, north or down, as well as some amount e.g. 10 N. We can use
Newton’s 2nd law to figure out what a newton (N) is in terms of mass, time and distance.
Since force is mass times acceleration, the units of a newton can be broken down to kg m/s2 .
Finally, notice the statement of Newton’s 2nd law says ’net force’. This means all the forces
added together acting on an object. Since they are vectors, we have to add all the forces
vectors.
We will investigate Newton’s second law and constant acceleration with the air table in
the configuration shown in figure 2.1. The hanging mass, m, will pull the puck across the
table as it falls. The mass of the puck, mp is 0.535 kg. The string connecting the hanging
mass and the puck has a tension, T. We assume there is no friction between the table and
the puck.
The force on the puck is simply the tension (pull of the hanging mass). Assuming no
friction, Newton’s 2nd law gives:
F orce = T = mp a
(2.6)
8
a
mp
T
T
a
mg
Figure 2.1: The air table arrangement for the Newton’s 2nd law experiment.
The forces on the hanging mass are T (upward) and the force of gravity, mg (downward). ’g’
is the acceleration of gravity (9.81 m/s2 ). The acceleration of the hanging mass is downward.
Newton’s second law gives:
F orce = T − mg = m(−a)
(2.7)
If we substitute equation 2.6 into equation 2.7 and solve for the acceleration we have:
a=
mg
mp + m
(2.8)
This equation shows that the acceleration is constant and proportional to the acceleration
of gravity through the ratio ( mpm+m )
2.3
Equipment:
Air table, spark timer, air pump, pulley, string, hanging mass, pucks, ruler.
2.4
Procedure
Special Caution:
• To avoid electrical shock, do not touch the pucks or air table while the spark timer is
on. Use an insulator such as a rolled up sheet of paper to push or hold the puck.
The air table apparatus is shown in figure 2.2. The white box at the back of the table is
the spark unit. Two pucks are shown connected to the rubber hoses which supply the air to
float the puck.
2.4.1
Constant Velocity
In this section of the procedure, the puck will be travel with constant velocity across the air
table. Since the puck travels without significant friction across the air table, the puck will
move with constant speed after it is set in motion with a small push.
9
Figure 2.2: The air table with various components used in the experiment.
• Level the air table by turning on the air pump and placing one of the pucks at the
table’s center. Adjust the table legs so the puck doesn’t move.
• Turn on the sparker power supply and set the spark timer to 20 Hz. Do not at this
time depress the sparker pad.
• The air table has a black graphite sheet lying on it. On top of the graphite sheet is a
sheet of white paper that will have, on its side adjacent to the graphite, black sparker
dots marking successive positions of a pin located at the puck’s center. Make several
folds in the white paper at an upper corner. On top of the folds place one of the two
pucks. This puck provides the return path for the sparker current. The folds keep the
puck from moving across the table.
• Turn on the air supply. Place the free puck near one side of the table. Using the
insulated tube, gently push the puck across the table and press the sparker pad. Release
the sparker pad before the puck collides with the side of the table.
• Turn off (for safety) the sparker power supply, remove both pucks, lift up the paper
sheet and turn it over to examine the sparker dots marking the puck trajectory.
• Using the ruler, measure the distance between the first dot on the paper and each
successive dot. Make a table of position and time. Consider the first dot at time zero
and each successive dot as 0.05 seconds after the previous dot.
• Plot position (y axis) vs time (x axis). Using the ruler, draw a line that comes closest
to all the points (’best fit’). Take the slope of the line. The slope of the line is the
velocity of the puck. Why?
2.4.2
Newton’s Second Law
In this section of the procedure, the puck will be pulled across the table by a hanging mass
connected to the puck by a string over a pulley.
10
• Attach the pulley to the side of the table if it is not already attached. Connect the
string to the collar and the other end to the hanging mass. Place the string over the
pulley wheel.
• Using the unmarked side the the white paper, configure the free puck on several folds
of the corner of the paper so it will not move. Hold the puck connected to the hanging
mass at the far end of the table with the insulated tube. Press the sparker pad and
release the puck. The puck should have zero initial speed so the puck should not be
released until the sparker has started. Stop the spark timer when the mass reaches the
end of the table.
• Turn off (for safety) the sparker power supply, remove both pucks, lift up the paper
sheet and turn it over to examine the sparker dots marking the puck trajectory.
• Qualitatively describe how the dots are spaced on the paper.
• Measure the distance between the first sparker dot an another dot approximately 0.20
meters from the first dot. Count the number of time intervals (0.05 seconds between
dots) to determine the total time between the first dot and the dot where you measure
the distance. Using equation 2.4, calculate the acceleration assuming v0 = 0.
• Using equation 2.8, calculate the acceleration of gravity. Calculate the percentage error
for your value with the standard value of 9.81 m/s2
2.5
Questions
• In the constant acceleration part of the experiment (Newton’s 2nd law), if the initial
velocity is not zero would the final result be serious effected?
• In the constant acceleration part of the experiment (Newton’s 2nd law) there may be
some friction between the puck and air table. Assuming the friction is constant, would
you still expect the acceleration to be constant? Would it change the value of g you
calculated?
11
12