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Solving problems is an essential part of the understanding

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the accelerated motion under gravity. The questions are categorized in terms of the characterizing features of the subject matter :

  • Motion plots
  • Equal displacement
  • Equal time
  • Displacement in a particular second
  • Twice in a position
  • Collision in air

Motion plots

Problem : A ball is dropped from a height of 80 m. If the ball looses half its speed after each strike with the horizontal floor, draw (i) speed – time and (ii) velocity – time plots for two strikes with the floor. Consider vertical downward direction as positive and g = 10 m / s 2 .

Solution : In order to draw the plot, we need to know the values of speed and velocity against time. However, the ball moves under gravity with a constant acceleration 10 m / s 2 . As such, the speed and velocity between strikes are uniformly increasing or decreasing at constant rate. It means that we need to know end values, when the ball strikes the floor or when it reaches the maximum height.

In the beginning when the ball is released, the initial speed and velocity both are equal to zero. Its velocity, at the time first strike, is obtained from the equation of motion as :

v 2 = 0 + 2 g h v = ( 2 g h ) = ( 2 x 10 x 80 ) = 40 m / s

Corresponding speed is :

| v | = 40 m / s

The time to reach the floor is :

t = v - u a = 40 - 0 10 = 4 s

According to question, the ball moves up with half the speed. Hence, its speed after first strike is :

| v | = 20 m / s

The corresponding upward velocity after first strike is :

v = - 20 m / s

Speed – time plot

The ball dropped from a height looses half its height on each strike with the horizontal surface.

Velocity – time plot

The ball dropped from a height looses half its height on each strike with the horizontal surface.

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

t = v - u a = 0 + 20 10 = 2 s

After reaching the maximum height, the ball returns towards floor and hits it with the same speed with which it was projected up,

v = 20 m / s

Corresponding speed is :

| v | = 20 m / s

The time to reach the floor is :

t = 2 s

Again, the ball moves up with half the speed with which ball strikes the floor. Hence, its speed after second strike is :

| v | = 10 m / s

The corresponding upward velocity after first strike is :

v = - 10 m / s

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

t = v - u a = 0 + 10 10 = 1 s

Problem : A ball is dropped vertically from a height “h” above the ground. It hits the ground and bounces up vertically to a height “h/3”. Neglecting subsequent motion and air resistance, plot its velocity “v” with the height qualitatively.

Solution : We can proceed to plot first by fixing the origin of coordinate system. Let this be the ground. Let us also assume that vertical upward direction is the positive y-direction of the coordinate system. The ball remains above ground during the motion. Hence, height (y) is always positive. Since we are required to plot velocity .vs. height (displacement), we can use the equation of motion that relates these two quantities :

v 2 = u 2 + 2 a y

During the downward motion, u = 0, a = -g and displacement (y) is positive. Hence,

v 2 = - 2 g y

This is a quadratic equation. As such the plot is a parabola as motion progresses i.e. as y decreases till it becomes equal to zero (see plot below y-axis).

Similarly, during the upward motion, the ball has certain velocity so that it reaches 1/3 rd of the height. It means that ball has a velocity less than that with which it strikes the ground. However, this velocity of rebound is in the upwards direction i.e positive direction of the coordinate system. In the nutshell, we should start drawing upward motion with a smaller positive velocity. The equation of motion, now, is :

v 2 = 2 g y

Again, the nature of the plot is a parabola the relation being a quadratic equation. The plot progresses till displacement becomes equal to y/3 (see plot above y-axis). The two plots should look like as given here :

Velocity – displacement plot

The nature of plot is parabola.

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Questions & Answers

anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
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Akash Reply
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characteristics of micro business
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for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
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what's the easiest and fastest way to the synthesize AgNP?
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Cied
types of nano material
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Source:  OpenStax, Kinematics fundamentals. OpenStax CNX. Sep 28, 2008 Download for free at http://cnx.org/content/col10348/1.29
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