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Time Travel Again !
Davide Writes: (The email is posted as such. SO, sorry for the telegraphic language and poor spelling and grammar)
“Hello there i have one more question,a crazy one:) according to Einstein we need a enormous amount of energy to push matter at a speed of light,now for me the real problem there is that an atoms at a speed of light will break a part right?so even if we will find a way to push it,we will still failed.I was wander are we looking to the right way?is it photon, so the world of small with no mass only work? now here my crazy question,lets speculate a bit,we have two red dwarf pretty close to earth Barnard’s Star” only “5.9 ly and Wolf 359 7.7ly a baby step compere the universe,if we can send a same kind of mission the we have done with Stardust Nasa’s comet sample and collect a piece the size of a soccer ball back to earth,we could work on it outside earth atmosphere in the space base.Since we can’t go in the first place at a speed of light let’s says that the all deal will last 150 year to go and back,(assuming that we use a new energy to go bit faster)in the mean time the space base will be pretty developed and maybe we even found a new power to push to the speed of light(i remind you that we are speculate)we could put our hands on the sample using electromagnetism so that we avoid any risks and then send it at a seed of light.What i try to say is that a single atom is weak but billions of it together is a different story,here an example, lets imagine to make a spherical shape out of Lego(the toy) and send it in to space at a great speed,it will break pretty soon,now we make another one but we glue the piece together,it will last bit longer,again another with glue and a put net around it will last even longer so instead of a Lego we use our sample,the glue in the inside is the immense mass that will have (piece of red dwarf) net outside the immense gravitational pull ,now do you think that it will be enough to keep it together at a speed of light?regardless the energy that we need to push it? after all even in the past we believed that going faster than a speed of sound a plane will brake a part or something bad could happen,instead only a big BUM wen we reach it and nothing more,so maybe if we can go faster than light nothing will happen,or maybe we find a new way to travel,we can even give a name for fun “a speed of matter” what do you think about it?i know it sound like sci fi but hey physics it self is blowing mind science right? Thank you kindly for your time:)”
I don’t like to respond now.
Responses from visitors invited
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Problem from Kinematics
The speed of a train changes from36km/h to 72km/h in 10sec. calculate the distance travelled during this time
Answer:
u=36km/h = 10 m/s
v=72 km/h=20 m/s
t=10 s
S=?
use v = u+ at to find a
a = 1 m/s2
Use v2-u2=2aS to find S
S = (400-100)/2 = 150 m
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Ron writes on Light and Relativity
Let me start by saying I am NOT a physicist. That much should be painfully clear soon enough. Please bear with me as I am not certain I have the language and in depth knowledge to explain myself adequately.
A few years ago I began experiencing parallels between the world of human behavior and physics. At the time it seemed simple enough. That is until I started teaching myself physics. But at its core it seemed to make sense, the rules of life recycle themselves in different forms. Out of simple ideas comes complexity. Newton’s laws, thermodynamics… pascals principle, snells law, red and blue shift, wave particle duality and on and on… all seem to have their parallels in human behavior and on the surface seem to follow the same mathematical equations. Then I ran into the most famous of physics equations E=MC^2. It makes sense in human behavior. The energy (work) we can get out of an individual is relative to that individuals’ mass. The more matter we attach to the individual, knowledge, life experience etc (therefore increasing their mass) the more we can get out of them in the form of energy… but then the problem of the speed of light squared. The only known constant (light speed) is a problem in human behavior. At least to this point I know of no known behavioral constant.
I juggled the idea of it being a relative constant. Constant for the individual but relative as it would differ for everyone. The speed of cognition, or thought speed, would remain the same in potential throughout life of a given individual. Or that it was just a theoretical potential that humanity had yet to obtain. I even juggled the idea of it being a collective ability, but all of these options cause problems with the original equation. This has forced me to contemplate the nature of light and left me with questions I simply lack the understanding of physics to answer.
Is the speed of light truly constant or is it only constant as it relates to the big picture? As in: our perception of time as it relates to all time that has ever existed would appear as a single point in time. The older we get the more time seems to “fly by” if we as humans could continue to live for 13 billion years would the perception of an hour become so perceptively small that we wouldn’t even know it has passed? Is it therefore possible we can only understand and therefore measure the speed of light at a specific point in time, even if we try to come back and remeasure and compare the speed of light now with the speed of light fifty years from now the difference between the two would be imperceptivity small as a result of the displacement in time as it relates to the whole of time?
If the four dimensionality of time/space is linked shouldn’t time expand as space expands? And vice versa… if the speed of light is to remain constant as measured under such conditions isn’t light actually slowing down/speeding up over time as it relates to the whole of existing time/space? It’s a distance displacement problem…. If points A and B are actually farther apart but light travels the same distance in the same time the “speed” may seem constant but time has actually expanded to give the light more “time” to cover that distance. The speed of light would therefore be constant as a relation to perception and not as it relates to physical principles. Like the fact that the perception of time changes as we get older even though the actual measure of time remains the same.
How can light exist forever? At the speed of light we theoretically freeze time for that photon but that would also require an infinite amount of energy to obtain and maintain. So even a subatomic particle with a lifespan of a nanosecond would appear to exist for all eternity but in actuality would still only exist for a nanosecond. Because quite simply…it can’t have infinite energy and if it can I don’t understand how. Thermodynamics: no system is a perfect system and will experience energy loss, Newton: equal and opposite reaction, if something begins it has to end to balance the equation.
Is it therefore possible that light is born of the fourth dimension… we experience it in the dimensionality of space as long as it loses its energy to the three dimensions. A photon folds and pushes its way through space/time The initial energy of the photon is high and generates bigger leading waves which resist the photon holding it from passing the “speed of light” as the energy dissipates/photon begins to die those folds restricts the photon less allowing it to maintain the speed of light. We experience light because of its “ripples” in three-dimensions light dissipates as the waves of space become less folded in front of the photon. But this would mean the photon eventually loses enough energy that it can no longer be perceived in the third dimension…. So what happens to it?? Imagine : A man running through a corn field has to exert the energy to push the stalks aside but over time if the corn stalks are slowly spaced out even as he loses energy he can maintain his pace, because there is less impeding his path. Our perception of light would be like being in a helicopter looking down on the field. In the begging the field is densely packed with cornstalks and they slowly spread out until there is none. We only know the man is there while he is running through the corn because he pushes the stalks aside and we see that movement but once there are no stalks or he lacks the energy to continue running we no longer have a way of measuring his presence.
That brings me for some reason to a theory of Dark matter. Why? Well what happens to photons that no longer move with enough force to be visible. Like that guy running from a cornfield into an empty field. How would we know he is still there? If it was an infinite number of guys all stopping in that empty field we would know they are there by the depression their weight leaves in the field. Or more accurately by the stones in the field rolling towards a depression we cant actually see. The thought: An infinite number of “massless” (or perceived mass less) subatomic particles would still have infinite mass. Infinite mass would supply more then enough gravitational pull even spread out over infinite distance to cause the continued and speeding expansion of space as more and more visible photons “die” contributing its “dark masslessness”. Any dark matter existing within the universe would act as force… a moving invisible mass existing only on the 4th dimension pushing upon any objects in its way. But because energy propagated internally expanding outward in all dimensions would compound on the outside as those energy’s converge.
I hope you followed those questions. I don’t know what these ideas would do to theoretical physics but it would allow for the relativity of cognitive speed between individuals and reopen the door to “what the hell is dark matter in relation to human behavior” but that’s something different all together.
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WAVE MOTION QUESTIONS AND ANSWERS FOR CLASS XI
One mark questions with answers
Q1.
Ans1.
(n – 1)/(1 + n) is the fraction of the momentum retained by the moving body so
(n – 1)/(1 + n) = (2 – 1)/(1 + 2) = 1/3.
Q2.
Ans2.
Q3.
Ans3.
Q4.
Ans4.
Q5.
Ans5.
Q6.
Ans6.
e = 0.2 = v2/50, so v2 = 10 m/sec.
Q7.
Ans7.
Q8.
Ans8.
Q9.
Ans9.
20 = (1/2)K (36 – 0)
W = (1/2)K (36 – 9)
Solving the above equations we get W = 15 J.
Two mark questions with answers
Q1.
Ans1.
Because the ball is dropped from rest, hence u = 0.
Hence, v2 = u2 + 2as
= 0 + (2 x 10 x 20) = 400
So, v = 20 m/s
Kinetic energy of the ball just before hiting the ground
= (1/2)mv2 = (1/2)m(400) = 200m Joule
Because the ball loses 30% of the kinetic energy on striking the ground, hencekinetic energy retained by the ball after striking the ground = 70% of 200m J
= 140m J
The energy loss is due to the inelastic collision with the ground.
Q2.
Ans2.height=82src=”./xi%20work,%20power%20and%20energy_files/image001.gif”alt=”workpowerandenergyf06q16i.gif (1336 bytes)” v:shapes=”_x0000_i1027″>
Q3.
Ans3.
Q4.
Ans4.
h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m
Q5.
Ans5.
If force of brakes be the same then m1a1 = m2a2……….(ii)
If truck stops over a distance S1 then v12 =2a1S1 ……..(iii)
If car stops over a distance S2 then v22 = 2a2S2………(iv)
From (i) and (ii)
(1/2)m1v12 = (1/2)m2v22………..(v)
From (ii) and (v)
v12/a1 = v22/a2…………(vi)
From (iii), (iv) and (vi)
2a1S1/a1 = 2a2S2/a2.
S1 = S2
Hence distances covered S1 and S2 are equal.
Q6.
Ans6.
If mass of water raised in one second = m kg.
Total work done in lifting water,W = mgh
Power P = W/t, but t = 1 minute = 60 sec.
4000 = mgh/60
4000 = (m x 10 x 20)/60
m = 1200 kg.
Q7.
Ans7.
m = 2 x 3 = 6 kg
If the rod is displaced through an angle qthen the work done on it, W = mg(l/2)(1 – Cosq).
The effective length of the rod is taken to be (l/2) because in uniformdistribution of mass the centre of mass is at the geometric centre so
W = 6 x 10 x (3/2)(1 – Cos60) =45 J.
Three mark questions with answers
Q1.
Ans1.
v2 = 2m1u1/(m1 + m2).
This is from the theory of conservation of momentum.
Momentum of the mass m2 after collision,
P2 = m2v2 = (2m1m2u1)/(m1+ m2)
Fraction of momentum transferred to m2.
= (2m1m2u1)/(m1 + m2)m1u1= 2m2/(m1 + m2)
= 2n/(1 + n) ….[because m2/m1 = n]
Q2.
Ans2.
W = F.S,
W = FS Cos q
If q
It remains +ve for the angle q between 0oto 90oheight=45src=”./xi%20work,%20power%20and%20energy_files/image002.gif”alt=”tutor2phyworkpowerandenergyf06q2i1.gif (1062 bytes)” v:shapes=”_x0000_i1037″>
or, qheight=59src=”./xi%20work,%20power%20and%20energy_files/image003.gif”alt=”tutor2phyworkpowerandenergyf06q2i2.gif (1097 bytes)” v:shapes=”_x0000_i1038″>
Thus work is +ve if Cos q
If q
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero
Q3.
Ans3.
= mg x 2 Joules = 2mg Joules
Kinetic energy at lowest position
= Potential energy at the highest position – the energy dissipitated againstair resistance or friction
= [mg x 2 – (10/100) x mg x2] Joule
= mg x 18/10 J
or, v = 1.9 ms-1.
Q4.
Ans4.
W = FS
(1) In S.I system,
If F = 1 kg weight or 1 kg force and S = 1m then,
W = (1 kg wt)(1m) = 1 kg m ……………(i).
Hence, one kgm is the gravitational unit of work in S.I (M.K.S) system and isdefined as the amount of work done if 1 kg force displaces a body through 1m inthe direction of the applied force.
(2) In C.G.S system,
F = 1 gmwt and S = 1cm,
W = (1 gm wt) (1 cm) = 1 gm cm ………………(ii).
Hence, one gm cm is the gravitational unit of work and is defined as the amountof work done, if 1 gm force displaces a body through 1 cm in the direction ofthe applied force.
1 gm cm = 980 ergs.
NOTE: 1 kg m = 9.8 Joules.
Q5.
Ans5.
One electron volt is the energy acquired by one elctron in moving it betweentwo point having a P.D of 1V.
Thus, 1eV = (1.6 x 10-19) C x1J/C = 1.6 x 10-19 Joules.
NOTE: The other practical units used are
1 Million electron volt = 1 MeV = 106 eV, 1 MeV = 106 x1.6 x 10-19 J, 1 MeV = 1.6 x10-13 Joules and
1 Billion eV = 109 eV, 1 BeV = 1.6 x 10-10joules.
Q6.
Ans6.
Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv3 (k is a constant and is equal to ‘ad’.)Taking log on both sides
log P = 3log v + log k
Differentiating on both sides
D
percentage change in power, DP/P x100 = 3 x 5%
= 15%.
Q7.
Ans7.
r = radius of pipe = 1.2m, average speed of water v = 12 m/s
V = 240 kV = 240 x 103 volt, density ofwater p = 103 kg/m3.
Now, kinetic energy of rushing water per second i.e.
Power P = (1/2)(mass flowing per sec) x v2
= (1/2)p
= (1/2)p
= (1/2) x 3.14 x (1.2)2x 103 x (12)3watt
= 3.9 x 106 watt
current = output power/voltage
= (60% of power P)/(240 x 1000)
= [(60/100) x 3.9 x 106]/(240x 1000) = 9.75 amp.
Five mark questions with answers
Q1.
(a) The velocity is given as a function of time by
v = (2Pt/m)1/2
(b) The position is given as a function of time by
s = (8P/9m)1/2t3/2.
(c) What is the shape of the graph between velocity and mass of the vehicle ifother factors remain same?
(d) What is the shape of the graph between displacement and power?
Ans1.
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
ò
(v2/2) = (P/m) x t + C1
Now as initially the body is at rest, i.e., v = 0 at t = 0, so C1= 0.
v = (2Pt/m)1/2 …………(1)
(b) By definition v = (ds/dt),
Using eq (1) above,
ds/dt = (2Pt/m)1/2
On integration we get
ò
s = (2P/m)1/2 x (2/3) xt3/2 + C2.
Now, as at t = 0, s = 0, so, C2 = 0
s = (8P/9m)1/2 t3/2.
(c)height=104src=”./xi%20work,%20power%20and%20energy_files/image004.gif”alt=”5a1.gif (1325 bytes)” v:shapes=”_x0000_i1046″>
(d)height=102src=”./xi%20work,%20power%20and%20energy_files/image005.gif”alt=”5a1i.gif (1265 bytes)” v:shapes=”_x0000_i1047″>
Q2.
Ans2.height=149src=”./xi%20work,%20power%20and%20energy_files/image006.gif”alt=”5a2.gif (2041 bytes)” v:shapes=”_x0000_i1048″>
Taking the reference level at the lowest point R, we have
hP = l – l cos q0
hQ = l – l cos q = l(1 – cos
So (a) potential energy at Q relative to R will be
PE = mghQ
PE = mgl(1 – cos q
(b) PE at P = mghP = mgl(1 – cos q0)
KE at P = 1/2 x mv2 = 0
so, total mechanical energy at P = mgl(1 – cos q0) …….(i)
Now, if KQ is the KE at Q,
then using eq. (i)
mechanical energy at Q = KQ + mgl(1 – cos
But by conservation of mechanical energy between P and Q
KQ + mgl(1 – cos q) = mgl(1 -cos q
i.e., KQ = mgl(cos q– cos q
(c) If v is the speed at point Q, from eq. (b)
1/2 x mv2 = mgl(cos
i.e., v = height=17src=”./xi%20work,%20power%20and%20energy_files/image007.gif”alt=”i2.gif (1049 bytes)” v:shapes=”_x0000_i1049″>.
(d) If ‘E’ is the energy at Ðq, then itis equal to mgl(1 – Cosq
Since the energy remains constant throughout, E = Eo.
mgl(1 – Cosq
or mv2 = 2mgl(Cosq – Cos
Therefore, tension ‘T’ at q would begiven by
T = mv2/l + mgCosq = mg Cos
or T = 3mgCosq
Q3.What do youmean by work in the language of physics? Give its absolute and gravitationalunits. Give two illustrations of zero work, negative work and positive work.
Ans.(Try yourself).
Q4.How will youfind work done by a variable force mathematically and graphically?
Ans.(Try yourself).
Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.
Ans.(Try yourself).
Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.
Ans.(Try yourself).
Q7.If a body iskept on the top of a rough inclined plane, find the expression for
(i) work done in bringing it down to the bottom of the plane with constantvelocity
(ii) work done in moving it up the plane with constant acceleration
(iii) work done in moving it down the plane with constant acceleration.
Ans.(Try yourself).
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Potential
Devna asked.
find the potential at A ….a prson brings a mass of 1 kg from infinity to the pt A .initially the mass was at rest but it moves at a speed of 2 m/sas it reches A .the work done by person on mass is -3 J
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A Few Problems from Kinematics
- A body falling freely from rest has velocity v after it falls a height h . Calculate the distance it should fall down further for its velocity to become double
- A particle in uniform acceleration in a straight line has a speed of v m/s at position x meter is given by √(25-16x). What is the acceleration of the particle
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Average Speed
Naman Asks:
“A car travels from A to B with a speed of 30km/h and returns back to A with a speed of 50km/h.Find distance,displacement,average speed and average velocity ifdistance between A and B is 100km.”
Answer:
Distance 200 m
Displacement 0
Average velocity 0
Average speed = = 37.5 kmph
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