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One mark questions with answers
(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.
e = 0.2 = v2/50, so v2 = 10 m/sec.
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
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.
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h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m
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.
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.
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
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]
W = F.S,
W = FS Cos q
It remains +ve for the angle q between 0oto 90o
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Thus work is +ve if Cos q
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero
= 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.
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.
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.
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
percentage change in power, DP/P x100 = 3 x 5%
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) 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
(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?
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
(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
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.
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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.
Q4.How will youfind work done by a variable force mathematically and graphically?
Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.
Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.
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.
How does a pressure cooker work? What is the Physics behind it?
A pressure cooker develops a very high pressure when water boils inside. At high pressure the boiling point of water also rises and therefore food gets cooked at a higher temperature. Therefore the food materials are cooked well and faster than at normal temperature and pressure.
A pressure cooker is extremely useful in hilly areas where the pressure is very low and the boiling point is less for water. AT a lower boiling point, cooking is very difficult as proper cooking will not take place. When a pressure cooker is used, it increases the pressure as well as the boiling point and the food materials can be cooked well.
Derive an expression for the minimum horizontal velocity to be given to a ball hanging vertically from a point so that it is able to just complete a vertical circular path.
While pushing, we are giving the ball some kinetic energy which will be converted to potential energy as it moves upward. If the whole kinetic energy is converted to potential energy at the top most point, it will fall straight down, resulting in only a semicircle. For the ball to continue the path, it should have a centrifugal force equal to weight of the body when it is at the topmost point.
let V be the initially applied velocity and v be the velocity at the topmost point
mv²/r = mg
v²/r = g
v² = rg
v = √(rg)
Potential Energy at the topmost point , Ep = 2mgr
Ep is the difference in kinetic energy between initial point(Ei) and at topmost point(Et).
Ei – Et = Ep
½mV² – ½mv² = 2mgr
½m(V²-v²) = 2mgr
V²-v² = 4gr
V² = 4gr + v²
V² = 4gr + rg
V² = 5rg
“What is meant by potential difference?”
“Potential Difference between two points is the difference in Potential between the two points and is equal to the work done per unit charge ion carrying a positive test charge from one point to another against the electric field and without any acceleration”
We know that it is multiverse and not universe.we also know that a universe expands continuously.if so then wont there be a point when all the expanding universes collide each other and stop expanding?
The velocity m, v m/s, of a particle is a straight line is given by
v = 6+pt+qt³
where t is the time in seconds after the particle passes through a fixed point O.
Given that t=2 s the distance of the particle from O is 16 m and its acceleration is 32m/s², calculate
a) the value of p and of q
b) the velocity of the particle at the instant when did acceleration is zero
If a helicopter takes off from a point on earth.(at certain altitude) and its still at that point in air.After certain time interval the observer whose position was same as that of helicopter’s take off point will see the helicopter right above him.
The question arises is that how it is possible if earth is rotating hence the observer is also, but the helicopter is in air.How is it possible?