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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.
A ball is dropped from a height of 10m. If the kinetic energy of the ball reduces by 40% after striking the ground, how high can the ball bounce back? (Elsy James asked)
Two stones of mass 5 kg and 10 kg are dropped from the same height above the ground level. Which one will hit the ground first? Why?
Answer: Both the stones will reach the ground simultaneously (provided we neglect the air resistance (the viscosity) and buoyant force.
This is because acceleration due to gravity is independent of mass of the freely falling body.
- A worker pulls a 200. N packing crate with an applied force of 55.0 N. The crate accelerates at a rate of 0.250 m/s(squared). What is the coefficient of kinetic friction between the crate and the factory floor?
- A student pulls a 150. N sled up a 28 degree slope at a constant speed by applying a force of 100. N. Near the top of the hill he releases the sled. With what acceleration does the sled go down the hill?
- A 200. N crate rests on a ramp; the maximum angle just before it slips is 25 degrees with the horizontal. What is the coefficient of static friction between the crate and the surface of the ramp?
- A man pulls a sled with a weight of 200. N with a constant velocity across a horizontal snow surface. If a force of 80N is being applied to the sled rope at an angle of 53 degrees to the ground, what is the coefficient of friction between the sled and the ground?
- A jet plane is flying with a constant speed along a straight line at an angle of 30degree above the horizontal. The weight of plane is 86 500N. Its engine provides a forward thrust T of 103 000N. The lift of force L(directed perpendicular to the wings)and the force R of air resistance (directed opposite to the motion)act on the plane. Find L & R
George’s favorite food was bananas. One day he was walking with his friend, the man in the yellow hat, as he was eating a banana. They were in New York City and they were walking up the Empire State Building. Finally they made it to the top! George spied a nice young woman walking down 5th avenue balancing this hat on her head. She was very noticeable for she was six feet tall! George, being the troublemaker he was, wanted to see if he could drop his banana peel in her hat. He noticed that she was jogging at a constant pace of 6 miles per hour. George was on the 102nd floor observatory, which is 1,224 feet from the ground.*
Assuming that George lands the banana peel in the hat……..
A) How long will the banana peel be in the air? (Assuming there is no wind and the banana will drop straight down from the empire state building)
B) What will the velocity of the banana peel be right before it lands in the woman’s hat?
C) How far away must the woman’s hat be from the landing point of the banana peel (horizontally) when George makes his scandalous move and drops the peel.
D) If George decides to throw another peel one second later to land it on the ground and trip the jogger, at what velocity must he throw it at so that it hits the ground at the same the original peel lands in the hat.
- A particle travels 20 m in 7th second and 24m in 9th sec. find initial velocity?
- in a projectile motion , a body thrown from the ground ,at what angle both the vertical height and range will be equal?
- The velocity v(cm/s) of a particle is given in terms of time t(in seconds) by the equation v = at+b/t+c. Dimension of a,b, and c are ?
A rifle at a height H aimed horizontally fires a bullet to the ground. At the same time , a bullet with the same mass in dropped from the same height. Neglecting air resistance, which one hits the ground first?Explain.
Both will hit the ground simultaneously.
When a body is projected horizontally, its initial vertical velocity is zero and vertical acceleration is g, the acceleration due to gravity.
The values of a velocity and acceleration of a freely falling body are also the same.
So, both will hit the ground simultaneously