One mark questions with answers

Q1. If a body of mass 2kg. is at restand is hit by a mass of 4kg. moving with 3m/s, find fraction of the momentumretained by the moving body assuming the collision to be elastic and head-on.

Ans1. If n = m₂/m₁
(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. If mass of the moving body ismuch greater than the mass of the body at rest than what is the approximatevilocity of the moving body after head-on collision?

Ans2. When the moving mass is muchgreater than the mass at rest then after the collision the heavier mass keepson moving with the same velocity and in the same direction.

Q3. At what point the potentialenergy of a body is taken to be zero?

Ans3. The potential energy of the bodyat the surface of the earth is taken to be zero (Potential energy = mgh, whereh is the height of the body from the surface.).

Q4. Does the work done on a body by aforce depend upon the path followed by it?

Ans4. May or may not be. If the force isconservative then it does not depend but if it is non-conservative (friction)then it depends.

Q5. If a body hits the ground from aheight h₁ and rebounds to a height h₂ after havinginelastic collision with the ground then what is the coefficient ofrestitution?

Ans5. e = Ö(h₂/h₁)

Q6. A body hits the ground with 50m/s velocity and has inelastic collision with the ground then with whatvelocity it will rebound if the coefficient of restitution is 0.2.

Ans6. Coefficient of restitution, e = v₂/v₁where v₁ is the velocity with which the body hits the ground and v₂is the velocity of rebound.
e = 0.2 = v₂/50, so v₂ = 10 m/sec.

Q7. A body at rest explodes in threefragments. Is it possible that two equal parts move in mutually perpendiculardirections with the same velocity and third mass moves midway between the two?

Ans7. No, it is not possible becausemomentum before the explosion is zero and after it also the momentum must bethe same. In the given situation the third particle must go in a directionopposite to the resultant of first two parts.

Q8. If the speed of a moving vehicleis increased by 200% then how much should be the change in the retarding forceto stop the vehicle over half the previous distance?

Ans8. (1/2)mv² = F.S where Fis the retarding force and S is the distance over which the vehicle comes torest. When ‘v’ is increased by 200% then K.E. increases by 800%. As S is halvedthen F should be made 16 times.

Q9. If 20 Joules of work is done in compressinga spring from 0 cm to 6 cm then find the work done in compressing the same from3cm to 6 cm.

Ans9. W.D = (1/2) K(x₂²– x₁²). This is the work done when the spring iscompressed from x₁ to x₂.
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. A ball is dropped from rest at aheight of 20m. If it loses 30% of its kinetic energy on striking the ground,what is the height to which it bounces? How do you account for this loss inkinetic energy?

Ans1. Suppose the ball acquires avelocity ‘v’ after falling through a height of 20m.
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. Why is no energy being consumedin planetary motion.

Ans2. A planet is a heavenly body whichrevolves round the star (the sun). The force which is responsible for circularmotion, is called centripetal force. The direction of the centripetal force isalways towards the centre. Thus, the angle between force F and displacement Sis q = 90^o at every point.Work done in moving planet, W = F.S = FSCosq = FS Cos90^o.So, W = 0. Hence,no energyis being consumed in planetary motion.
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Q3. How will you justify thathydro-electric power plant is an illustration of law of conservation of energy?

Ans3. A hydroelectric power-plant isused in generating electric energy (Power).The potential energy of water storedat a height is converted into K.E. when water is made to rush down. This fallof water is used to rotate the turbine and the coil and armature of generatoris rotated and electricity is produced . Thus, the K.E. of the fall of water isconverted into electrical form of energy. Hence the hydroelectric power-plantis an example of law of conservation of energy.

Q4. If a body is dropped from aheight of 40m then after 3 inelastic collisions with the ground to which heightthe body will rise? (given: Coefficient of restitution = 0.5)

Ans4. If the body is dropped from aheight of ‘H’ and ‘e’ is the coefficient of restitution then after ‘n’inelastic collisions with the ground the body rises to a height ‘h’ given by
h = H.e²ⁿ.
h = 40 x (1/2)^{2 x 3}= 40 x (1/2⁶) = 40/64 = 0.625 m

Q5. A truck and a car are moving withthe same K.E. on a straight line road. If their engines are made off at thesame time, which one of them will stop at a lesser distance.

Ans5. Given : (1/2)m₁v₁²= (1/2)m₂v₂² ………….(i)
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. The power of a pump motor is 4KW. How much water  in kg/minute can it raise a height of 20m? (g = 10 m/s²)

Ans6. Given power of motor P = 4KW =4000 W
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. A rod of length 3m is suspendedvertically from a fixed point. It is given an angular displacement of 60^oin the vertical plane. If its mass per unit length is 2 kg then find the workdone?

Ans7. Let ‘m’ be the mass of the rod and’l’ be its length then
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. When a mass m₂is at rest and mass m₁ moving with velocity u₁ hits itelastically, show that the fraction of the momentum transferred to the mass atrest is 2n/(1 + n) where n is ratio of the masses.

Ans1. v₁ = (m₁ -m₂)u₁/(m₁ + m₂).
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. Is it possible for work to bepositive negative or zero? "Explain with example.

Ans2. Work is a scalar quantity which isgiven by the scalar product of the force applied F and the displacement S movedby body i.e.
W = F.S,
W = FS Cos q ……….(1).
If q = 0. The work is maximum.
It remains +ve for the angle q between 0^oto 90^o
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or, q lying between 270^o and360^o i.e., if  the displacement is in a directionoppisite to which the force is applied.
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Thus work is +ve if Cos q is +ve. The work done will be -veif Cosq is -ve i.e. q lies between 90^o to 270^o.
If q = 90^o then Cos 90^o= 0.
Hence work done W = FS Cos 90^o = 0.
Thus, W may be +ve, -ve or zero

Q3. The bob of a simple pendulum isreleased from a horizontal position. If the length of the pendulum is 2m, whatis the speed with which the bob arrives at the lowermost point? Given that itdissipates 10% of its initial energy against air resistance?

Ans3. Gravitational potential energy atthe highest position
= 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
(1/2)mv² = mg x18/10
or, v = 1.9 ms^-1.

Q4. Give the various gravitationalunits of work. Give their relations also.

Ans4. We know that the physical workdone is given by
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. What is 1 e.v (electron volt)?Give its values in Joules. Give values of 1 MeV and 1 BeV also.

Ans5. One electron volt is the unit ofenergy in Atomic and Nuclear Physics.
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. When water is flowing through apipe then its velocity changes by 5%, find the change in the power of water?

Ans6. Power = Force xVelocity = Rate of change of momentum x velocity ={(mass/time) x velocity} xvelocity = {(adv) x v} x v =adv³ where ‘a’ is area of cross section, ‘d’ is the density of waterand ‘v’ is the velocity of flow of water.
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
DP/P = 3.Dv/v
percentage change in power, DP/P x100 = 3 x 5%
= 15%.

Q7. The kinetic energy of rushing outwater from a dam is used in rotating a turbine. The pipe through which water isrushing is 2.4 meters and its speed is 12 m/sec. Assuming that whole of kineticenergy of the water is used in rotating the turbine, calculate the currentproduced if efficiency of the dynamo is 60% and the station transmits power at240 kV. Density of water = 10³ kg/m³.

Ans7. Given that
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)pr²(l/t) rv²
= (1/2)pr² rv³
= (1/2) x 3.14 x (1.2)²x 10³ x (12)³watt
= 3.9 x 10⁶ watt
Current in the transmissioncables
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 vehicle of mass m isaccelerated from rest when a constant power P is supplied by its engine; showthat :
(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)^1/2t^3/2.
(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. (a) Given that Power = Fv = P =constant
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
òv dv = ò(P/m) x dt
(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
òds = ò(2Pt/m)^1/2 dt
s = (2P/m)^1/2 x (2/3) xt^3/2 + C₂.
Now, as at t = 0, s = 0, so, C₂ = 0
s = (8P/9m)^1/2 t^3/2.
(c)
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(d)
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Q2. A simple pendulum of mass m and lengthl swings back and forth up to a maximum angle q₀ with the vertical. When at an angle q, what is its (a) potential energy, (b) kinetic energy, (c) speed, and(d) tension?

Ans2.
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Taking the reference level at the lowest point R, we have
h_P = l – l cos q₀ = l(1 – cos q₀)
h_Q = l – l cos q = l(1 – cos q)
So (a) potential energy at Q relative to R will be
PE = mgh_Q
PE = mgl(1 – cos q) ………(a)
(b) PE at P = mgh_P = 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 K_Q is the KE at Q,
then using eq. (i)
mechanical energy at Q = K_Q + mgl(1 – cos q) ……….(ii)
But by conservation of mechanical energy between P and Q
K_Q + mgl(1 – cos q) = mgl(1 -cos q₀)
i.e., K_Q = mgl(cos q– cos q₀) ……..(b)
(c) If v is the speed at point Q, from eq. (b)
1/2 x mv² = mgl(cos q – cos q₀)
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) + (1/2)mv².
Since the energy remains constant throughout, E = E_o.
mgl(1 – Cosq) + (1/2)mv² = mgl(1 -Cosq_o)
or mv² = 2mgl(Cosq – Cosq_o)
Therefore, tension ‘T’ at q would begiven by
T = mv²/l + mgCosq = mg Cosq + 2mg(Cosq– Cosq_o)
or T = 3mgCosq – 2 mgCosq_o.

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).