**One mark questions with answers**

**Q1.**

**Ans1.**_{2}/m_{1}

(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.**_{1} and rebounds to a height h_{2} after havinginelastic collision with the ground then what is the coefficient ofrestitution?

**Ans5. **_{2}/h_{1})

**Q6.**

**Ans6. **_{2}/v_{1}where v_{1} is the velocity with which the body hits the ground and v_{2}is the velocity of rebound.

e = 0.2 = v_{2}/50, so v_{2} = 10 m/sec.

**Q7.**

**Ans7. **

**Q8.**

**Ans8. **^{2} = 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.**

**Ans9.**_{2}^{2}- x_{1}^{2}). This is the work done when the spring iscompressed from x_{1} to x_{2}.

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, v^{2} = u^{2} + 2as

= 0 + (2 ** x** 10

**20) = 400**

*x*So, v = 20 m/s

Kinetic energy of the ball just before hiting the ground

= (1/2)mv

^{2}= (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.**^{o} at every point.Work done in moving planet, W = **F.S** = FSCos^{o}.So, W = 0. Hence,no energyis being consumed in planetary motion.

**Q3.**

**Ans3.**

**Q4.**

**Ans4.**

h = H.e^{2n}.

h = 40 ** x** (1/2)

^{2 x 3}= 40

**(1/2**

*x*^{6}) = 40/64 = 0.625 m

**Q5.**

**Ans5.**_{1}v_{1}^{2}= (1/2)m_{2}v_{2}^{2} .............(*i*)

If force of brakes be the same then m_{1}a_{1} = m_{2}a_{2}..........(*ii*)

If truck stops over a distance S_{1} then v_{1}^{2} =2a_{1}S_{1} ........(*iii*)

If car stops over a distance S_{2} then v_{2}^{2} = 2a_{2}S_{2}.........(*iv*)

From (*i*) and (*ii*)

(1/2)m_{1}v_{1}^{2} = (1/2)m_{2}v_{2}^{2}...........(*v*)

From (ii) and (v)

v_{1}^{2}/a_{1} = v_{2}^{2}/a_{2}............(*vi*)

From (*iii*), (*iv*) and (*vi*)

2a_{1}S_{1}/a_{1} = 2a_{2}S_{2}/a_{2}.

S_{1} = S_{2}

Hence distances covered S_{1} and S_{2} are equal.

**Q6.**^{2})

**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

**20)/60**

*x*m = 1200 kg.

**Q7. **^{o}in the vertical plane. If its mass per unit length is 2 kg then find the workdone?

**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

**10**

*x***(3/2)(1 - Cos60) =45 J.**

*x*

Three mark questions with answers

**Q1.*** _{2}*is at rest and mass m

_{1}moving with velocity u

_{1}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.**_{1} = (m_{1} -m_{2})u_{1}/(m_{1} + m_{2}).

v_{2} = 2m_{1}u_{1}/(m_{1} + m_{2}).

This is from the theory of conservation of momentum.

Momentum of the mass m_{2} after collision,

P_{2} = m_{2}v_{2} = (2m_{1}m_{2}u_{1})/(m_{1}+ m_{2})

Fraction of momentum transferred to m_{2}.

= (2m_{1}m_{2}u_{1})/(m_{1} + m_{2})m_{1}u_{1}= 2m_{2}/(m_{1} + m_{2})

= 2n/(1 + n) ....[because m_{2}/m_{1} = n]

**Q2.**

**Ans2. ***i.e.*

W = F.S,

W = FS Cos q

If q

It remains +ve for the angle q between 0^{o}to 90^{o}

**or**, q^{o} and360^{o} *i.e.*, if the displacement is in a directionoppisite to which the force is applied.

Thus work is +ve if Cos q*i.e.* ^{o} to 270^{o}.

If q^{o} then Cos 90^{o}= 0.

Hence work done W = FS Cos 90^{o} = 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

**2 - (10/100)**

*x***mg**

*x***2] Joule**

*x*= mg

**18/10 J**

*x*\

^{2}= mg

**18/10**

*x*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

**1J/C = 1.6**

*x***10**

*x*^{-19}Joules.

**NOTE:**The other practical units used are

1 Million electron volt = 1 MeV = 10

^{6}eV, 1 MeV = 10

^{6}

**1.6**

*x***10**

*x*^{-19}J, 1 MeV = 1.6

**10**

*x*^{-13}Joules and

1 Billion eV = 10

^{9}eV, 1 BeV = 1.6

**10**

*x*^{-10}joules.

**Q6.**

**Ans6.**** x**Velocity = Rate of change of momentum

**velocity ={(mass/time)**

*x***velocity}**

*x***velocity = {(adv)**

*x***v}**

*x***v =adv**

*x*^{3}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 = Kv

^{3}(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

**100 = 3**

*x***5%**

*x*= 15%.

**Q7.**^{3} kg/m^{3}.

**Ans7.**

r = radius of pipe = 1.2m, average speed of water v = 12 m/s

V = 240 kV = 240 ** x** 10

^{3}volt, density ofwater p = 10

^{3}kg/m

^{3}.

Now, kinetic energy of rushing water per second i.e.

Power P = (1/2)(mass flowing per sec)

**v**

*x*^{2}

= (1/2)p

^{2}(l/t)

^{2}

= (1/2)p

^{2}

^{3}

= (1/2)

**3.14**

*x***(1.2)**

*x*^{2}

**10**

*x*^{3}

**(12)**

*x*^{3}watt

= 3.9

**10**

*x*^{6}watt

\

current = output power/voltage

= (60% of power P)/(240

**1000)**

*x*= [(60/100)

**3.9**

*x***10**

*x*^{6}]/(240

**1000) = 9.75 amp.**

*x*

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)^{1/2}t^{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.**

*i.e.*, m ** x** (dv/dt)

**v =P [as F = ma = m**

*x***(dv/dt)]**

*x*After rearranging and integrating on both sides

ò

**dt**

*x*(v

^{2}/2) = (P/m)

**t + C**

*x*_{1}

Now as initially the body is at rest,

*i.e.*, v = 0 at t = 0, so C

_{1}= 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

^{1/2 }dt

s = (2P/m)

^{1/2}

**(2/3)**

*x***t**

*x*^{3/2}+ C

_{2}.

Now, as at t = 0, s = 0, so, C

_{2}= 0

s = (8P/9m)

^{1/2}t

^{3/2}.

(c)

(d)

**Q2.**_{0} with the vertical. When at an angle q, what is its (a) potential energy, (b) kinetic energy, (c) speed, and(d) tension?

**Ans2.**

Taking the reference level at the lowest point R, we have

h_{P} = l - l cos q_{0}_{0}

h_{Q} = l - l cos q = l(1 - cos

So (a) potential energy at Q relative to R will be

PE = mgh_{Q}

PE = mgl(1 - cos q

(b) PE at P = mgh_{P} = mgl(1 - cos q_{0})

KE at P = 1/2 ** x** mv

^{2}= 0

so, total mechanical energy at P = mgl(1 - cos q

_{0}) .......(

*i*)

Now, if K

_{Q}is the KE at Q,

then using eq. (

*i*)

mechanical energy at Q = K

_{Q}+ mgl(1 - cos

*ii*)

But by conservation of mechanical energy between P and Q

K

_{Q}+ mgl(1 - cos q) = mgl(1 -cos q

_{0}

*i.e.*, K

_{Q}= mgl(cos q- cos q

_{0}

(c) If v is the speed at point Q, from eq. (b)

1/2

**mv**

*x*^{2}= mgl(cos

_{0})

*i.e.*, v = .

(d) If 'E' is the energy at Ðq, then itis equal to mgl(1 - Cosq

^{2}.

Since the energy remains constant throughout, E = E

_{o}.

mgl(1 - Cosq

^{2}= mgl(1 -Cosq

_{o}

or mv

^{2}= 2mgl(Cosq - Cos

_{o}

Therefore, tension 'T' at q would begiven by

T = mv

^{2}/l + mgCosq = mg Cos

_{o}

or T = 3mgCosq

_{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)**.