CH-9 (Centre of Mass)

Chapter # 9

1.

Centre of Mass, Linear Momentum, Collision

Objective - I

Consider the following two equations : fuEu nks lehdj.kksa ij fopkj dhft;s %    1  F m r (A) and rFkk (B) . ii aCM = R = M i M



In a non-inertial frame (A) both are correct (C*) A is correct but B is wrong

Sol.

fdlh vtM+Roh; funsZ'k ra=k esa (A) nksuksa gh lR; gSA (C*) A lR; gS fdUrq B vlR; gSA C

(B) both are wrong (D) B is correct but A is wrong (B) nksuksa gh vlR; gS

(D) B lR; gS] fdUrq A vlR; gSA

In a noninertial frame, positing the particle is not change. 2.

Sol.

Consider the following two statements : (a) linear momentum of the system remains constant (b) centre of mass of the system remains at rest. (A) a implies b and b implies a (B) a does not imply b and b does not imply a (C) a implies b but b does not imply a (D*) b implies a but a does not imply b

fuEu nks dFkuksa ij fopkj dhft, & (A) fdlh fo{ks; dk jSf[kd laosx fu;r jgrk gSA (B) fudk; dk laosx nzO;eku dsUnz fojkekoLFkk esa jgrk gSA (a) A,B dks O;Dr djrk gS rFkk B,A dks O;Dr djrk gSA (b) A,B dks O;Dr ugha djrk gS rFkk B,A dks O;Dr djrk ugha djrk gSA (c) A,B dks O;Dr djrk gS fdUrq B,A dks O;Dr ugha djrk gSA (d) B,A dks O;Dr djrk gS fdUrq A,B dk O;Dr ugha djrk gSA D

Centre of mass of the system   1 R   m i ri M

 1 dR  =0=  mi vi M dt It means linear momentum of the system remain constant. 3.

Consider the following two statements : (a) Linear momentum of the system of particles is zero (b) Kinetic energy of a system of particles is zero (A) a implies b and b implies a (B) a does not imply b and b does not imply a (C) a implies b but b does not imply a (D*) b implies a but a does not imply a

fuEu nks dFkuksa ij fopkj dhft, & (a) d.k&ra=k dk jSf[kd laosx 'kwU; gSA (b) d.k&ra=k dh xfrt ÅtkZ 'kwU; gSA (A) a,b dks O;Dr djrk gS rFkk b,a dks O;Dr djrk gSA (B) a,b dks O;Dr djrk gS fdUrq b,a dks O;Dr ugha djrk gSA (C) a,b dks O;Dr djrk gS fdUrq b,a dks O;Dr ugha djrk gSA (D) b,a dks O;Dr djrk gS fdUrq a,b dk O;Dr ugha djrk gSA

manishkumarphysics.in

[1]

Chapter # 9 Sol. D


[2]

Kinetic energy of the system = 1/2 m1v12 + 1/2 m2v22 + .............. = zero mean v1 = v2 = v3 = ................ = 0   Linear momentum of the system = m1v1 + m 2 v 2 + .................   v1  v 2  0 In this case   m1v1 = – m 2 v 2 (If two particle system) It is possible that 4.

Sol.

Consider the following two statements : (a) the linear momentum of a particle is independent of the frame of reference (b) the kinetic energy of a particle is independent of the frame of reference (A) both a and b are true (B) a is true but b is false (C) a is false but b is true (D*) both a and b are false

fuEu nks dFkuksa ij fopkj dhft, & (a) fdlh d.k dk jSf[kd laosx funsZ'k&ra=k ij fuHkZj ugha djrk gSA (b) d.k dh xfrt ÅtkZ funsZ'k&ra=k ij fuHkZj ugha djrh gSA (A) a rFkk b nksuksa lR; gSA (B) a lR; gS fdUrq b vlR; gSA (C) a vlR; gS fdUrq b lR; gSA (D) a o b nksuksa gh vlR; gSA D

Linear momentum & kinetic energy of a particle is dependent of the frame of reference. Because velocity is dependent on the frame of reference Linear momentum = mv K.E. = 1/2 mv2 5.

Sol.

All the particles of a body are situated at a distance R from the origin. The distance of the centre of mass of the body from the origin is fdlh oLrq ds leLr d.k ewy fcUnq ls R nwjh ij fLFkr gSA oLrq ds nzO;eku dsUnz dh ewy fcUnq ls nwjh & (A) = R (B*)  R (C*) > R (D)  R B

Distance of centre of mass from the origin 1 S  mr M  11 Half particle lies in y-axis   m3R j  m 4R j  ..........  ycom = m1  m 2  m3  m 4  ...........

R' =

 Rj = 2

R R Coordinate of com is  ,  2 2

Distance centre mass from the origin =

R ......... (1) 2 R' < R

When conclude Distance of centre of mass from the origin  R


Chapter # 9 Centre of Mass, Linear Momentum, Collision [3] 6. A circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be d O;kl okyh ,d o`Ùkkdkj IysV] d Hkqtk okyh oxkZdkj IysV ds lEidZ esa fp=kkuqlkj j[kh gqbZ gSA çR;sd LFkku ij inkFkZ dk ?kuRo ,oa eksVkbZ ,d leku gSA la;qDr fudk; dk nzO;eku dsUnz gksxk -

Sol.

(A) inside the circular plate (C) at the point of contact (A) o`Ùkkdkj IysV ds vUnj (B) oxkZdkj IysV ds vUnj

(B) inside the square plate (D*) outside the system (C) lEidZ fcUnq ij (D) fudk; ds ckgj

B

m1 = (d/2)2 t {t  thichness} 2 m2 = d t {mass = desily x volume} Centre of mass of circular plate at centre. Lett the centre of circular plate is origin.  r1 = 0 Centre of mass of square plate at centre.  m1 m2 r2 = 2d   (0,0) m1 r1  m 2 r2 m1  0  d 2 t  2d R= m m = 2 d / 2  t  d 2 t 1 2 d d 2d R= = (1.12) d  / 4 1 R>d Centre of mass of the system lies in the square plate. 7.

Consider a system of two identical particles. One of the particles is at rest and the other has an  acceleration a . The centre of mass has an acceleration.

nks ,d leku d.kksa ds fudk; ij fopkj dhft,A ,d d.k fLFkj gS rFkk nwljs d.k dk Roj.k gSA nzO;eku dsUnz dk Roj.k gksxkA (A*) zero 'kwU; Sol.

(B)

1  a 2

 (C) a

 (D*) 2 a

B

m rest

m

a

  m1a1  m 2a 2 Acceleration of centre of mass = m  m 1 2   m  0  ma a = = 2m 2 8.

Sol.

Internal forces can change vkUrfjd cy ifjofrZr dj ldrk gS & (A) the linear momentum but not the kinetic energy (B*) the kinetic energy but not the linear momentum (C) linear momentum as well as kinetic energy (D) neither the linear momentum nor the kinetic energy (A) jSf[kd laosx fdUrq xfrt ÅtkZ ugha (B*) xfrt ÅtkZ fdUrq jSf[kd laosx ugha (C) jSf[kd laosx o lkFk gh xfrt ÅtkZ Hkh (D) u rks jSf[kd laosx u gh xfrt ÅtkZ B

Internal forces can change the kinetic energy but not the linearmomentum. Because the centre of mass of the system remains at rest.


Chapter # 9 Centre of Mass, Linear Momentum, Collision [4] 9. A bullet hits a block kept at rest on a smooth horizontal surface and gets embedded into it. Which of the following does not change ? (A) linear momentum of the block (B) kinetic energy of the block (C*) gravitational potential energy of the block (D) temperature of the block

Sol.

{kSfrt ,oa fpdus ry ij j[ks gq, ,d xqVds ls canwd dh xksyh Vdjkdj blesa gh Qal tkrh gSA fuEu esa ls D;k ifjofrZr ugha gksxk & (A) xqVds dk jSf[kd laosx (B) xqVds dh xfrt ÅtkZ (C) xqVds dh xq:Roh; fLFkfrt ÅtkZ (D) xqVds dk rki C

Initial position

final position

(rest) m2 v m1 Linear momentum conservation m1 × 0 + m2v = (m1 + m2) v'

m1+m2

v’

m2v v' = m  m 1 2



Initial linear momentum of the block = m1 × 0 = 0 Final linear momentum of the block = m1v'  m1m 2  =  m  m v   1 2  Change in kinetic energy of the block = K.Ef – K.Ei = 1/2 m1v'2 – 0 2

 m1m 2  v  = 1/2 m1   m1  m 2  In this process energy temperature of the block increases. But gravitation potential energy of the block is remain same. Because height of the block is not change. 10.

A uniform sphere is placed on a smooth horizontal surface and a horizontal force F is applied on it at a distance h above the surface. The acceleration of the centre (A) is maximum when h = 0 (B) is maximum when h = R (C) is maximum when h = 2R (D*) is independent of h ,d le:i xksyk fpduh lery lrg ij j[kk gqvk gS rFkk bldh lrg ls h Åpk¡bZ ij ,d {kSfrt cy F yxk;k tkrk gSA dsUnz dk

Roj.k & Sol.

(A) h = 0 gksus ij vf/kdre gksxkA (C) h = 2R gksus ij vf/kdre gksxkA D

sphere

(B) h = R gksus ij vf/kdre gksxkA (D) h ij fuHkZj ugha djsxkA

F h

smooth On smooth surface, sphere is only slip. Acceleration of centre of sphere is = F/m which is independent of h. 11.

A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together shifts horizontally towards (A) heavier piece (B) lighter piece (C*) does not shift horizontally (D) depends on the vertical velocity at the time of breaking


Chapter # 9

Sol.


[5]

xq:Roh; cy ds çHkko esa m/oZ fn'kk esa uhps dh vksj fxj jgh ,d oLrq nks vleku nzO;ekuksa esa foHkDr gks tkrh gSA nksuksa VqdM+ks ds ,d lkFk ysus ij nzO;eku dsUnz {kSfrt fn'kk esa foLFkkfir gksxk & (A) Hkkjh VqdM+s dh vksj (B) gYds VqdM+s dh vksj (C) {kSfrt fn'kk esa foLFkkfir ugha gksxkA (D) m/okZ/kj osx rFkk VwVus ds le; ij fuHkZj C

A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the system does not shift horizontally.As external horizontal force is zero. 12.

Sol.

A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of mass (A) of the box remains constant (B*) of the box plus the ball system remains constant (C) of the ball remains constant (D) of the ball relative to the box remains constant

,d can ckWDl esa fLFkr xsan] ckWDl dh nhokjksa ls Vdjkrh gqbZ xfr dj jgh gSA ckWDl ,d fpduh lrg ij j[kk gqvk gS & (A) ckWDl ds nzO;eku dsUnz dk osx fu;r jgrk gSA (B) ckWDl ,oa xsan ds fudk; ds nzO;eku dsUnz dk osx fu;r jgrk gSA (C) xsan ds nzO;eku ds dsUnz dk osx fu;r jgrk gSA (D) ckWDl ds lkis{k xsan ds nzO;eku dsUnz dk osx fu;r jgrk gSA B

The velocity of the centre of masss of the system is ramains constant. (assume that collision will taken as perfectly elastic). 13.

Sol.

A body at rest breaks into two pieces of equal masses. The parts will move (A) in same direction (B) along different lines (C*) in opposite directions with equal speeds (D) in opposite directions with unequal speeds

fojkekoLFkk esa fLFkr ds nks Hkkxksa esa VwV tkrh gSA VqdM+s xfr djsaxs & (A) leku fn'kk esa (B) fHkUu&fHkUu js[kkvksa ds vuqfn'k ls (C) foijhr fn'kkvksa esa leku pky ls (D) foijhr fn'kkvksa esa vleku pky ls C

By linear momentum conservation pi = pf m×0=

m m v1 + v 2 2 2

v1 = – v2 14.

A heavy ring of mass m is clamped on the periphery of a light circular disc. A small particle having equal mass is clamped at the centre of the disc. The system is rotated in such a way that the centre moves in a circle of radius r with a uniform speed v. We conclude that an external force (A)

(C*)

mv 2 must be acting on the central particle r 2mv 2 must be acting on the system r

(B)

2mv 2 must be acting on the central particle r

(D)

2mv 2 must be acting on the ring. r

m nzO;eku dh ,d oy;] ,d gYdh pdrh dh ifjf/k ij dhyfdr dh x;h gSA leku nzO;eku dk ,d NksVk d.k pdrh ds dsUnz ij dhyfdr gSA ;g fudk; bl çdkj ?kqek;k tkrk gS fd bldk nzO;eku dsUnz r f=kT;k ds o‘Ùkkdkj iFk ij v pky ls ?kwerk gSA ge fu"d'kZ

fudkyrs gS fd ckº; cy & (A) ] dsUnzh; d.k ij yx jgk gSA (C) ] fudk; ij yx jgk gSA

(B) ] dsUnzh; d.k ij yx jgk gSA (D) ] oy; ij yx jgk gSA


Chapter # 9 Sol. C

Centre of mass =


m0  mR R = 2m 2

[6]

m R

mv 2 mv2 External force acting on the system = = r R/2 2mv 2 = R 15.

Sol.

m

The quantities remaining constant in a collision are (A) momentum, kinetic energy and temperature (B) momentum and kinetic energy but not temperature (C) momentum and temperature but not kinetic energy (D*) momentum, but neither kinetic energy nor temperature

la?kV~V esa ¼VDdj esa½ fu;r jgus okyh jkf'k;k¡ gS & (A) loasx] xfrt ÅtkZ rFkk rki (C) laosx rFkk rki fdUrq xfrt ÅtkZ ugha D

(B) laosx] xfrt ÅtkZ fdUrq rki ugha (D) laosx fdUrq u rks xfrt ÅtkZ u gh rki

After collision, linear momentun of the system is remain same. Kinetic energy is not remain same because it depend upon the collision is perfectly elastic or in elastic. 16.

Sol.

 A nucleus moving with a velocity v emits an -particle. Let the velocities of the -particle and the   remaining nucleus be v1 and v 2 must be parallel to be m 1 and m 2.    (A) v , v1 and v 2 must be parallel to each other    (B) None of the two of v , v1 and v 2 should be parallel to each other    (C) v1 and v 2 must be parallel to v    (D*) m 1 v1 + m 2 v 2 must be parallel to v

osx ls xfr'khy ukfHkd ,d –d.k mRlftZr djrk gSA ekukfd –d.k rFkk ks k ukfHkd ds osx Øe'k% rFkk gS ,oa buds nzO;eku m1 rFkk m2 gS & (A) rFkk ijLij lekukUrj gksxhA (B) ,oa esa ls dksbZ Hkh nks ijLij lekukUrj ugha gksaxsA (C) ds lekukUrj gksxkA (D) ds lekukUrj gksxkA D

By linear momentum conservation pi = pf   = m1v1  m 2 v 2   m1v1  m 2 v 2 must be parallel to v . 17.

A shell is fired from a canon with a velocity V at an angle  with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces it path to the cannon. The speed of the other piece immediately after the explosion is ,d rksi ls {kSfrt ls  dks.k ij v pky ls xksyk nkxk tkrk gSA blds iFk ds mPpre fcUnq ij ;g foLQksVd ds rqjUr i'pkr~ nwljs VqdM+s

dh pky gksxh &

(A*) 3V cos Sol.

(B) 2V cos

(C)

3 V cos 2

(D) V cos

A

Linear momentum is conserved in horizontal direction vcos pi = pf vsin v m m  mv cos  = – v cos  + v' m vcos 2 2 v' = 3 v cos 

m/2


m/2

v’

xcom

Chapter # 9 Centre of Mass, Linear Momentum, Collision 18. In an elastic collision (A*) the initial kinetic energy is equal to the final kinetic energy (B) the final kinetic energy is less than the initial kinetic energy (C) the kinetic energy remains constant (D) the kinetic energy first increases then decreases.

Sol.

[7]

fdlh çR;kLFk VDdj esa & (A*) çkjfEHkd xfrt ÅtkZ] vafre xfrt ÅtkZ ds cjkcj gksrh gSA (B) vafre xfrt ÅtkZ çkjfEHkd xfrt ÅtkZ ls de gksrh gSA (C) xfrt ÅtkZ fu;r jgrh gSA (D) xfrt ÅtkZ igys c<+rh gS i'pkr~ de gksrh gSA A

In an elastic collision, no energy is loss into heat. So the final K.E. is euqal than the initial K.E. 19.

Sol.

In an inelastic collision (A) the initial kinetic energy is equal to the final kinetic energy (B*) the final kinetic energy is less than the initial kinetic energy (C) the kinetic energy remains the constant (D) the kinetic energy first increases then decreases

fdlh vizR;kLFk VDdj esa (A) izkjfEHkd xfrt ÅtkZ] vafre xfrt ÅtkZ ds cjkcj gksrh gSA (B*) vafre xfrt ÅtkZ] izkjfEHkd xfrt ÅtkZ ls de gksrh gSA (C) xfrt ÅtkZ fu;r jgrh gSA (D) xfrt ÅtkZ igys c<+rh gS] rr~i'pkr~ de gksrh gSA B

In an elastic collision, same energy is loss into heat. So the final K.E. is loss than the initial K.E.

Objective - II 1.

Sol.

The centre of mass of a system of particles is at the origin. It follows that (A) the number of particles to the right of the origin is equal to the number of particles to the left (B) the total mass of the particles to the right of the origin is same as the total mass to the left of the origin (C) the number of particles on X-axis should be equal to the number of particles on Y-axis. (D) if there is a particle on the positive X-axis, there must be at least one particle on the negative Xaxis.

fdlh d.k&ra=k dk nzO;eku dsUnz ewy fcUnq ij gSA bldk vfHkçk; gS & (A) ewy fcUnq ds nka;h vksj d.kksa dh la[;k] nka;h vksj d.kksa dh la[;k ds cjkcj gksxhA (B) ewy fcUnq ds nka;h vksj okys d.kksa dk dqy nzO;eku ewy fcUnq nka;h vksj okys d.kksa ds dqy nzO;eku ds cjkcj gksxkA (C) x–v{k ij d.kksa dh dqy la[;k] Y–v{k ij d.kksa dh dqy la[;k ds cjkcj gksxhA (D) ;fn x–v{k ij dksbZ d.k gksxk rks de ls de ,d d.k _.kkRed x–v{k ij vo’; gksxkA D (none)

Product of the miri of left side particle equal to the product of the of the mjrj of right side particle. R=

1 M miri = mjrj

O=  2.

1 M

mkrk mkrk =

1 ( M

miri + mjrj)

A body has its centre of mass at the origin. The x-coordinates of the particles (A) may be all positive (B) may be all negative (C*) may be all non-negative (D*) may be positive for some cases and negative in other cases


Chapter # 9

Sol.


[8]

,d oLrq dk nzO;eku dsUnz ewy&fcUnq ij gSA d.kksa ds x–funsZ'kkad & (A) lkjs /kukRed gks ldrs gSA (B) lkjs _.kkRed gks ldrs gSA (C*) lkjs v&_.kkRed gks ldrs gSA (D*) dqN ds fy;s /kukRed o dqN vU; ds fy;s _.kkRed gks ldrs gSA CD

rCOM (0, 0) XCOM = 0 =

(Given) m1x1  m 2 x 2  m3x 3 ....... m1  m 2  m3  .........

m1x1 + m2x2 + m3x3 + ............. = 0 May be all mom-negative mean its cansider –ve and zero.All –ve is not possible but all position is zero is posible. 3.

Sol.

In which of the following cases the centre of mass of a rod is certainly not at its centre ? (A*) the density continuously increases from left to right (B*) the density continuously decreases from left to right (C) the density decreases from left to right upto the centre and then increases (D) the density increases from left to right upto the centre and then decreases

fuEu es ls fdu fLFkfr;ksa ds fy;s NM+ dk nzO;eku dsUnz fuf'pr :i ls blds dsUnz ij ugha gksxk & (A*) ?kuRo cka;h ls nka;h vksj fujarj c<+rk jgsA (B*) ?kuRo cka;h ls nka;h vksj fujUraj de gksrk jgsaA (C) ?kuRo cka;h ls nka;h vksj fujUrj de gksrk jgs rRi'pkr~ de gksrk jgsA (D) ?kuRo cka;h ls nka;h vksj dsUnz rd c<+rk jgs rRi'pkr~ de gksrk jgsA AB

Centre of mass of a rod is not at its centre means mass of the rod varies according to length. Or we can say that density is continuously change according to length from left to right or right to left. 4.

Sol.

If the external forces acting on a system have zero resultant, the centre of mass (A) must not move (B*) must not accelerate (C*) may move (D) may accelerate

;fn fdlh fudk; ij yx jgs ckº; cyksa dk ifj.kkeh 'kwU; gks] nzO;eku dsUnz & (A) xfr ugha djsxkA (B*) Rofjr ugha gksxkA (C*) xfr dj ldrk gSA BC

(D) Rofjr gks ldrk gSA

Centre of mass of acceleration is zero. But it may be more uniformaly with costant velocity.  Fnet Acceleration of centre of mass = =0 M 5.

Sol.

A nonzero external force acts on a system of particles. The velocity and the acceleration of the centre of mass are found to be v0 and a0 at an instant t. It is possible that ,d v'kwU; cy fdlh d.k&ra=k ij yx jgk gSA fdlh {k.k t ij nzO;eku dsUnz dk osx v0 rFkk Roj.k a0 gSA ;g lEHko gS fd& (A) v0 = 0, a0 = 0 (B*) v0 = 0, a0  0 (C) v0  0, a0 = 0 (D*) v0  0, a0  0 BD

A non zero external force acts on a system, that causes acceleration of centre of mass at an instant 't' is 'a0' But velocity of centre of mass is v0 or zero. 6.

Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls while in air (A) depends on the direction of the motion of the balls (B) depends on the masses of the two balls (C) depends on the speeds of the two balls (D*) is equal to g

nks xsans ,d lkFk ok;q esa QSadh x;h gSA nksuksa xsanksa dk nzO;eku dsUnz dk Roj.k tc os ok;q esa gS & (A) xsanks dh xfr dh fn'kk ij fuHkZj djrk gSA (B) nksuksa xsanks ds nzO;ekuksa ij fuHkZj djrk gSA (C) nksuksa xsanksa dh pkyksa ij fuHkZj djrk gSA (D*) g ds cjkcj gSA


Chapter # 9 Sol. D


[9]

  m1a1  m 2a 2 Acceleration of centre of mass = m  m 1 2 m1g  m 2g = m m 1 2 =g

7.

Sol.

A block moving in air breaks in two parts and the parts separate (A*) the total momentum must be conserved (B) the total kinetic energy must be conserved (C) the total momentum must change (D*) the total kinetic energy must change

gok esa xfr’khy ,d xqVdk nks Hkkxksa esa VwV tkrk gS rFkk VqdM+sa vyx&vyx gks tkrs gS & (A) dqy laosx lajf{kr jgsxkA (B) dqy xfrt ÅtkZ lajf{kr jgsxhA (C) dqy laosx ifjofrZr gks tk;sxkA (D) dqy xfrt ÅtkZ ifjofrZr gks tk;sxhA AD

A block moving in air & breaks in two parts then the total momentum must be conserved. But the total kinetic energymust change. 8.

Sol.

In an elastic collision (A) the kinetic energy remains constant (B*) the linear momentum remains constant (C*) the final kinetic energy is equal to the initial kinetic energy (D*) the final linear momentum is equal to the initial linear momentum

fdlh izR;kLFk VDdj esa (A) xfrt ÅtkZ fu;r jgrh gSA (B*) jSf[kd laosx fu;r jgrk gSA (C*) vafre xfrt ÅtkZ] izkjfEHkd xfrt ÅtkZ ds cjkcj gksrh gSA (D*) vafre jSf[kd laosx] izkjfEHkd jSf[kd laosx ds cjkcj gksrk gSA BCD

In a elastic collision  The linear momentum remain constant  The final linear momentum is equal to the initial linear momentum  The final kinetic energy is equal to the initial kinetic energy. 9.

Sol.

A ball hits a floor and rebounds after an inelastic collision. In this case (A) the momentum of the ball just after the collision is same as that just before the collision (B) the mechanical energy of the ball remains the same during the collision (C*) the total momentum of the ball and the earth is conserved (D*) the total energy of the ball and the earth remains the same

,d xsan Q'kZ esa Vdjkrh gS rFkk vçR;kLFk VDdj ds i'pkr~ iqu% ykSV tkrh gSA bl fLFkfr esa & (A) VDdj ds rqjUr i'pkr~ xsan dk laosx] blds VDdj ls iw.kZ laosx ds cjkcj gksrk gSA (B) VDdj ds nkSjku xsan dh ;kaf=kd ÅtkZ leku jgrh gSA (C) xsan rFkk i`Foh dk dqy laosx lajf{kr jgrk gSA (D) xsan rFkk i`Foh dh dqy ÅtkZ leku jgrh gSA CD

The total momentum of the system (earth + ball) is conserved. The total energy of the ball is change but the total energy of the ball and the earth remains the same. 10.

A body moving towards a finite body at rest collides with it. It is possible that (A) both the bodies come to rest (B*) both the bodies move after collision (C*) the moving body comes to rest and the stationary body starts moving (D) the stationary body remains stationary, the moving body changes its velocity.


Chapter # 9

Sol.


fojkekoLFkk esa fLFkfr ,d ifjfer oLrq dh vksj xfr djrh gqbZ ,d oLrq blls Vdjkrh gSA ;g lEHko gS fd & (A) nksuksa oLrq,sa fojkekoLFkk esa vk tk;saA (B) VDdj ds i'pkr~ nksuksa oLrq,sa xfr'khy gks tk;saA (C) xfr'khy oLrq fLFkj gks tk;s rFkk fLFkj oLrq xfr çkjEHk dj nsaA (D) fLFkj oLrq fLFkj jgs] xfr'khy oLrq dk osx ifjofrZr gks tk;sA BC

Before collision (rest) u m1 m2 By linear momentum conservation  pi = pf m1u = m1v1 + m2v2  v1 = v2  0  11.

Sol.

[10]

after collision

m1

v1

v2 m2

m1u may be posible v1 = 0 & v2 = m 2

In head on elastic collision of two bodies of equal masses (A*) the velocities are interchanged (B*) the speeds are interchanged (C*) the momenta are interchanged (D*) the faster body slows down and the slower body speeds up.

leku nzO;eku okyh nks oLrqvksa dh lEeq[k çR;kLFk VDdj esa & (A) osx ijLij ifjofrZr gks tkrs gSA (B) pkysa ijLij ifjofrZr gks tkrs gSA (C) laosx ijLij ifjofrZr gks tkrs gSA (D) fLFkj oLrq fLFkj jgsa] xfr'khy oLrq dk osx ifjofrZr gks tk;saA ABCD

Before collision m After elastic collision u2 m

   

u2 m

u1

u1 m

{Let

u1 > u2}

(By linear momentum conservation)

velocities are inter changed speeds are inter changed momenta are inter changed faster body slows down and the slower body speed up.


CH-9 (Centre of Mass)

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