3. System of Particles

7/29/2019 3. System of Particles

1/19

Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 31

EXERCISE # 1

1.1

ycm

=

526)(2

252)2(0=

11

10

4 11

10=

11

34

1.2 COM can lie anywhere within the radius r.

1.4.

COM of brick 1 and 5 2

L1 o 5 bZV dk nzO;eku dsUnz

2

L

COM of brick 2 and 4 2

L+

5

L2 o 4 bZV dk nzO;eku dsUnz

2

L+

5

L

COM of brick 3 2L

+5

L23 bZV dk nzO;eku dsUnz

2L

+5

L2

Xcm

=m5

10

L2

2

Lm

5

L

2

Lm2

2

Lm2

=50

L33

2.1 Using21

2211

mm

xmxm

= Xcm

dk mi;ksx djrs gSa

By putting values Xcm

= 10m from base.

eku j[kus ij Xcm = vk/kkj ls 10m

nzO;eku dsUnz dh

izkjfEHkd fLFkfr10 m

B

4kg

15m/s

vcm

2kg

A

SYSTEM OF PARTICLES


2/19


Initial velocity of COM 21

2211

mm

vmvm

= 10 m/s upward

nzO;eku dsUnz (COM) dk izkjfEHkd osx 21

2211

mm

vmvm

= 10 m/s ij dh vksj

Also, acceleration of COM is g thus using

nzO;eku dsUnz dk Roj.k g gS] fuEu dk mi;ksx djrs gq,v2 = u2 + 2as for COM

0 = 102 + 2 (10)x hmax

hmax

= 5m from the initial position of COM

hmax

= 5m nzO;eku dsUnz dh izkjfEHkd fLFkfr lsHence total height from base = 10 + 5m = 15m.

vr% vk/kkj ls dqy pkbZ = 10 + 5m = 15m.

2.2 a = mnm

)mnm(

g

=)1n(

)1n(

g

a1

= a2

= a

acm

=)mnm(

manma 21

=a

)1n(

)1n(

acm

= g)1n(

)1n(2

2

.

2.3 Let the displacement of shell along horizontal direction be x. Then the displacement of ball along horizontal

is2

R3 x.

Since centre of mass of two body system does not shift along horizontal direction.

mx = m

x

2

R3or x =

4

R3[ Ans.

4

R3]

2.5 Initially the centre of mass is at

4

Ldistance from the vertical rod.

4

L

mm

)0(m)(mx,As 2

1

cm

centre of mass does not move in x-direction as Fx = 0.After they lie on the floor, the pin joint should be at L/4 distance from the origin shown inorder to keep the

centre of mass at rest.

Finally x-displacement of the pin is4

Land y-displacement of the pin is obviously L.

Hence net displacement =

4

L17

16

LL

22


3/19


3.2 The resultant force can be accelerating or decelerating, hence the momentum can increase or decrease. Hence

(A) is wrong.

Since Fnet

= M acm

acm

0 ;hence v

cmmust change

Hence (B)In case of a circular motion of centre of mass about a point the distance of centre of mass will remain constant.

Hence (C)

4.1 The spring pulls both the blocks with same force. Hence force on both blocks is equal and opposite.

since no net external force acts on system and its initial momentum is zero, therefore net momentum of

system is always zero. Hence momentum of blocks are equal and opposite.

fLizax nksuksa CykWdksa dks leku cy ls [khaprh gSA blfy;s nksuksa CykWdks ij cy cjkcj rFkk foijhr gSApwfd fudk; ij dksbZ usV ck; cy dk;Z ugh djrk gS vkSj bldk izkjfEHkd laosx 'kwU; gS] blfy;s fudk; dk usV laosx

ges'kk 'kwU; gksxkA blfy;s CykWdksa dk laosx cjkcj rFkk foijhr gSA

4.3 Pi= mv

1+ mv

2

Pf= (m + M) v

Pi= P

f v =

)Mm(

Mvmv 21

By energy conservation mtkZ laj{k.k ls

2

1mv

12 +

2

1Mv

22 =

2

1(M + m) v2 +

2

1kx2

mv12 + Mv

22 = (M + m)

2

2

221 kx

)mM(

)Mvmv(

solving gy djus ij x = (v1

v2)

k)mM(mM .

5.1 Force on table due to collision of balls :

xsan }kjk est ij yxk;s tkus okyk cy

Fdynamic = dt

dp= 2 20 20 103 5 0.5 = 2 N

Net force on one leg =4

1(2 + 0.2 10) = 1 N

,d Vkax ij cy =4

1(2 + 0.2 10) = 1 N

6.2 As )say(P|P||P||P| 021

From conservation of linear momentum

PPP 21

cosPP2PPP 2122

21

or cosP2PPP 222

2

1cos

= 120 ]


4/19


6.5 8 sin = v cos

sinv2

cos8

2 tan = cot tan = 2

1

90V

2g3.2=8 m/s

v =2

8= 24 m/s

k =

2

28241

2

1= 16 J

Projectile never travels vertically downward.

6.9 As shown in figure

+ 90 = 150 = 60

6.11 For the striker :

02 = (2)2 2(0.2) (10) s

s = 1 m

From A to B :

SAB

=

22

22

1

22

1

=

4

1=

2

1m

Similarly : SBC

=2

1m

The striker stops at the point C having coordinates

2

1,

22

1.

7.2 Neglecting gravity,

xq:Ro ux.; ekurs gq;s

v = un

t

0

m

m;

u = ejection velocity w.r.t. balloon. m0

= initial mass mt= mass at any time t.

u = xSl dk osx xqCkkjs ds lkis{k. m0

= izkjfEHkd nzO;eku mt= fdlh le; t ij nzO;eku

= 2n

2/m

m

0

0= 2n2.


5/19


EXERCISE # 2PART- I1. x & y coordinates of all the particles are positive. Hence x & y coordinates of centre of mass of the system has

to be positive.

2. This disc can be assumed to be made of a complete uniform disc and a square plate with same negativemass density.

bl pdrh dks] ,d leku ?kuRo okyh iwjh pdrh rFkk leku _.kkRed nzO;eku ?kuRo okyh oxkZdkj IysV ds la;kstuls cuh gqbZ eku ldrs gSA

Ycm

=21

2211

mm

ymym

=

)(r

)2/r()()0()r(22

22

=)r(2

r22

2

=

2

14

r

)2

rr(2

2

r

22

3

3. KEA/CM

=2

1.1.(v

A/CM)2 = 2 Joules

VA/CM

= 2 m/s.Let ; COM move towards +ve x-direction.

Then, CM/Av

= i2

vB/CM

= i ( Use ; CMv

=21

2211

mm

vmvm

)

KESystem

= 2CMvi2.1.2

1 + 2CMvi2.

2

1

= CM2

CMCM

2

CM v.i.2v1.2.2

1

v.i

2.2v42

1

= )v.i241()v.i222( CMCM

= 9 J Ans.

4. As both the balls are released simultaneously, at any instant before the lower balls reaches the ground both

have the same velocity ; v = gt i.e. v vs. t is a straight line graph.

nksuksa xsan ,d lkFk NksM+h tkrh gS] uhps okyh xsan ds tehu ls igqpus ls igys nksuksa xsan leku osx es gksxh ;v = gt vr% v vs. t dk xzkQ ,d lh/kh js[kk gksxh

VCM

=m2

)t(mv)t(mv = v(t) ; v(t) being the instantaneous velocity. (v(t) rkRdkfyd osx

Just after the lower ball strikes ground and comes to rest :

igyh xsan dk tehu ls Vdjkus ds ckn og fojke esa vk tkrh gSA

VCM

=m2

)t(vm=

2

)t(v

i.e. the velocity suddenly drops to half its value. vr%] osx vpkud vk/kk gks tk;sxkHence graphs (A) & (B) are chosen. vr% AB xzkQ lgh gSaAfter collision :

VDdj ds ckn aCM

=mm

)0(m)g(m

=

2

g

i.e. the slope (of vt curve) should decrease to half.

vr% vt xzkQdh


6/19


6. From figure, only point 'P' will move in a circle

9. Velocity of the ball on striking Vdjkus ij xasn dk osx = gh2

After that ball goes to height less than (h) due to inelastic collission = )dh(g2 .

blds i'pkr~ xsan vizR;kLFk VDdj ds dkj.k (h) ls de pkbZ = )dh(g2 rd mNyrh gSA

)dh(g2 = e gh2

h d = e2h dh = 2e1

1 .

10. Let u be the velocity of the ball

w.r.t. wedge when it reaches the f loor.

Then, the x-component of velocity of the ball

w.r.t. ground wil l be

2

uv towards right.

By momentum conservation :

0 = m(v) + m

2

uv u = 2 2 v

Therefore, the y-component of velocity of the ball after the elastic collision with the floor will be u cos45

=2

u= 2v (upward)

Maximum height =g

v2

g2

)v2( 22

11. If we treat the train as a ring of mass 'M' then its COM will be at a distanceR2

from the centre of the circle.

Velocity of centre of mass is :

;fn Vsu dks 'M' nzO;eku dk oy; ekus rks mldk nzO;eku dsUnz

R2 nwjh ij gksxkA nzO;eku dsUnz dk osxVCM = RCM .

=R2

. =

R

V.

R2( =

R

V)

VCM = V2

MVCM = MV2

As the linear momentum of any system = MVCM

The linear momentum of the train =

MV2Ans.

fdlh Hkh fudk; dk js[kh; laosx = MVCM

vr% Vsu dk js[kh; laosx =

MV2


7/19


12. M1

is very large as compared to M2. Hence for collision between M

1and M

2, M

1can be considered

equivalent to a wall and M2

as a small block. Thus the velocity of M2

will be 2vo

after collision with M1

.

Similarly after collision between M2

and M3, the velocity of M

3will be 2(2v

o). In sequence, the velocity of M

4

shall be 2(2(2vo)) = 8 v

oafter collision with M

3.

M2dh rqyuk esa M

1cgqr cM+k gSA vr% M

1o M

2ds e/; VDdj ds fy, M

1dks ,d nhokj dh rjg ekuk tk ldrk gS

rFkk M2dks ,d NksVs CykWd dh rjg ekuk tk ldrk gSA vr% M

1ds lkFk VDdj ds ckn M

2dk osx 2v

ogksxkA blh rjg

M2o M

3dh VDdj ds ckn M

3dk osx 2(2v

o) gksxkA blh rjg ekxr :i ls M

3ds lkFk VDdj ds ckn M

4dk osx

2(2(2vo)) = 8 v

ogksxkA

14. use momentum conservation equation laosx laj{k.k lsm.u. = 5mv

V =5

u

Impulse imparted by tension force to block of mass 3m.

3m nzO;eku ds CykWd dks ruko cy }kjk izkIr vkosx

= 3mx v =5

mu3.

17. I. Since velocity of both R and S is positive they will move in same direction.

II. At mid point velocities of R and S are same.

III. Change velocity of R is small compair to change in velocity of S. But change in momentum is same for

both in magnitude. Hence mass of R should be greater than S.

Hence all three are correct.

22.

When the string becomes tight, both particles begin to move with velocity components v in the direction AB.

Using conservation of momentum in the direction AB

tc Mksjh ruh gqbZ gksxh] nksuksa d.kksa dk AB fn'kk esa osx dk ?kVd v gksxkA AB fn'kk esa laosx laj{k.k yxkus ijmu cos 30 = mv + mv

or;k4

3uv

Hence the velocity of ball A just after the jerk is

4

3uv .

>Vdk (jerk) yxus ds Bhd ckn A xsan dk osx4

3uv gSA


8/19


Mathod II. fof/k II

extF

= 0, using conservation of linear momentum along the string.

extF

= 0, Mksjh dh fn'kk esa js[kh; laosx laj{k.k yxkus ij

m u cos 30 = mv + mv

2v = u.

2

3 v =

4

3u

23.

During 1st collision perpendicular component of V, V becomes e times, while IInd component IIV remains

unchanged and similarly for second collision. The end result is that both IIV and V becomes e times their

initial value and hence V"= eV (the ( ) sign indicates the reversal of direction).

24. Let v be the initial velocity. Tangential velocity

remains same during collision and equal to

v cos60 = v/2

Let v be the normal component of velocity after impact.

In OAB : tan60 =

v

2/v

v =

32

v

Then : e =

30cosv

v=

)2/v3(

)32/v(=

3

1


9/19


PART - II

1. Take an elementary ring of radius y and width dy

Then y-coordinate of centre of mass of half disc is

y =

a

0

a

0cm

dm

dmy

=

a

0

a

0

dyy

dyy

y2

=

a

0

a

0

dyy)ky(

dyy)ky(

y2

=a

2

3

[Ans: 3a/2]3. When the cube reaches the bottom most point on the block, then by momentum conservation in horizontal

direction ; mv = MV

V =M

mV............(1)

Now by energy conservation ;

mgR =2

1mv 2 +

2

1MV2

Put V from (1)

v =

M

m1

gR2

.

4. Force F on plate = force exerted by dust particles

= force on dust particles by the plate

= rate of change of momentum of dust particles= mass of dust particles striking the plate per unit time change in velocity of

dust particles.

=A (v+ u) (v + u)= A (v + u)2

Ans. A(u + v)2

5. Since e =5

1

Final normal component of velocity =5

37cosv 0.

As the angle of rebound is equal to the angle before impact.

Therefore, both normal & tangential components of velocities must change by the same factor.

Tangential velocity after impact becomes5

37sinv 0.

Let the time of impact be t.

N =

t

5

37cosv37cosvm

00

=t5

37cosmv6 0

where N is the normal force imparted on the ball by the wall.


10/19


Frictional force = N =t

37cosmv

5

6 0

Also frictional force =

t

5

37sinv37sinvm

00

t

5

37sinv37sinvm

00

=t

37cosmv

5

6 0

=3

2tan370 =

2

1

4

3.

3

2 Ans.

pwafd e =5

1

osx dk vfUre yEcor~ ?kVd= 537cosv0

.

tSlk fd ijkorZu dks.k vkiru dks.k ds cjkcj gS blfy, osx ds yEcor~ rFkk Li'kZjs[kh; ?kVd leku xq.kd ls ifjofrZr gksaxsA

la?kV~V ds ckn Li'kZ js[kh; osx gksxk5

37sinv 0.

ekuk la?kV~V dk le;t gSA

N =

t

5

37cosv37cosvm

00

=t5

37cosmv6 0

tgk N nhokj }kjk xsan ij yxk;k x;k vfHkyEcor~ cy gSA

?k"kZ.k cy = N =t

37cosmv

5

6 0

?k"kZ.k cy =t

5

37sinv37sinvm

00

t

5

37sinv37sinvm

00

=t

37cosmv

5

6 0

=3

2tan370 =

2

1

4

3.

3

2 Ans.

6.

(figure - 1)

Let u and v be the speed of wedge A and block B at just after the block B gets off the wedge A. Applying

conservation of momentum in horizontal direction, we get.

mu = mv ............(1)

Applying conservation of energy between initial and final state as shown in figure (1), we get


11/19


mgh =2

1mu2 +

2

1mv2 .............(2)

solving (1) and (2) we get

v = gh .............(3)

At the instant block B reaches maximum height h on the wedge C (figure 2), the speed of block B and wedgeC are v.Applying conservation of momentum in horizontal direction, we get

mv = (m + m) v .............(4)

Applying conservation of energy between initial and final state

2

1mv2 =

2

1(m + m) v2 + mgh ............(5)

Solving equations (3), (4) and (5) we get

h=4

hAns.

ostAls CykWd B ds fxjus ds rqjUr ckn ekuk u rFkk v e'k% ostArFkk CykWd B dh pkysa gSaA {kSfrt fn'kk esa laosx laj{k.k yxkusij] ge izkIr djrs gS &mu = mv ............(1)

fp=k (1) esa fn[kkbZ xbZ izkjfEHkd rFkk vfUre voLFkk ds chp tkZ laj{k.k yxkus ij

mgh =2

1mu2 +

2

1mv2 .............(2)

(1) rFkk (2) dks gy djus ij ge izkIr djrs gSa &

v = gh .............(3)

ost C ij CykWd B }kjk vf/kdre pkbZ h (fp=k&2) igqpus ds {k.kij CykWd B rFkk ost C dh pky vgSA {kSfrt fn'kk esa laosx laj{k.k

yxkus ij] ge izkIr djrs gSaAmv = (m + m) v .............(4)


12/19


EXERCISE # 4PART- I

3.

300

ux

uy

u=100m/s

(i) 100 m/s velocity of the ball is relative to ground.

100 m/s xsan dk Hkwfe ds lkis{k osx gS[Unless and until it is mentioned in the question,the velocity is always relative to ground.]

[tc rd iz'u esa crk;k uk tk;s] osx lnSo Hkwfe ds lkis{k ekfu;s]Horizontal component of its velocity, u

x= u cos300

osx dk {ksfrt ?kVd ux= u cos300

or;k ux= (100)

2

3m/s = 50 3 m/s

and vertical component of its veloctiy o osx dk m/oZ ?kVd u = u sin 300

uy = 100 2

1m/s = 50 m/s

Vertical displacement of the ball when it strikes the carriage isxsan ds xkMh ls Vdjkus rd m/oZ foLFkkiu 120 m or

Sy= u

yt +

2

1a

yt2

120 = (50 t) +2

1(-10) t2

t2 10 t 24 = 0

t = 12 s or ;k 2sIgnoring the negative time, we have _.kkRed le; dks vizHkkoh ekuus ij

t0

= 12s Ans 12

(ii) When it strikes the carriage, its horizontal component of velocity is still 50 3 m/s. It sticks to the carriage.

Let V2be the velocity of (carriage + ball) system after collision. Then applying conservation of linear momentum

in horizontal direction

xkMh lss Vdjkus ij Hkh bldk {kSfrt osx ?kVd 50 3 m/s gh jgsxkA ;g xkMh ls fpid tkrh gSA ekuk VDdj ds i'pkr~(xkMh+ xsan ) fudk; dk osx V

2gSA {ksfrt fn'kk eas jsf[kd laosx lajf{kr djus ij

(mass of ball) (horizontal component of its velocity before collision) = (mass of ball + carriage) (V2)

(xsan dk nzO;eku) (VDdj ls iwoZ osx dk {kSfrt ?kVd ) = (xsan + xkMh ) dk nzO;eku (V2)

(1 Kg)(50 3 m/s) = (10 Kg) (V2)


13/19


V2

= 5 3 m/s

The second ball is fired when the first ball strikes the carriage i.e.after 12 seconds.In these 12 seconds the car

will move forward a distance of 12V1

or 60 3 m. The second ball also takes 12 second to travel a vertical

displacement of 120m. This ball will strike the carriage only when the carriage also covers the same distance

of 60 3 m in these 12 seconds. This is possible only when resistive forces are zero because velocity of car (V1)

= Velocity of carriage after first coll ision (V2) = 5 3 m/s.

nwljh xsan rc nkxh tkrh gS tc igyh xsan xkMh ls Vdjkrh gS]vFkkZr~ 12 lSd.M i'pkr~A bu 12 lSd.M esa dkj 12V1;k

60 3 m. nwj vkxs py tk;sxhA nwljh xsan Hkh 120m.ds m)Z foLFkkiu ds fy, 12lSd.M ckn xkMh ls Vdjk;sxhA ;g nwljh

xsan xkMh ls Vdjk;sxh ;fn dkj Hkh bu 12lsd.M esa60 3 m dh nwjh r; djsA ;g rHkh laHko gS tc ?k"kZ.k cy 'kwU; gks] D;ksfd

dkj dk osx (V1) = xkMh dk izFke VDdj ds i'pkr~ osx (V

2) = 5 3 m/s.

Hence at the time of second collision :

vr% nwljh VDdj ds le;Horizontal component of velocity of ball = 50 3 m/s and horizontal veloctiy of carriage + f irst ball = 5 3 m/s.

Let V be the desired velocity of carriage after second collision. Then conservation of linear momentum in hori-

zontal direction gives

xsan ds osx dk {kSfrt ?kVd = 50 3 m/s o xkMh + igyh xsan dk {ksfrt osx = 5 3 m/s. ekuk f}rh; VDdj i'pkr~ xkMh

dk vfHk"V osx V gSA {ksfrt fn'kk esa jsf[kd laosx lja{k.k fu;e ls

11 V = (1) (50 3 ) + (10) (5 3 ) = 100 3

V =11

3100m/s or V 15.75 m/s Ans 14(ii)

In this particular problem values are so adjusted that even it we take the velocity of ball with respect to car, we

get the same results of both the parts, although the method will be wrong.Refer Q.No.1 of Year 1997 (new paper)

in which it was clearly mentioned that the velocity of stone is with respect to cart.

bl iz'u esa eku ,sls pqus x, gS fd ;fn xsan dk osx dkj ds lkis{k Hkh ys rks Hkh nksuks Hkkxksa dk mRrj leku gh izkIr gksxk ;|firjhdk xyr gksxkA o"kZ 1997 iz'u la[;k 1 (U;w isij) dks ns[ks ftlesa Li"Vr% crk;k x;k gS fd iRFkj dk osx xkMh ds lkis{k

gSA

4. )vmv(m)vmvm( 22112211

= | Change in momentum of the two particles |

= | nksuks d.kksa ds laosx esa ifjorZu |= | External force on the system | t ime interval

= | fudk; ij ck cy | le; vUrjky= (m

1+ m

2) g(2 t

0)

= 2(m1

+ m2) gt

0


14/19


5. vCOM

=21

2211

mm

vmvm

///////////////////////////////////////////////

v = 02v = 14 m1

m = 10 m1 m = 4 m2

+ ve

=410

1041410

= 10 m/s.

6. Angular speed of particle about centre of the circle,

ok dsUnz ds lkis{k d.k dh dks.kh; pky

=R

v2, = t =

R

v2t

pv

= ( v2sin i + v2 cos j ) or;k pv

=

jtR

vcosvit

R

vsinv

2

2

2

2

and rFkkmv

= v1j

linear momentum of particle w.r.t. man as a fanction of time is

O;fDr ds lkis{k d.k dk jSf[kd laosx le; Qyu :i esapmL

= (m pL

mL

)

= m

jvtR

vcosvit

R

vsinv 1

2

2

2

2

7. (i) X1

= 0t A (1 cost)

Xcm

=22

2211

mm

xmxm

=

0t X

2=

0t +

2

1

m

mA (1cos t) Ans.

(ii) a1

=2

12

dt

xd= 2 A cos t

The separation X2 X

1between the two blocks will be equal to

0when a

1= 0 or cos t = 0

x2

x1

=2

1

m

mA (1cos t) + A (1 cos t)

0

=

1

m

m

2

1A (cos t = 0)

Thus the relation between 0and A is,

0

=

1

m

m

2

1AA Ans.

(i) X1

= 0t A (1 cost)

Xcm

=22

2211

mm

xmxm

= 0t X

2=

0t +

2

1

m

mA (1cos t) Ans.

(ii) a1

= 21

2

dt

xd= 2 A cos t

nks CykWd ds e/; nwjh X2

X1

, 0ds cjkcj gS tc a

1= 0 ;k cos t = 0

x2

x1

=2

1

m

mA (1cos t) + A (1 cos t)

0

=

1

m

m

2

1A (cos t = 0)

blfy, 0o A esa laca/k ,

0=

1

m

m

2

1AA Ans.


15/19


8. According to Newtons Law U;wVu fu;ekuqlkj

e =21

12

uu

vv

For elastic collision cofficient of restitution izR;kLFk VDdj ds fy, izR;koLFkku xq.kkad e = 1 so vr%

12 vv = 21 uu Statement - 1 is correct dFku 1 lR; gSALinear momentum is conserved in both elastic & non elastic collision but its not the explanation of

statement -1 so it is not the correct explanation of the statement A.

jsf[kd laosx nksuks gh izR;kLFk o vizR;kLFk VDdj esa ljaf{kr jgrk gS fdUrq ;g dFku 1 dh lgh O;k[;k ugh gSA

9. ipP1

ipP2

as there is no external force so momentum will remain conserved

ck cy 'kwU; gS vr% laosx lajf{kr jgsxkA

2121

PP'PP

0PP 21

Now from option vc fodYi ls

(A)21 PP

= kcj)bb(i)aa( 12121

(B) 21 PP

= k)cc( 21

(C) 21 PP

= j)bb(i)aa( 2121

(D) 21 PP

= jb2i)aa( 121

and it is given that a1

b1

c1

, a2, b

2, c

2, 0

in case of A and D it is not possible to get 21 PP

= 0Hence Ans. (A) and (D)

rFkk fn;k x;k gS a1

b1

c1

, a2, b

2, c

2, 0

A rFkk D izj.k esa 21 PP

= 0 izkIr gksuk lEHko ugha gSA

vr% Ans. (A) o (D) lgh gkssaxsA

10. At point B there is perfectly inelast ic col lision so component of v elocity to incline plane becomes zeroand component parallel to second surface is retained

fcUnq B ij iw.kZr;k vizR;kLFk VDdj gS blfy;s ur ry ds yEcor~ osx dk ?kVd 'kwU; gks tkrk gSA ,oa nwljh lrg ds lekUrj?kVd vifjofrZr jgrk gSA

velocity immediately after it strikes second inclineblds nwljs ur ry ij Vdjkus ds Bhd ckn osx

V = 30cosgh2 = 3102 2

3=

4

9102

V = 45 m/s

11. At point C fcUnq C ij

gh2VV 2B2C

VC

2 = 45 + 2 10 3

VC

= 105 m/s


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12. The block coming down from incline AB makes an angle 30 with incline BC. If the block collides with

incline BC elastically, the angle of block after collision with the incline shall be 30.

Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately af ter the

block strikes second inclined, its vertical component of velocity will be zero.

ur ry AB ls uhps vk jgk xqVdk ur ry BC ds lkFk 30 dk dks.k cukrk gSA ;fn xqVdk urry BC ls izR;kLFk :i lsVdjkrk gS rks urry ls Vdjkus ds ckn xqVds dk dks.k 30 gksxkvr% ur ry BC ls VDdj ds Bhd ckn xqVds dk osx {kSfrt gksxkA xqVds ds nwljs urry ls Vdjkus ds Bhd ckn blds osxdk /okZ/kj ?kVd 'kwU; gksxkA

13.

ycm

=54321

5544332211

mmmmm

ymymymymym

ycm

=m6mmmm

)a(m)a(m)0(m)a(m)0(m6

=10

a.

14. Since masses of particles are equal, collisons are elastic, so particles will exchange velocities after each

collision. The first collision will be at a point P and second at point Q again and before third collision the

particles will reach at A.

pwafd d.kksa dk nzO;eku leku gS VDdj izR;kLFk gS vr% d.k izR;sd VDdj ds i'pkr~ osx ifjofrZr djsaxsA izFke VDdj fcUnqP ij gksxh rFkk f}rh; VDdj iqu% Q ij gksxhS rFkk rrh; VDdj ds igys d.k A ij igqpsxsaA

15.

from momentum conservation :

9m = (2m) V1 (m)V

2

9 = 2V1

V2

..... (1)

e = 19

VV 21

......(2)

from eqn(1) and eqn(2) V1

= 6 m/sec.

for second collision between second block and third block :

(2m) 6 + m(0) = (2m + m) VC

VC

= 4 m/sec.


17/19


laosx laj{k.k ls:9m = (2m) V

1 (m)V

2

9 = 2V1

V2

..... (1)

e = 19

VV 21

......(2)

lehdj.k (1) o (2) ls V1

= 6 m/sec.

nwljs fi.M o rhljs fi.M ds e/; nwljh VDdj ds fy, :(2m) 6 + m(0) = (2m + m) V

C

VC

= 4 m/sec. Ans. 4

16*.

Since collision is elastic, so e = 1

Velocity of approach = velocity of separation

So, u = v + 2 .............(i)

By momentum conservation :

1 u = 5v 1 2

u = 5v 2

v + 2 = 5v 2

So, v = 1 m/s

and u = 3 m/s

Momentum of system = 1 3 = 3 kgm/s

Momentum of 5kg after collision = 5 1 = 5 kgm/s

So, kinetic energy of centre of mass =2

1(m1 + m2)

2

21

1

mm

um

=2

1(1 + 5)

2

6

31

= 0.75 J

Total kinetic energy =2

1 1 32

= 4.5 J.


18/19


D;ksfd VDdj izR;kLFk gS vr% e = 1

lkehI; osx = vyxko osx

vr% u = v + 2 .............(i)

laosx lja{k.k ls

1 u = 5v 1 2

u = 5v 2

v + 2 = 5v 2vr% v = 1 m/s

vkSj u = 3 m/s

fudk; dk laosx = 1 3 = 3 kgm/s

5kg dk VDdj ds ckn laosx = 5 1 = 5 kgm/s

vr% nzO;eku dsUnz dh xfrt tkZ =2

1(m1 + m2)

2

21

1

mm

um

=2

1(1 + 5)

2

631

= 0.75 J

dqy xfrt tkZ =2

1 1 32

= 4.5 J.

17. 5 =1

2gt2 t = 1 sec

so final velocity of bullet is 100 m/s

& of ball is 20m/s

Applying momentum conservation 0.01 V = 0.01 100 + 0.2 20

V = 100 + 400 = 500 m/s

PART - II

1. By the conservation of linear momentum

jsf[kd laosx laj{k.k ls(m + m)v' = m. 2 v mv

2mv' = mv

v' =

2

v


19/19


3. vmax

= a = aT

2

=m

ka

k

M2

a2

Hence vr% 2

1

2

1

max

max

k

k

a

a

v

v

2

1

1max

v =2max

v (given fn;k x;k gS)

1

2

k

k.

4. Initial thrust of the blast izkjfEHkd mR{ksi cy = ma= 3.5 104 10= 3.5 105 N

5. Before breaking, the centre of mass of system is moving under gravity. Thus, acceleration of the centre of massis gravitational acceleration. During breaking, internal forces come into play which are not responsible for theacceleration of the centre of mass.This indicates that, the acceleration of centre of mass remains the same (equal to gravitational acceleration).Thus, the centre of mass of system continues its original path.

VwVus ls igys fudk; dk nzO;eku dsUnz xq:Ro ds vUrxZr xfr dj jgk gSA vr% nzO;eku dsUnz dk Roj.k xq:Roh; Roj.k gh gksxkAVwVus ds nksjku vkUrfjd cy mRiUu gksrs gS] tks fd nzO;eku dsUnz ds Roj.k ds fy, ftEesnkj ugh gSA;g n'kkZrk gS fd nzO;eku dsUnz dk Roj.k leku (xq:Roh; Roj.k ds rqY;) jgsxkAvr% fudk; dk nzO;eku dsUnz izkjfEHkd iFk ij gh xfr djsxkA

6. According to conservation of energytkZ laj{k.k vuqlkj

22 M

2

1kL

2

1 kL2 =

M

M2

MkL2 = p2 (p = m) p = L Mk

7. In x-direction x- fn'kk esamu

1+ 0 = 0 + m

x

KL2 = p2 (p = m)

x=

In y-direction y- fn'kk esa

0 + 0 = m

3 m

y

y = 3

Velocity of second mass after collision la?k ds ckn nwljs nzO;eku dk osx

=22

2

3

4

3

=3

2

14. If initial momentum of particles is zero, then they loss all their energy in inelastic collision but here initialmomentum is not zero.Principle of conservation of momentum holds good for all collision.

;fn d.k ds izkjfEHkd laosx 'kwU; gS] rks vizR;kLFk VDdj esa ;s viuh laiw.kZ tkZ O;; dj nssxsa] fdUrq bl izdj.k easizkjfEHkd laosx 'kwU; ugh gSAlHkh izdkj dh VDdjksa ds fy, laosx laj{k.k fl)kar oS/k gSA

3. System of Particles

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