center of mass & collisions/solution target iit-jee 2019 ...1 com exe. 2, 3, 4... · center of...

Center of Mass & Collisions/SOLUTION Target IIT-JEE 2019

Page # 1

CENTER OF MASS & COLLISIONSSOLUTION

Exercise #2 (Multiple Answer)Q.1 ABCSol. Multi correct

(A)

21 m/s

1kgA

4 m/s

2kgB

1 m/sA

v

B

Using conservation of linear momentum1 × 1 + 2 × v = 1 × 21 – 2 × 4 v = +6 m/sMulti correct

(B) e = 42116

= 0.2

(C) Loss in K.E. = 21

21

21mm

mm (u1 – u2)

2 (1 – e2)

= 21

32

(25)2 (1 – 0.04) = 200 J

(D) I = pf – pi = 2(+6) – 2(–4) = 20 Ns

Q.2 BDSol. C.M. of system lie on straight line of slope 1 and lie between centre of circle and centre of rod.

Q.3 CSol. Factual

Q.4 DSol. (i) From graph final velocity of both body is positive.

(ii) K.E. is minimum when velocity of each body becomes equal.(iii) Magnitude of change in velocity of R is less.

Q.5 ABCD

Sol. (A), (D) e =21

12uuvv

(B) energy is dissipated.

(C) 0Fextsystem


Page # 2

Q.6 AC

Sol. t = 0u u

t = v2L

u u

t = v2L

uu

t = vL u u

t > vL

Rest Rest

Q.7 BCDSol. Use C.M. frame

Q.8 BDSol. 12 vv

= )uu(e 21

for 1-D collision.

Q.9 ACSol. (A) Separation between them increases with increasing.

(B) F = dtdp

= veve

= +ve

(C) extsystemF = 0

(D) Internal energy is consumed.

Q.10 ACSol. Take µ high and low.


Page # 3

Exercise #3 (Neet/AIIMS Special)

Q.1 DSol. If the linear momentum of a particle is changing then work done (i.e. change in kinetic energy may or may

not equal to zero.

and dtpdF

Q.2 BSol. Centoid is not the point about which body is symmetric.

Q.3 C

Q.4 CSol. During collision potential energy is stored

Q.5 DSol. At the time of maximum defermation some potential energy is storedQ.6 DSol. momentum is a vector equantity

Q.7 DSol. C.M. frame is a zero momentum frome

Q.8 DSol. Linear momentum of the system is changed by net external force but internal force can do some non zero

work.

Q.9 A

Sol. Ksystem = Ksystem/CM +M2

P2system

Q.10 CSol. In any inertial frome force on body remain same

Momentum = vmp


Page # 4

Exercise #4 (Matrix Match)Q.1 (A)-Q, (B)-Q, (C)-P,R , (D)-Q,SSol. Use C.M. frame

Q.2 (A)-Q (B)-S (C)-P]

Sol. Evaluate e =21

12uuvv

for each case alongwith using conservation of linear momentum.


Page # 5

Exercise #5 (Passage Based)Q.1 ASol. ( + x)g – T = ( + x) a ...(i)

T – ( – x) g = ( – x) a ...(ii) TT

(L–x)ga

(L+x)g

x

x

aafter solving we get.

a = Lx

g ]

Q.2 C

Sol. dxdvv = L

xg

v

0

dvv = L

0

dxxLg

v = gL

Q.3 B

Sol. 0Fextsys

sysp

= constant]

Q.4 B

Sol.Q–v''

v'

10

20 kgv

180 kg

(2–v)20kg

2–v''

v'

Using conservation of linear momentum180 v + 20 [– (2–v)] = 200 (10) 200 v = 2040 v = 10.2]

Q.5 BSol. He will give impulse to initial direction.

Q.6 ASol. Using conservation of linear momentum

160 v' + 20v' = 180 v v' = v = 10.2v

180 kg

v'

v''

160 v'' + 20 {– (2– v'' )} = 0 v'' 0.2

vnet = 202.10''v'v 22 ]


Page # 6

Q.7 A

[Sol.

(A) from conservation of linear momentum along x-axis speed of cart will increase(B) boy has velocity only in y - direction (figure)(C) cont be correct because (a) is correct(D) figure]

Q.8 ASol. m1 x1 + m2 x2 = 0

mx + m {– (Lsin – x} = 0

x = 2sinL

Q.9 ASol. Using conservation of mechaincal energy

)cos–1(mgmv21mv

21 22

v

v

v = )cos–1(g

Q.10 BSol. xCM = 0 (because Fx = 0 )

yCM =21

2211

mmymym

= mm)}cos1(L{m)0(m

= 2L

(1 – cos)


Page # 7

Exercise #6 (Previous Year’s Question)SECTION-A

(IIT JEE Previous Year's Questions)Q.1Sol. (a) 100 m/s velocity of the ball is relative to ground

[Unless and until it is mentioned in the question, the velocity is always relative to ground]

30º

u = 100 m/s

ux

uyy (vertical)

y (horizontal)

120 m

C

A

Horizontal component of velocity of cannon ball,ux = u cos 30º

or ux = (100) ×23

= 50 3 m/s

and vertical component o its velocityuy = u sin 30º

uy = 100× 21

= 50 m/s

and vertical displacement of the ball when it strikes the carriage is – 120 m or

sy = uy t + 21

ay t2

– 120 = (50t) +

21

(–10)t2

t2 – 10t – 24 = 0 t = 12 s or – 2sIgnoring the negative time, we have(b) When it strikes the carriage, its horizontal component of velocity is still 50 3 m/s. It strikes to thecarriage. Let v2 be the velocity of (carriage + ball) system after collision. The applying conservation oflinear momentum in horizontal direction(mass of ball) (horizontal component of its velocity before collision) = (mass of ball + carrigage) (v2) (1 kg) (50 3 m/s) = (10 kg) (v2)

v2 = 35 m/sThe second ball is fired when the first ball strikes the carrige i.e. after 12s. In these 12x the car will moveforward a distance of 12 v1 or 60 3 m.


Page # 8

The second ball also take 12s to travel a vertical displacment of –120m. This ball will strike the carriage60 3 m in these 12s. This is possible only when resistive carriage after first collision. (v2) = 5 3 m/sHence, at the time of second collisionHorizontal component of velocity of ball = 50 3 m/s and horizontal velocity of carriage + first ball =

5 3 m/s. Let v be the disired velocity of carriage after second collision. Then conservation of linearmomentum in horizontal direction gives

1kg 50 m/s35 m/s3

v

Before collision

11 v = (1) (50 3 ) + (10) ( 35 ) = 100 3

v =11

3100 m/s or v = 15.75 m/s

In this particular problem values are so adjusted that even it we that the velocity of ball with respect tocar, we get the same results of both parts, although the method will be worng. ]

Ans t0 = 12 sec, v = 1003/11

Q.2 C

Sol. vCM =21

2211

mmvmvm

= 410041410

= 14140

= 10 m./s

v = 14 m/s1v = 02

m =10 kg2 m =4 kg2

Q.3 A, DSol. Initial momentum of the system 21 PP

= 0

Final momentum 21 'P'P

should also be zero.

option (b) is allowed becaure if we put c1 = – c2 0, 21 'P'P

will be zero. Similary, we can check otheroptions. correct option are (a) and (d).

Q.4 BQ.5 BQ.6 CSol 1. Between A and B, height fallen by block

h1 = 3 tan60º = 3m Speed of blcok just before striking the second incline,

v1 = 1gh2 = 3102 60 ms–1

In perfectly inelastic collision, component of v1 perpendicular to BC become zero, while component ofv1 parallel to BC will remain unchanged. Speed of block B immediately after it strikes the second incline is


Page # 9

30º30º

30º30ºB

Cv1

v2 = component of v1 along BC

= v1 cos 30º = ( 60 )

23

= 45 ms–1

correct option (B).2. Height fallen by the block from B to C

h2 = 3 3 tan 30º = 3 mLet v3 be the speed of block, at point C, just before it leaves the second incline then :]

Q.7 4

Sol. m 2m m

A B C

9m/s

for collision between A & Bm(9) + 0 = mVA + 2mVB (conservation of linear momentum)9 = VA + 2VB

& Vsep = Vapp (elastic collision)VB – VA = 9 – 0

VA = – 3m/s& VB = 6m/sfor collision between B & C

6(2m) = (m + 2m) V (conservation of linear momentum)

V = m3m12

= 4m/s

Q.8 C

Sol.

• •1 2Av 2v

12

••v

2v

I collision

v2v•

•

II collision

v2v• •

i.e. 2 collisions.


Page # 10

Q.9 ASol. C2(0, 0) m1 = 6m

C2 (– a, a) m2 = mC3 (a, a) m3 = mC4 (0, – a) m4 = mC5 (0, 0) m5 = m

ycm = m10mamama

ycm = 10a

Q.10 ACSol. v = 5v2 – 2 (1)

v = v2 + 92---------------2v = 6 v2v2 = v/3v = 5v/3 – 2-2v / 3 = – 2v = 3 m/sv2 = 1 m/s

Q.11 DSol. VBall xt = 20

t = g/H2 = 1 secVBall = 20 m/sVBullets × t = 100 m

sec1gH2t

VBullet = 100 m/sm1V1 = m1v1' + m2v20.01 × V = 0.01 × 100 + 0.2 × 20V = 500 m/s

SECTION-B (AIEEE Previous Year's Questions)

Q.12 B

Sol. 0.5×2 = (1 + 0.5) v

v = 32

5.11

Energy loss = Ki – kf


Page # 11

= 21

× 0.5 × 22 – 21

× 1.5 ×2

32

= 1 – 186

= 1 – 31

= 32

= 0.67 J

Q.13 D

Sol. xcm =

dm

xdm =

dx

Lxk

dxLxkx

n

n

n

n

=

L

0

n

L

0

1n

dxx

dxx

=

1nL

2nL

1n

2n

x =

2n1n

L

at n = 0 xcm = 2L

at n xcm = L

Q.14 CSol. First velocity is negetive and increasing. At the time of collision it becomes +ve from –ve again and again

hence graph CQ.15 AQ.16 B

Sol. Impulse = p = mV = m 2 V= 0.4× 2 × slope

= 0.4 × 2 × 22

= 0.8Q.17 C

Sol. Energy loss = 21

µ U2rel (1 – e2)

= 21

MmmM

V2 (1 – e2)


Page # 12

=

2mv

21

f

f = )e–1(Mm

M

Hence Statement I is fallse

center of mass & collisions/solution target iit-jee 2019 ...1 com exe. 2, 3, 4... · center of...

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