GUIDED REVISION
JEE (Advanced) 2024 2024
JEE (Advanced)
ENTHUSIAST
ENTHUSIAST COURSE COURSE
STAR BATCH
STAR BATCH
PHYSICS GR # CENTRE OF MASS AND MOMENTUM
SOLUTION
SECTION-I
Single Correct Answer Type 19 Q. [3 M (–1)]
1. Ans. (C)
Sol. To remain in equilibrium C must be CM of semicircular disc.
4R
(\ IMg = 0) So OC = = 1.33 m
3p
2. Ans. (C)
u u
Sol. In (i) & (ii) will be no impulse by string & A will not move v B = but in (iii), v B = , A will also move
2 3
u
vA =
3
3. Ans. (C)
4. Ans. (B)
5. Ans. (C)
6. Ans. (A)
7. Ans. (C)
u/3 u/3
B A
Sol. 2m m
mu = 2mV + mV
V = u/3
1
2
(
k 02 - x 2 )
1 1 1
m ( u / 3 ) + 2m ( u / 3 ) - mu 2
2 2
=
2 2 2
8. Ans. (B)
1
Sol. F (x1 + x2) = k (x1 + x2)2
2
2F
(x1 + x2) =
K
9. Ans. (A)
2 m×v
v0
Sol.
q
mv 1
m × v = mv1 sin q
2mv0 = mv1 cos q
Physics / GR # Centre of mass and Momentum E-1/9
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
v
Þ = tan q
2v 0
1
´ 2mv 20 = 3 ´ 10 9
2
1
´ mv 2 = 2 ´ 10 9
2
3 ´ 10 9 4 ´10 9
v0 = v=
m m
4 1
2´ = tan q =
3 3
q = 30°
4 ´ 10 9 10 9
v1 = 2 =4
m m
1 1
E f = ´ mv 2 + mv12
2 2
1 10 9
= + ´ ´ ´
2 m 16 = 10GJ
2 m
10. Ans. (D)
11. Ans. (A)
12. Ans. (C)
Sol. Let bigger prism moves a distance X towards left then in horizontal direction
2L
5MX = M (4L–x) Þ 6MX = 4ML Þ X =
3
5M
M
5L
13. Ans. (D)
e(vcos37°)
vsin37° vsin37°
vcos37°
Sol. v=Ö2gh
Before collision After collision
Þ After collision, net velocity in vertical direction will be
vy = (e v cos 37°) cos 37° – (v sin 37°) sin 37°
E-2/9 Physics / GR # Centre of mass and Momentum
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
2 2
3 æ4ö æ3ö
= v ´ç ÷ - vç ÷
4 è5ø è5ø
3v
Þ vy =
25
v 2y 9v 2 9 9h
Þ h max = = = ´ 2gh =
2g 1250g 1250g 625
14. Ans. (B)
ò (dm)(r) x
Sol. x cm =
ò dm dx
a
ò rdm = ò 2kx ´ rxdx
2
a
2kra 4
ò0
3
= 2kp x dx =
4
ò dm = ò 2kx rdx
2
2kra 3 3
= ; x cm = a
3 4
ycm = 0 as object is symmetric about x = 0
15. Ans. (C)
Sol. Initial mass = 4000 kg
µvrel – mg = ma
µ × 980 – 4000 × 9.8 = 4000 × 19.6
µvrel
mg
Þ µ = 40 + 80 = 120
16. Ans. (D)
v
Sol. M
Energy loss will be maximum when collision will be perfectly elastic
mM v'
(By momentum) mv = (m + M)v'
mv
Þ v' =
m+M
Maximum energy loss = Ki – Kf
1 1
= mv2 – (m + M)v'2
2 2
Physics / GR # Centre of mass and Momentum E-3/9
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
1 1 m 2v 2
= mv – (m + M)
2
2 2 (m + M)2
1 é m ù
= mv 2 ê1 - ú
2 ë m + Mû
æ M ö1
=ç ÷ mv2
èm +Mø2
statement 1 is false.
17. Ans. (C)
Sol. Area of F-t graph gives change in momentum and above area is taken positive and down area is negative.
F(t)
4N
q 3s 4.5s
t
O q
x
4 x
tan q = = Þ x =2
3 1.5
1 1
Þ Area = ´ 3 ´ 4 + ( -2 )(1.5 ) = mv
2 2
Þ 6 – 1.5 = 2v
4.5
Þ v= = 2.25
2
1 1
mv 2 = 2 ´ ( 2.25 ) = 5.06 J
2
Þ K.E. =
2 2
18. Ans. (B)
Sol. When a particle of mass m, moving with velocity u, collides with stationary particle of mass km elastically,
2u
then velocity of second particle is , here velocity of 4kg particle after all collision is
km + 1
æ 20 ö
2ç ÷
è m +1 ø = 40
æ4 ö æ 4ö
ç m + 1÷ ç5 + m + m ÷
è ø è ø
This is maximum when m = 2kg
19. Ans. (A)
Sol.
From frame of plate, particles come at v and go at v.
Dp = 2mv
E-4/9 Physics / GR # Centre of mass and Momentum
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
dp dN
F= = ´ 2mv
dt dt
dN dv
= ´ ´ 2mv
dv dt
= n × A × 2mv2
dv
Mg – 2mnAv2 = Mv
dx
vdv dx
ò Mg - 2mnAv M 2
=ò
Mg – 2mnAv = z – 4mnAvdv = dz
2
v 4mnAx
ln(Mg - 2mnAv 2 ) = -
0 M
4mnAx
Mg - 2mnAv 2 -
= -e M
Mg
4mnAx
-
1-e M
v=
2mnA
4´10 -3 ´104 ´10-2 ´10
-
1-e 1
æ 1 ö
= -3 4 -2
= 5 ç1 - 4 ÷
2 ´ 10 ´ 10 ´ 10 è e ø
Multiple Correct Answer Type 4 Q. [4 M (–1)]
20. Ans. (A, C)
21. Ans. (A,B,C)
Sol. m1u1 + m2u2 = m1v1 + m2v2
v2 = 6 m/s
æ v - v1 ö æ 6 - 1 ö 5
e = -ç 2 =ç ÷= = 0.2
è u2 - u1 ø÷ è 4 + 21ø 25
1 1 1 1
´ 1 ´ ( 21) + ´ 2 ´ ( 4) - ´ 1 ´ 1 - ´ 2 ´ 62 = 200J , mDv = 1 × (21–1) = 20 N-s
2 2
Dk =
2 2 2 2
22. Ans. (A,B,C,D)
dm
Sol. = 5kg / s
dt f
vC = 0.75 m/s
Fext – f = 0 f
Fext = f Fext
r mdvr r dm
f= - v rel
dt dt
(
f iˆ = - 0 - vC iˆ ) dm
dt
f = 0.75 × 5
Physics / GR # Centre of mass and Momentum E-5/9
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
f = 3.75 N
f
W = ò Fext dS = 3.75 (S) = 3.75 (v t)
C
i
W = 3.75 (0.75 × 1)
= 2.81 J
dk 1 æ dm ö 2
= v
dt 2 çè dt ÷ø
1
( 5) ´ ( 0.75 )
2
=
2
= 1.41 J
23. Ans. (A, B, D)
Sol.
u cos 60° = v1 cos 30°
u
v1 =
3
v1 cos60° 1
e= =
u cos30° 3
1 v sin f v sin f
= 2 °
= 2
6 v1 cos30 u 3
´
3 2
v1 cos 60° = v2 cos f
u 1
´ = v 2 cos f
3 2
u
= v 2 sin f
12
1 1 4 3
v2 = u + =
144 12 12
1 1
tan f = ´2 3 =
12 2 3
E-6/9 Physics / GR # Centre of mass and Momentum
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
SECTION-II
Numerical Answer Type Question 1Q.[3M(0)]
(upto second decimal place)
24. Ans. 1.25 m
SECTION-III
Numerical Grid Type (Ranging from 0 to 9) 2 Q. [4 M (0)]
25. Ans. 4
26. Ans. 3 N
Sol. For collision in normal direction : eucos30° = vcos60°
u 30° 60°
1 v
and in tangential direction usin 30° = vsin60° Þ we get e = 30° 60°
3
SECTION-IV
Matrix Match Type (4 × 5) 2 Q. [8 M (for each entry +2(0)]
27. Ans. (A)®(P,R,T); (B)®(Q,S); (C)®(P,S,T); (D)®(P, S, T)
Sol. No external force along horizontal in A, C, D so center of mass will remain at rest and linear momentum
will always remain constant.
In B work is done by F so mechanical energy increases.
In C, D internal forces of man does work so mechanical energy increases
28. Ans. (A)-P,Q,R,S; (B)-P,R,S; (C)-P,Q; (D)-P,Q
Sol. 2 3 6
3´ 2 2
v CM = =
9 3
centre of mass frame
4/3 2/3
3 6
(A) no deformation
velocity in ground frame
2 3 6 v=0
centre of mass frame
v=0 v=0
3 6
(B) no deformation
velocity in ground frame
2/3 3 2/3 6
centre of mass frame
4/3 2/3
3 6
(C) Deformation is zero
velocity in ground frame
3 2/3 6 v=0
Physics / GR # Centre of mass and Momentum E-7/9
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
centre of mass frame
v=0 v=0
3 6
(D) Deformation is maximum
velocity in ground frame
2/3 3 2/3 6
Subjective Type 4Q. [4 M (0)]
10
29. Ans. v1 = m/s, t = 2/3 sec, solved
3
13
30. Ans. (a) 20 , (b) , (c) 15
2
r
Sol. v G = 4 ĵ
r
v P = 2î
r
(a) v G / P = 4ˆj - 2î = - 2î + 4ˆj
r
\ v G / P = 20
r 2 ´ 4ˆj + 6 ´ 2î 3î
(b) v CM = = + ĵ
8 2
r 9 13
v CM = +1 =
4 2
r r r 3
(c) v G / CM = v G – v CM = - î + 3ˆj
2
r r r 1
v P / CM = v P – v CM = î - ˆj
2
1 æ9 ö 1 æ1 ö 45 15
KEsys/CM = (2)ç + 9 ÷ + ´ 6ç + 1÷ = + = 15
2 è4 ø 2 è4 ø 4 4
r
31. Ans. PPM = m vPM
= - mv2 sin w t iˆ + m(v2 cos w t - v1 ) ˆj
v2 v2cosq
q
v2sinq
v1
q=wt
Sol.
E-8/9 Physics / GR # Centre of mass and Momentum
GUIDED REVISION
JEE (Advanced) 2024
ENTHUSIAST COURSE
STAR BATCH
v2
w=
R
r
v rel = ( -v 2 × sin q ) iˆ + ( v 2 cos q - v1 ) ˆj
= ( - v 2 × sin wt ) iˆ + ( v 2 cos wt - v1 ) ˆj
r r
Prel = m × v rel
= -mv 2 sin wt iˆ + m ( v 2 cos wt - v1 ) ˆj
32. Ans: vB = 55/36m/s, vA = 11/6 m/s, solved
Sol. Given vD/C = 6 m/sec
Case Ist
C.O.L.M.
r r
O = MD v D + mB v B
r r
v D = -6i + v B
\ i.e. vD = – 5î toword A
Case II In this condition when dog & trolly A will move with common speed again
nd
C.O.L.M. – 5i × 4 = (20 + 4) vcommon
vcommon = – 5/6 î
so will jumping from A ® B
r r
v D = 6î + v A
î = ( 6î + vr × 4 + vr A . 20
( )
5
– (24) ×
6 A
11
vA = – î
6
Case IIIrd
11 25
vD = 6 – î = î
6 6
r r
MD v D + M B v B = (MB + MD) vcommon
11
\ vcommon = m/sec Ans.
6
Physics / GR # Centre of mass and Momentum E-9/9
