Solutions Slot – 2 (Physics) Page # 1

COM
EXERCISE – I SINGLE CORRECT
1. D

a'
COM lie on this line
C' C''

–
5. B
C'  r 4r  C''
   M
 2 3 
M/3
 r 4r 
2
(2r )  
 2 3  L–x x
a' = r 2 M L
2r 2  (L – x) = M – x  x=
2 3 4
a' = 2(3r  8r) 6. C
3(4  )

4r m
Required Ans = a' + COM 2m
3 x1
R/2
2(3r  8r) 4r 2r
= + = m.R R
3(4  ) 3 3(4  ) = 2mx1  x1 =
2 4
2. C
R 5R
 Distance = R + =
4 4
a 7. C
M1 M2

a/3 COM of triangle

a/3

20m

M2   a2 a
r1 = M  M = 2 2 × 3
1 2 a  a x 8–x
2 80 (8 – x) = 200x
a a a
= 2 = 640 – 80x = 200x
3a (  1) 3(  1)
x = 2.3 m
3. B Now, Required distance
= 20 – (8 – x)
= 20 – (8 – 2.3) = 20 – 5.7
= 14.3 m
8. [C]

4  6
= 8cm
3

T T
Center of mass coincides a a
4. D m nm
nmg – T = nma

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com
 Page # 2 Solutions Slot – 2 (Physics)

T – mg = ma 16. (a) [B] (b) [C]

After solving (a) It could be non-zero, but it must be
constant.
(n  1)g
a= (b) It could be non-zero and it might not
n1 be constant.
  17. (a) [C] (b) [B]
m1a1  m2 a2
aCOM = Mv0 = (Nm + M)v'
m1  m2
Mv0
ma  nma v' =
a  na (Nm  M)
= = 18. A
(n  1)m n1
2 L sin 
n  1
=  g
n  1 v2
9. A L/2 cos
 x1
CMi
L /2 x2

L –x 2x x L(1 – cos )
L+ x a= g= g CMf
2L L
mx1 = mx2 [ Fx = 0]
x1 = x2
10. C
Now x1 + x2 = L sin 
L sin 
Initial  CMf =
 1 2
2L mg = mv2  u = g 19. D
2 2
VCMx = 0 and Fx = 0
11. B mv1 = mv2  v1 = v2 = v(let)
When internal force acts. Now E.C.
Net force is zero.
1 2
dP mg (1 – cos ) = 2  mv 
F= Momentum conserved.  2 
dt
v2 = g (1 – cos )
Internal force will not change the linear
momentum. 
Distance from centre of mass = R =
But due to force, K.E. increases. 2
12. D
Speed constant K.E.  Constant mv2 mg(1  cos )
So T = =
Gravitational potential energy change. R  /2
 T = 2mg (1 – cos )
 Momentum = mv
 Direction of v changes 20. A
 Momentum changes vmax = V = g(1  cos )1 / 2
13. D
21. B
P2 Only in vertical direction
= K.E. [ fx = 0 always]
2m
L L
P2 So displacement = – cos 
ln = ln K.E. 2 2
2m
L
2ln P – ln (2m) = ln K.E. = [1 – cos ]
Straight line with intercept. 2
22. D
14. D
Positive Negative
m1(g)  m2 (g)
acom = m1  m2 = –g urel–v'

15. C
C1 will move but C2 will be stationary with M1 M M2
respect to the ground. A B
v'

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com
 Solutions Slot – 2 (Physics) Page # 3
By momentum conservation
O = m1 (urel – v') – (m2v' + Mv') 4F/3 2M M 4F/3
m1(urel – v') = m1v' + Mv'
x2 x1
m1urel
v' = m  m  M
1 2
4F 4F 1
23. D x1 + x2 = k (x1 + x2)2
3 3 2
A B (u–v')
rel

8F
v' = (x1 + x2)
3K
m2(urel – v') = (m1 + M)v' 28. [B]
m2 urel – m2v' = m1v' + Mv'
v'(m1 + M + m2) = m2 urel
m2urel
v' = m  M  m
1 2
2d
24. A Time between two collisions = v
  0
Fnet  0 Vcom  0 v0
 COM is at rest. So no. of collision/sec =
2d
u u Impulse in one collision = v0 – (–mv0)
= 2mv0
v'
v0 mv20
–m1u + m2u + Mv = 0 F = 2mv0 × =
2d d
(m1  m2 ) 29. B
v' = u
M (i) From M.C. mv = 2mv'
25. [A] v' = v/2
m1 m2 (ii) from M.C. mv = 2mv'
urel+v'
v' = v/2
(iii) Impulse = mv = 3mv'
urel–v' v'
v
m2(Urel + v') + Mv' = m1(urel – v') v' =
3
| m1  m2 | Urel 30. B
v' = m1  m2  M –I = –m2v – mv
I = 3mv
26. D
1 1
W.D. = m(2v)2 – mv
v2
(urel –v') 2 2
1 3 2
m(u – v') = (M + m)v' = m u
2
murel
v' = u
(M  2m) Iu
W.D. =
2 2u I
(Urel + v'')
31. B
v''
m
(M  m)murel
m(urel + v'') + Mv'' =  
(M  2m)
27. B
P = 2mv cos 
2F 2M M F Favg unit volume
F = (2mv cos ) (nv) = 2mnv2 cos 
aCOM = F
3M
Pressure = = 2mnv2 cos  cos 
w.r. to COM area

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com
 Page # 4 Solutions Slot – 2 (Physics)

32. C 20 m/s

 F.dt = P 25 m/s

Given  F.dt  J e=
V1  20
Now 45

F.2dt  J'  J' = 2J V1

33. D v1 = 65 m/s
39. C
e=1
So K = 0
(at the time 1
T of collision) kf = ki = m u12  u 22 
2
v v 40. C
3m 2m
So, –Tt = 2mv – mu (for bullet) N N N

I = Tt = 3mv (for mass 3m)
3mv = 2mv – mu 3R/2 3R/2
v = u/5
u'
3mu
I= 
5 2R
34. B
N N
If e = 1 then angle = 45°
If 0 < e < 1 then angle is less than
45° with the horizontal.
35. B
 2Nsin  .dt  mv 0 ..........(i)
N
 Ncos  . dt  mv 0

mv 0
Vf  N.dt =
2 5
.3
3

(3 / 2)R 2  R 2
si
n

sin  =
30

3 / 2R
°

3 cos 30°
=
1.

5 2
5

3 m/sec
sin  = ; cos
3 3
Vf cos 30° = 1.5
vf = 3 m/s mv 0 3 2 v0
= mv'  v' =
36. D 2 5 3 5
1 41. C
0.25 × 0.45 × 10 = 1 + (0.25)v2 mv 0
2 2N sin dt =
v = 1 m/s  2
Ball B is heavy
37. B
From momentum conservation
 N. cos dt = mv'
1 × 21 – 2 × 4 = 1 × 1 + 2 × v' mv 0
1m/s 6m/s
 2N sin dt  2
v' = 6 m/sec. A B  N cos dt  mv'
1 mv 0
e= = = 0.2 5
38. A
5  2N  3
dt 
2
2N
 dt  mv'
3
394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671
IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com
 Solutions Slot – 2 (Physics) Page # 5
On dividing 46. A
2N  5 3 v0
× = 1
3 2N 2v' mgh = mv2 h
2
v0
v' = v= 2gh
2 5
By momentum conservation
42. D
m 2gh + 0 = 2mv'
O
h/2

hOO n 2gh
t= v' =
g 2
gt
By energy conservation
2 1 (2gh)
h h 1  h (2m)v'2 = 2mgh', m = 2mgh'
=v – g   2 4
2 g 2  g
 
h
h' =
h 4
h=v , v= hg 47. B
g
v
h 2m m v 2mv  em(0  v)
vf = hg  g = e=1
g 3 3m
1 2
vf = 0 48. D
Now mg.t = 3mv' –1 before
gt/3 = v' v(ms) collision collision after collision
1.0
1 h 1 0.8
 3m (gt/3)2 + 3mg = 3m.v12
2 2 2

10gh
v1 =
3 0.2
43. A
1 2 t(s)
u v1 v2
u=0 (i)  v is +ve for both.
A B  A B (ii) Yes (when maximum compression)
(iii)  S have greater velocity after collision then R
v 2  v1 have before collision and K.E. of S will be less
c=  cu = v2 – v1 ....... (1) then initial K.E. of R
u
Now mu = mv1 + mv2 1 1
msVs2 < mR (VR)2
u = v1 + v2 ....... (2) 2 2
v1 but VS > VR So ms < mR
1e
from (1) and (2) v = 49. C
2 1e v2 – u2 = 2aS
44. D
Infinite
45. A h
4
0 v v'  2gh
5
1 st m m 4m
v 2gh
A B C
2
4 
mv  4m(0  v) 3m 0–  2gh  = 2x – g × h
2nd v= = 5 
5m 5
16
3v/5 × 2gh = 2gh
25
m m
16
h' = h
 3v/5 m m 25

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com
 Page # 6 Solutions Slot – 2 (Physics)

16 dm
or h' = e2h = h [e = 4/5] Ft = Vr
25 dt
50. A dm
Vr – mg = m + a
2gh dt

1 2 1 dm
2
m  2gh + mgh =
2
mv2 980 ×
dt
– 4000 × 9.8 = 4000 + 196

1 v  2gh
dm
v = 2 gh  e , = 120 kg/sec.
2 dt
51. C 57. D
P = mv
2  10  5 = 10 m/sec. dP dv dm
=m +v
dt dt dt
10 2  e  10 2  e2  10
 + + + ....
10 10 10 dv
Fnet = (m0t) + v
1 + 2 [e + e2 + ....] dt
2e
1+ = 3 sec.
1e
52. B

2v cos 
g

53. D
dm
Fthurst = Vrel.
dt
dm
= (10 × (10–2)3 × 103) = 0.1 kg/sec.
dt
Volume
AVrel = = 10 × (10–2)3
sec .

10 5 20
= 2 =
 1  10  3  
 
 2 
 

20
Fthrust = × 0.1 = 0.127 N

54. C
5 × 103 × 1.2 = 6 × 103 × v
v = 1 m/sec.
55. D
dm
Fthurst = Vrel .
dt
dm
210 = 300 .
dt
dm
= 0.7 kg/sec.
dt
56. C

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www.motioniitjee.com, info@motioniitjee.com