Page 1
29. When the spider eats up the moth and
travels towards A with velocity
v
2
relative
to rod.
24
2
48 0 m
v
v mv
R R
+
æ
è
ç
ö
ø
÷ + =
[v
R
= velocity (absolute) of rod]
Þ v
v
R
= 
6
Option (c) is correct.
30. Time taken by spider to reach point A
starting from point B
= +
4 8
2
L
v
L
v/
= = =
20 20
20
L
v
L
L T
T
/
= ´ = 20 4 80 s
31. Form CM not to shift
Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48
= ´ 64 48 m
i.e., x
L
¢ = 
8
3
Option (a) is correct.
More than One Cor rect Op tions
1. Along vertical : 2
2
45 mV m
v
sin sin q = × °
i.e., 2
2 2
V
v
sin q = …(i)
Along horizontal :
mv m
v
mV = ° +
2
45 2 cos cos q
i.e., 2 1
1
2 2
V v cos q = 
æ
è
ç
ö
ø
÷
…(ii)
Squaring and adding Eq. (i)and (ii),
8
2 2 2 2
2
2 2
V
v
v
v
=
æ
è
ç
ö
ø
÷
+ 
æ
è
ç
ö
ø
÷
= + +  × ×
v
v
v
v
v
2
2
2
8 7
2
2 2
= 
5
4 2
2 2
v v
Dividing Eq. (ii) by Eq. (i)
tan q =

1
2 2
2 2 1
2 2
=

<
1
2 2 1
1
\ q < ° 45
Thus, the divergence angle between the
particles will be less than
p
2
.
Option (b) is correct.
Initial KE =
1
2
2
mv
Final KE =
æ
è
ç
ö
ø
÷ +
1
2 2
1
2
2
2
2
m
v
mV
= + ×  ×
é
ë
ê
ù
û
ú
1
2
1
4
2
5
32
2
1
8 2
2
mv
As Final KE < Initial KE
Collision is inelastic.
Option (d) is correct.
2. v
m m
m m
v
2
2 1
1 2
2
¢ =

+
=

+
m m
m m
v
5
5
2
= 
2
3
2
v
= 
2
3
2gl
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  197
48 m
48 m
x'
8L M + S
v/2
m
2m
q
45°
L
Rest
2m m
v
v
2
v'
2
v'
1
v = 0
1
m = 5m
1
m = m
2
Page 2
29. When the spider eats up the moth and
travels towards A with velocity
v
2
relative
to rod.
24
2
48 0 m
v
v mv
R R
+
æ
è
ç
ö
ø
÷ + =
[v
R
= velocity (absolute) of rod]
Þ v
v
R
= 
6
Option (c) is correct.
30. Time taken by spider to reach point A
starting from point B
= +
4 8
2
L
v
L
v/
= = =
20 20
20
L
v
L
L T
T
/
= ´ = 20 4 80 s
31. Form CM not to shift
Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48
= ´ 64 48 m
i.e., x
L
¢ = 
8
3
Option (a) is correct.
More than One Cor rect Op tions
1. Along vertical : 2
2
45 mV m
v
sin sin q = × °
i.e., 2
2 2
V
v
sin q = …(i)
Along horizontal :
mv m
v
mV = ° +
2
45 2 cos cos q
i.e., 2 1
1
2 2
V v cos q = 
æ
è
ç
ö
ø
÷
…(ii)
Squaring and adding Eq. (i)and (ii),
8
2 2 2 2
2
2 2
V
v
v
v
=
æ
è
ç
ö
ø
÷
+ 
æ
è
ç
ö
ø
÷
= + +  × ×
v
v
v
v
v
2
2
2
8 7
2
2 2
= 
5
4 2
2 2
v v
Dividing Eq. (ii) by Eq. (i)
tan q =

1
2 2
2 2 1
2 2
=

<
1
2 2 1
1
\ q < ° 45
Thus, the divergence angle between the
particles will be less than
p
2
.
Option (b) is correct.
Initial KE =
1
2
2
mv
Final KE =
æ
è
ç
ö
ø
÷ +
1
2 2
1
2
2
2
2
m
v
mV
= + ×  ×
é
ë
ê
ù
û
ú
1
2
1
4
2
5
32
2
1
8 2
2
mv
As Final KE < Initial KE
Collision is inelastic.
Option (d) is correct.
2. v
m m
m m
v
2
2 1
1 2
2
¢ =

+
=

+
m m
m m
v
5
5
2
= 
2
3
2
v
= 
2
3
2gl
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  197
48 m
48 m
x'
8L M + S
v/2
m
2m
q
45°
L
Rest
2m m
v
v
2
v'
2
v'
1
v = 0
1
m = 5m
1
m = m
2
T
mv
l
mg 
¢
=
2
2
or T
mv
l
mg =
¢
+
2
= +
m g
mg
8
9
=
17
9
mg
Option (a) is correct.
Velocity of block
v
m
m m
v
1
2
1 2
2
2
¢ =
+
=
+
×
2
5
2
m
m m
gl
=
1
3
2gl
Option (c) is correct.
Maximum height attained by pendulum
bob
=
¢
= =
v
g
gl
g
l
g
2
2
2
8 9
2
4 /
Option (d) is correct.
3. v u cos cos f = q Þ
v
u
=
f
cos
cos
q
and e
v
u
=
f sin
sinq
=
f
´
f cos
cos
sin
sin
q
q
=
tan
tan
f
q
Option (b) is correct.
Change in momentum of particle
=  f  + ( sin ) ( sin ) mv mu q
\ Impulse delivered by floor to the particle
= mv mu sin sin f + q
=
f
+
é
ë
ê
ù
û
ú
mv
u
v
sin
sin
sin
q
q
= +
é
ë
ê
ù
û
ú
mv
u
v
e
u
v
sinq
= + mu e sin ( ) q 1
Option (d) is correct.
u e 1 1
2 2
  ( ) sin q
=  + u e 1
2 2 2
sin sin q q
= + u e cos sin
2 2 2
q q
= + f u
v
u
cos sin
2
2
2
2
q
= + f u v
2 2 2 2
cos sin q
= f + f v v
2 2 2 2
cos sin
=v
Option (c) is correct.
cos sin
2 2 2
q q + e
= +
f
cos
tan
tan
sin
2
2
2
2
q
q
q
= + cos ( tan )
2 2
1 q f
= f cos
2 2
q sec
=
f
cos
cos
2
2
q
= =
v
u
2
2
FinalKE
InitialKE
Option (d) is correct.
4. u i j
®

= + ( )
^ ^
3 2
1
ms
Impulse received by particle of mass m
=  +
® ®
m m u v
=  + +  + m m ( ) ( )
^ ^ ^ ^
3 2 2 i j i j
=  + m( )
^ ^
5 i j unit
Option (b) is correct.
Impulse received by particle of mass M
=  (impulse received by particle of
mass m)
= + m( )
^ ^
5i j
Option (d) is correct.
198  Mechanics1
v
v cos f
v sin f
u
f
u sin q
q
+
u cos q
m
u
M
v = (–2 i + j) m/s
^ ^
®
m
Page 3
29. When the spider eats up the moth and
travels towards A with velocity
v
2
relative
to rod.
24
2
48 0 m
v
v mv
R R
+
æ
è
ç
ö
ø
÷ + =
[v
R
= velocity (absolute) of rod]
Þ v
v
R
= 
6
Option (c) is correct.
30. Time taken by spider to reach point A
starting from point B
= +
4 8
2
L
v
L
v/
= = =
20 20
20
L
v
L
L T
T
/
= ´ = 20 4 80 s
31. Form CM not to shift
Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48
= ´ 64 48 m
i.e., x
L
¢ = 
8
3
Option (a) is correct.
More than One Cor rect Op tions
1. Along vertical : 2
2
45 mV m
v
sin sin q = × °
i.e., 2
2 2
V
v
sin q = …(i)
Along horizontal :
mv m
v
mV = ° +
2
45 2 cos cos q
i.e., 2 1
1
2 2
V v cos q = 
æ
è
ç
ö
ø
÷
…(ii)
Squaring and adding Eq. (i)and (ii),
8
2 2 2 2
2
2 2
V
v
v
v
=
æ
è
ç
ö
ø
÷
+ 
æ
è
ç
ö
ø
÷
= + +  × ×
v
v
v
v
v
2
2
2
8 7
2
2 2
= 
5
4 2
2 2
v v
Dividing Eq. (ii) by Eq. (i)
tan q =

1
2 2
2 2 1
2 2
=

<
1
2 2 1
1
\ q < ° 45
Thus, the divergence angle between the
particles will be less than
p
2
.
Option (b) is correct.
Initial KE =
1
2
2
mv
Final KE =
æ
è
ç
ö
ø
÷ +
1
2 2
1
2
2
2
2
m
v
mV
= + ×  ×
é
ë
ê
ù
û
ú
1
2
1
4
2
5
32
2
1
8 2
2
mv
As Final KE < Initial KE
Collision is inelastic.
Option (d) is correct.
2. v
m m
m m
v
2
2 1
1 2
2
¢ =

+
=

+
m m
m m
v
5
5
2
= 
2
3
2
v
= 
2
3
2gl
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  197
48 m
48 m
x'
8L M + S
v/2
m
2m
q
45°
L
Rest
2m m
v
v
2
v'
2
v'
1
v = 0
1
m = 5m
1
m = m
2
T
mv
l
mg 
¢
=
2
2
or T
mv
l
mg =
¢
+
2
= +
m g
mg
8
9
=
17
9
mg
Option (a) is correct.
Velocity of block
v
m
m m
v
1
2
1 2
2
2
¢ =
+
=
+
×
2
5
2
m
m m
gl
=
1
3
2gl
Option (c) is correct.
Maximum height attained by pendulum
bob
=
¢
= =
v
g
gl
g
l
g
2
2
2
8 9
2
4 /
Option (d) is correct.
3. v u cos cos f = q Þ
v
u
=
f
cos
cos
q
and e
v
u
=
f sin
sinq
=
f
´
f cos
cos
sin
sin
q
q
=
tan
tan
f
q
Option (b) is correct.
Change in momentum of particle
=  f  + ( sin ) ( sin ) mv mu q
\ Impulse delivered by floor to the particle
= mv mu sin sin f + q
=
f
+
é
ë
ê
ù
û
ú
mv
u
v
sin
sin
sin
q
q
= +
é
ë
ê
ù
û
ú
mv
u
v
e
u
v
sinq
= + mu e sin ( ) q 1
Option (d) is correct.
u e 1 1
2 2
  ( ) sin q
=  + u e 1
2 2 2
sin sin q q
= + u e cos sin
2 2 2
q q
= + f u
v
u
cos sin
2
2
2
2
q
= + f u v
2 2 2 2
cos sin q
= f + f v v
2 2 2 2
cos sin
=v
Option (c) is correct.
cos sin
2 2 2
q q + e
= +
f
cos
tan
tan
sin
2
2
2
2
q
q
q
= + cos ( tan )
2 2
1 q f
= f cos
2 2
q sec
=
f
cos
cos
2
2
q
= =
v
u
2
2
FinalKE
InitialKE
Option (d) is correct.
4. u i j
®

= + ( )
^ ^
3 2
1
ms
Impulse received by particle of mass m
=  +
® ®
m m u v
=  + +  + m m ( ) ( )
^ ^ ^ ^
3 2 2 i j i j
=  + m( )
^ ^
5 i j unit
Option (b) is correct.
Impulse received by particle of mass M
=  (impulse received by particle of
mass m)
= + m( )
^ ^
5i j
Option (d) is correct.
198  Mechanics1
v
v cos f
v sin f
u
f
u sin q
q
+
u cos q
m
u
M
v = (–2 i + j) m/s
^ ^
®
m
5. T m a =
1
and m g T m a
2 2
 =
Solving, a
m
m m
g =
+
2
1 2
( ) a
m a m
m m
x CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
1
1 2
=
+
m m
m m
g
1 2
1 2
2
( )
Option (b) is correct.
( )
( )
a
m m a
m m
y CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
2
1 2
=
+
æ
è
ç
ö
ø
÷
m
m m
g
1 2
2
Option (c) is correct.
6. As the block comes down, the CM of the
system will also come down i.e., it does not
remain stationary.
a
mg
m M
g
CM
=
+
¹
a
CM
is downwards and also a g
CM
< .
Option (d) is correct.
As no force acts along horizontal direction,
the momentum of the system will remain
conserved along horizontal direction.
Option (c) is correct.
7. Velocity of B after collision :
v
e
v
1 2
1
2
¢ =
+
æ
è
ç
ö
ø
÷
=
3
4
v [as e =
1
2
and v v
2
= (given)]
¹
v
2
Impulse given by A to B
= change in momentum of B
=
æ
è
ç
ö
ø
÷  × m v m
3
4
0
=
3
4
mv
Option (b) is correct.
Velocity of A after collision
v
e
v
2 2
1
2
¢ =

æ
è
ç
ö
ø
÷
=
v
4
Loss of KE during collision
=  ¢ + ¢
1
2
1
2
2
2
1
2
2
2
mv m v v ( )
= 
æ
è
ç
ö
ø
÷ 
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
1
2 4
3
4
2
2 2
m v
v v
=
3
16
2
mv
Option (c) is correct.
8. As the mass of the system keeps on
decreasing momentum of the system does
not remain constant.
Thrust force is developed on the rocket
due to Newton’s 3rd law of motion.
Option (b) is correct.
As, a
dv
dt
v
m
dm
dt
g
i
= = 
æ
è
ç
ö
ø
÷

The value of a will remain constant if v
i
and 
dm
dt
are constant.
Option (c) is correct.
F F
t net
= (Thrust force due to gas ejection)
 W (weight of rocket)
a
F
m
=
net
Thus, Newton’s 2nd law is applied.
Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  199
m
1
m
2
mg
2
a
T
T T
T
x
y
a
v
2
Before collision
v'
2
m
2
m
1
m
1
m
2
v = 0
1
A B
After collision
v'
1
B A
Page 4
29. When the spider eats up the moth and
travels towards A with velocity
v
2
relative
to rod.
24
2
48 0 m
v
v mv
R R
+
æ
è
ç
ö
ø
÷ + =
[v
R
= velocity (absolute) of rod]
Þ v
v
R
= 
6
Option (c) is correct.
30. Time taken by spider to reach point A
starting from point B
= +
4 8
2
L
v
L
v/
= = =
20 20
20
L
v
L
L T
T
/
= ´ = 20 4 80 s
31. Form CM not to shift
Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48
= ´ 64 48 m
i.e., x
L
¢ = 
8
3
Option (a) is correct.
More than One Cor rect Op tions
1. Along vertical : 2
2
45 mV m
v
sin sin q = × °
i.e., 2
2 2
V
v
sin q = …(i)
Along horizontal :
mv m
v
mV = ° +
2
45 2 cos cos q
i.e., 2 1
1
2 2
V v cos q = 
æ
è
ç
ö
ø
÷
…(ii)
Squaring and adding Eq. (i)and (ii),
8
2 2 2 2
2
2 2
V
v
v
v
=
æ
è
ç
ö
ø
÷
+ 
æ
è
ç
ö
ø
÷
= + +  × ×
v
v
v
v
v
2
2
2
8 7
2
2 2
= 
5
4 2
2 2
v v
Dividing Eq. (ii) by Eq. (i)
tan q =

1
2 2
2 2 1
2 2
=

<
1
2 2 1
1
\ q < ° 45
Thus, the divergence angle between the
particles will be less than
p
2
.
Option (b) is correct.
Initial KE =
1
2
2
mv
Final KE =
æ
è
ç
ö
ø
÷ +
1
2 2
1
2
2
2
2
m
v
mV
= + ×  ×
é
ë
ê
ù
û
ú
1
2
1
4
2
5
32
2
1
8 2
2
mv
As Final KE < Initial KE
Collision is inelastic.
Option (d) is correct.
2. v
m m
m m
v
2
2 1
1 2
2
¢ =

+
=

+
m m
m m
v
5
5
2
= 
2
3
2
v
= 
2
3
2gl
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  197
48 m
48 m
x'
8L M + S
v/2
m
2m
q
45°
L
Rest
2m m
v
v
2
v'
2
v'
1
v = 0
1
m = 5m
1
m = m
2
T
mv
l
mg 
¢
=
2
2
or T
mv
l
mg =
¢
+
2
= +
m g
mg
8
9
=
17
9
mg
Option (a) is correct.
Velocity of block
v
m
m m
v
1
2
1 2
2
2
¢ =
+
=
+
×
2
5
2
m
m m
gl
=
1
3
2gl
Option (c) is correct.
Maximum height attained by pendulum
bob
=
¢
= =
v
g
gl
g
l
g
2
2
2
8 9
2
4 /
Option (d) is correct.
3. v u cos cos f = q Þ
v
u
=
f
cos
cos
q
and e
v
u
=
f sin
sinq
=
f
´
f cos
cos
sin
sin
q
q
=
tan
tan
f
q
Option (b) is correct.
Change in momentum of particle
=  f  + ( sin ) ( sin ) mv mu q
\ Impulse delivered by floor to the particle
= mv mu sin sin f + q
=
f
+
é
ë
ê
ù
û
ú
mv
u
v
sin
sin
sin
q
q
= +
é
ë
ê
ù
û
ú
mv
u
v
e
u
v
sinq
= + mu e sin ( ) q 1
Option (d) is correct.
u e 1 1
2 2
  ( ) sin q
=  + u e 1
2 2 2
sin sin q q
= + u e cos sin
2 2 2
q q
= + f u
v
u
cos sin
2
2
2
2
q
= + f u v
2 2 2 2
cos sin q
= f + f v v
2 2 2 2
cos sin
=v
Option (c) is correct.
cos sin
2 2 2
q q + e
= +
f
cos
tan
tan
sin
2
2
2
2
q
q
q
= + cos ( tan )
2 2
1 q f
= f cos
2 2
q sec
=
f
cos
cos
2
2
q
= =
v
u
2
2
FinalKE
InitialKE
Option (d) is correct.
4. u i j
®

= + ( )
^ ^
3 2
1
ms
Impulse received by particle of mass m
=  +
® ®
m m u v
=  + +  + m m ( ) ( )
^ ^ ^ ^
3 2 2 i j i j
=  + m( )
^ ^
5 i j unit
Option (b) is correct.
Impulse received by particle of mass M
=  (impulse received by particle of
mass m)
= + m( )
^ ^
5i j
Option (d) is correct.
198  Mechanics1
v
v cos f
v sin f
u
f
u sin q
q
+
u cos q
m
u
M
v = (–2 i + j) m/s
^ ^
®
m
5. T m a =
1
and m g T m a
2 2
 =
Solving, a
m
m m
g =
+
2
1 2
( ) a
m a m
m m
x CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
1
1 2
=
+
m m
m m
g
1 2
1 2
2
( )
Option (b) is correct.
( )
( )
a
m m a
m m
y CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
2
1 2
=
+
æ
è
ç
ö
ø
÷
m
m m
g
1 2
2
Option (c) is correct.
6. As the block comes down, the CM of the
system will also come down i.e., it does not
remain stationary.
a
mg
m M
g
CM
=
+
¹
a
CM
is downwards and also a g
CM
< .
Option (d) is correct.
As no force acts along horizontal direction,
the momentum of the system will remain
conserved along horizontal direction.
Option (c) is correct.
7. Velocity of B after collision :
v
e
v
1 2
1
2
¢ =
+
æ
è
ç
ö
ø
÷
=
3
4
v [as e =
1
2
and v v
2
= (given)]
¹
v
2
Impulse given by A to B
= change in momentum of B
=
æ
è
ç
ö
ø
÷  × m v m
3
4
0
=
3
4
mv
Option (b) is correct.
Velocity of A after collision
v
e
v
2 2
1
2
¢ =

æ
è
ç
ö
ø
÷
=
v
4
Loss of KE during collision
=  ¢ + ¢
1
2
1
2
2
2
1
2
2
2
mv m v v ( )
= 
æ
è
ç
ö
ø
÷ 
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
1
2 4
3
4
2
2 2
m v
v v
=
3
16
2
mv
Option (c) is correct.
8. As the mass of the system keeps on
decreasing momentum of the system does
not remain constant.
Thrust force is developed on the rocket
due to Newton’s 3rd law of motion.
Option (b) is correct.
As, a
dv
dt
v
m
dm
dt
g
i
= = 
æ
è
ç
ö
ø
÷

The value of a will remain constant if v
i
and 
dm
dt
are constant.
Option (c) is correct.
F F
t net
= (Thrust force due to gas ejection)
 W (weight of rocket)
a
F
m
=
net
Thus, Newton’s 2nd law is applied.
Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  199
m
1
m
2
mg
2
a
T
T T
T
x
y
a
v
2
Before collision
v'
2
m
2
m
1
m
1
m
2
v = 0
1
A B
After collision
v'
1
B A
Match the Columns
1. If x
0
is the compression made in the
spring, the restoring force on B will
decrease from kx
0
to zero as the spring
regains its original length. Thus, the
acceleration of B will also decrease from
kx
m
B
0
to zero.
So, the a
CM
will also decrease from
kx
m m
A B
0
+
to zero.
\ (a) ® (r)
When spring is released after compressing
it, the restoring on B will accelerate it
towards right while the reaction force on
A will apply a force on the wall which in
turn will apply equal and opposite force on
A and consequently A will travel towards
right. As both travel towards right the
velocity of CM will be maximum in the
beginning.
After this A will start compressing the
spring and at a certain instant when the
spring is compressed to maximum value
both the blocks will travel towards right
with a constant velocity and then the
velocity of CM will become constant.
\ (b) ® (q)
As the blocks will never move along
yaxis, the ycomponent of the CM of the
two blocks will not change.
\ (d) ® (p)
As the two blocks will keep on moving
towards right (surface below being
smooth) the xcoordinate of the CM of the
blocks will keep on increasing.
\ (c) ® (s)
2. Initial a
m g m g
m m
CM
=
+ + +
+
( ) ( )
= + g
= + 10 SI unit
\ (a) ® (q)
Initial v
m m
m m
CM
=
 + ´
+
( ) 20 0
=  10
\   v
CM
= 10 SI unit
\ (b) ® (q)
For the time taken by the first particle to
return to ground
s ut at = +
1
2
2
0 20 5
2
=  + ( ) t t
Þ t = 4 s
Now, as the collision of the first particle
with the ground is perfectly inelastic, the
first particle will remain on ground at
rest.
Now, let us find the position of 2nd
particle at t = 5 s
s = + ( ) ( ) 0 5
1
2
10 5
2
= 125 m
The particle (2nd) will still be in space
moving downwards.
a
m m g
m m
CM
=
× + ×
+
0
= =
g
2
5 (SI unit)
\ (c) ® (p)
Velocity of 2nd particle at t = 5 s
v = + ´ 0 10 5
= 50 ms
1
200  Mechanics1
B A
m
B
m
A
k
B A
x
0
kx
0
kx
0
–1
20 ms
m
1st particle
+
180 m
2nd particle
u = 0
Page 5
29. When the spider eats up the moth and
travels towards A with velocity
v
2
relative
to rod.
24
2
48 0 m
v
v mv
R R
+
æ
è
ç
ö
ø
÷ + =
[v
R
= velocity (absolute) of rod]
Þ v
v
R
= 
6
Option (c) is correct.
30. Time taken by spider to reach point A
starting from point B
= +
4 8
2
L
v
L
v/
= = =
20 20
20
L
v
L
L T
T
/
= ´ = 20 4 80 s
31. Form CM not to shift
Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48
= ´ 64 48 m
i.e., x
L
¢ = 
8
3
Option (a) is correct.
More than One Cor rect Op tions
1. Along vertical : 2
2
45 mV m
v
sin sin q = × °
i.e., 2
2 2
V
v
sin q = …(i)
Along horizontal :
mv m
v
mV = ° +
2
45 2 cos cos q
i.e., 2 1
1
2 2
V v cos q = 
æ
è
ç
ö
ø
÷
…(ii)
Squaring and adding Eq. (i)and (ii),
8
2 2 2 2
2
2 2
V
v
v
v
=
æ
è
ç
ö
ø
÷
+ 
æ
è
ç
ö
ø
÷
= + +  × ×
v
v
v
v
v
2
2
2
8 7
2
2 2
= 
5
4 2
2 2
v v
Dividing Eq. (ii) by Eq. (i)
tan q =

1
2 2
2 2 1
2 2
=

<
1
2 2 1
1
\ q < ° 45
Thus, the divergence angle between the
particles will be less than
p
2
.
Option (b) is correct.
Initial KE =
1
2
2
mv
Final KE =
æ
è
ç
ö
ø
÷ +
1
2 2
1
2
2
2
2
m
v
mV
= + ×  ×
é
ë
ê
ù
û
ú
1
2
1
4
2
5
32
2
1
8 2
2
mv
As Final KE < Initial KE
Collision is inelastic.
Option (d) is correct.
2. v
m m
m m
v
2
2 1
1 2
2
¢ =

+
=

+
m m
m m
v
5
5
2
= 
2
3
2
v
= 
2
3
2gl
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  197
48 m
48 m
x'
8L M + S
v/2
m
2m
q
45°
L
Rest
2m m
v
v
2
v'
2
v'
1
v = 0
1
m = 5m
1
m = m
2
T
mv
l
mg 
¢
=
2
2
or T
mv
l
mg =
¢
+
2
= +
m g
mg
8
9
=
17
9
mg
Option (a) is correct.
Velocity of block
v
m
m m
v
1
2
1 2
2
2
¢ =
+
=
+
×
2
5
2
m
m m
gl
=
1
3
2gl
Option (c) is correct.
Maximum height attained by pendulum
bob
=
¢
= =
v
g
gl
g
l
g
2
2
2
8 9
2
4 /
Option (d) is correct.
3. v u cos cos f = q Þ
v
u
=
f
cos
cos
q
and e
v
u
=
f sin
sinq
=
f
´
f cos
cos
sin
sin
q
q
=
tan
tan
f
q
Option (b) is correct.
Change in momentum of particle
=  f  + ( sin ) ( sin ) mv mu q
\ Impulse delivered by floor to the particle
= mv mu sin sin f + q
=
f
+
é
ë
ê
ù
û
ú
mv
u
v
sin
sin
sin
q
q
= +
é
ë
ê
ù
û
ú
mv
u
v
e
u
v
sinq
= + mu e sin ( ) q 1
Option (d) is correct.
u e 1 1
2 2
  ( ) sin q
=  + u e 1
2 2 2
sin sin q q
= + u e cos sin
2 2 2
q q
= + f u
v
u
cos sin
2
2
2
2
q
= + f u v
2 2 2 2
cos sin q
= f + f v v
2 2 2 2
cos sin
=v
Option (c) is correct.
cos sin
2 2 2
q q + e
= +
f
cos
tan
tan
sin
2
2
2
2
q
q
q
= + cos ( tan )
2 2
1 q f
= f cos
2 2
q sec
=
f
cos
cos
2
2
q
= =
v
u
2
2
FinalKE
InitialKE
Option (d) is correct.
4. u i j
®

= + ( )
^ ^
3 2
1
ms
Impulse received by particle of mass m
=  +
® ®
m m u v
=  + +  + m m ( ) ( )
^ ^ ^ ^
3 2 2 i j i j
=  + m( )
^ ^
5 i j unit
Option (b) is correct.
Impulse received by particle of mass M
=  (impulse received by particle of
mass m)
= + m( )
^ ^
5i j
Option (d) is correct.
198  Mechanics1
v
v cos f
v sin f
u
f
u sin q
q
+
u cos q
m
u
M
v = (–2 i + j) m/s
^ ^
®
m
5. T m a =
1
and m g T m a
2 2
 =
Solving, a
m
m m
g =
+
2
1 2
( ) a
m a m
m m
x CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
1
1 2
=
+
m m
m m
g
1 2
1 2
2
( )
Option (b) is correct.
( )
( )
a
m m a
m m
y CM
=
+
+
1 2
1 2
0
=
+
m
m m
a
2
1 2
=
+
æ
è
ç
ö
ø
÷
m
m m
g
1 2
2
Option (c) is correct.
6. As the block comes down, the CM of the
system will also come down i.e., it does not
remain stationary.
a
mg
m M
g
CM
=
+
¹
a
CM
is downwards and also a g
CM
< .
Option (d) is correct.
As no force acts along horizontal direction,
the momentum of the system will remain
conserved along horizontal direction.
Option (c) is correct.
7. Velocity of B after collision :
v
e
v
1 2
1
2
¢ =
+
æ
è
ç
ö
ø
÷
=
3
4
v [as e =
1
2
and v v
2
= (given)]
¹
v
2
Impulse given by A to B
= change in momentum of B
=
æ
è
ç
ö
ø
÷  × m v m
3
4
0
=
3
4
mv
Option (b) is correct.
Velocity of A after collision
v
e
v
2 2
1
2
¢ =

æ
è
ç
ö
ø
÷
=
v
4
Loss of KE during collision
=  ¢ + ¢
1
2
1
2
2
2
1
2
2
2
mv m v v ( )
= 
æ
è
ç
ö
ø
÷ 
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
1
2 4
3
4
2
2 2
m v
v v
=
3
16
2
mv
Option (c) is correct.
8. As the mass of the system keeps on
decreasing momentum of the system does
not remain constant.
Thrust force is developed on the rocket
due to Newton’s 3rd law of motion.
Option (b) is correct.
As, a
dv
dt
v
m
dm
dt
g
i
= = 
æ
è
ç
ö
ø
÷

The value of a will remain constant if v
i
and 
dm
dt
are constant.
Option (c) is correct.
F F
t net
= (Thrust force due to gas ejection)
 W (weight of rocket)
a
F
m
=
net
Thus, Newton’s 2nd law is applied.
Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  199
m
1
m
2
mg
2
a
T
T T
T
x
y
a
v
2
Before collision
v'
2
m
2
m
1
m
1
m
2
v = 0
1
A B
After collision
v'
1
B A
Match the Columns
1. If x
0
is the compression made in the
spring, the restoring force on B will
decrease from kx
0
to zero as the spring
regains its original length. Thus, the
acceleration of B will also decrease from
kx
m
B
0
to zero.
So, the a
CM
will also decrease from
kx
m m
A B
0
+
to zero.
\ (a) ® (r)
When spring is released after compressing
it, the restoring on B will accelerate it
towards right while the reaction force on
A will apply a force on the wall which in
turn will apply equal and opposite force on
A and consequently A will travel towards
right. As both travel towards right the
velocity of CM will be maximum in the
beginning.
After this A will start compressing the
spring and at a certain instant when the
spring is compressed to maximum value
both the blocks will travel towards right
with a constant velocity and then the
velocity of CM will become constant.
\ (b) ® (q)
As the blocks will never move along
yaxis, the ycomponent of the CM of the
two blocks will not change.
\ (d) ® (p)
As the two blocks will keep on moving
towards right (surface below being
smooth) the xcoordinate of the CM of the
blocks will keep on increasing.
\ (c) ® (s)
2. Initial a
m g m g
m m
CM
=
+ + +
+
( ) ( )
= + g
= + 10 SI unit
\ (a) ® (q)
Initial v
m m
m m
CM
=
 + ´
+
( ) 20 0
=  10
\   v
CM
= 10 SI unit
\ (b) ® (q)
For the time taken by the first particle to
return to ground
s ut at = +
1
2
2
0 20 5
2
=  + ( ) t t
Þ t = 4 s
Now, as the collision of the first particle
with the ground is perfectly inelastic, the
first particle will remain on ground at
rest.
Now, let us find the position of 2nd
particle at t = 5 s
s = + ( ) ( ) 0 5
1
2
10 5
2
= 125 m
The particle (2nd) will still be in space
moving downwards.
a
m m g
m m
CM
=
× + ×
+
0
= =
g
2
5 (SI unit)
\ (c) ® (p)
Velocity of 2nd particle at t = 5 s
v = + ´ 0 10 5
= 50 ms
1
200  Mechanics1
B A
m
B
m
A
k
B A
x
0
kx
0
kx
0
–1
20 ms
m
1st particle
+
180 m
2nd particle
u = 0
\ At t = 5 s
v
m m
m m
CM
=
× + ×
+
0 50
=25 (SI unit)
\ (d) ® (s)
3. Initial KE of block B = 4 J
\
1
2
4
2
´ ´ = 0.5 u
Þ u = 4 ms
1
\ Initial momentum of B = ´ 0.5 4
=

2
1
kg ms
\ (a) ® (r)
Initial momentum
p p p
A B CM
= +
= + 0 2
=

2
1
kgms
\ (b) ® (r)
Velocity given to block B will compress the
spring and this will gradually increase the
velocity of A. When the spring gets
compressed to its maximum both the
blocks will have the same velocities i.e.,
same momentum as both have same mass.
p p
A B
=
(at maximum compression of the spring)
But, p p
A B
+ = initial momentum of B.
\ p p
A A
+ = 2
i.e., p
A
= 1 kgms
1
\ (c) ® (q)
After the maximum compression in the
spring, the spring will gradually expand
but now the velocity of block A will
increase and that of B will decrease and
when the spring attains maximum
expansion the velocity of B will be zero
and so will be its momentum.
\ (d) ® (p)
4. If collision is elastic, the two blocks will
interchange there velocities (mass of both
balls being equal).
Thus, velocity of A after collision = v
\ (a) ® (r)
If collision is perfectly inelastic, the two
balls will move together (with velocities V).
\ mv m m V = + ( )
Þ V
v
=
2
\ (b) ® (s)
If collision is inelastic with e =
1
2
,
v
e
v
1 2
1
2
¢ =
+
×
=
+
×
1
1
2
2
v [Q v v
2
= (given)]
=
3
4
v
\ (c) ® (p)
If collision is inelastic with e =
1
4
,
v v v
1
1
1
4
2
5
8
¢ =
+
æ
è
ç
ö
ø
÷
=
\ (d) ® (q).
5. If A moves x towards right
Let plank (along with B) move by x¢ to the
right.
\
x x ´ + ¢ +
+ +
=
30 60 30
30 60 30
0
( )
( )
i.e., x
x
¢ = 
3
=
x
3
, towards left.
\ (a) ® (r)
If B moves x towards left
Let plank (along with A) move x¢ to the
left
\
x x × + ¢ +
+ +
=
60 30 30
60 30 30
0
( )
( )
i.e., x x ¢ = 
= x, towards right
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  201
B A
m = 0.5 kg m
C 30 kg
B A
50 kg 60 kg
Smooth
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