Page 1
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m
v
1 1 1
1
0 5
3
+ ´ = +
m
v
m g l
1
1
2
3
5 × =
\ v
m
m
g l
1
1
3
2
5 =
Option (b) is correct.
2. mu mv = 2
Þ v
u
=
2
0 2
2 2
= +  v g H ( )
Þ H
v
g
=
2
2
=
( / ) u
g
2
2
2
=
u
g
2
8
or H
gl
g
=
2
8
or l
l
( cos ) 1
4
 = a
or 1
1
4
 = cos a
or a =

cos
1
3
4
Option (c) is correct.
3. Here     v v
2 1
® ®
= = v say
Net momentum of the two elements as
shown in figure = =
®
dp   dp
=2vdmsinq
=
æ
è
ç
ö
ø
÷ 2v
M
R
Rd
p
q q ( )sin
=
2Mv
d
p
q q sin
p
Mv
d =
ò
2
0
2
p
q q
p /
sin
= 
2
0
2
Mv
p
q
p
[cos ]
/
=  
2
0 1
Mv
p
[ ] =
2Mv
p
Option (b) is correct.
4.
1
2
2
1
2
2
0
2
( ) m v kx =
or v
k
m
x =
2
0
or 2 2
0
mv mk x =
or F t mkx
av
´ = D 2
0
or F
mk
t
x
av
=
2
0
D
Option (b) is correct.
190  Mechanics1
l
v
1
m
1
m
Rest
u = Ö5gl
v /3
1
m
H
l
m
a
2m
m
v
m m
Rest
u = Ö2gl
R
q
q
d q
d q
2q
dm
v
2
v
1
dp
dm
v
1
®
®
®
®
dl
kx
0
kx
0
Page 2
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m
v
1 1 1
1
0 5
3
+ ´ = +
m
v
m g l
1
1
2
3
5 × =
\ v
m
m
g l
1
1
3
2
5 =
Option (b) is correct.
2. mu mv = 2
Þ v
u
=
2
0 2
2 2
= +  v g H ( )
Þ H
v
g
=
2
2
=
( / ) u
g
2
2
2
=
u
g
2
8
or H
gl
g
=
2
8
or l
l
( cos ) 1
4
 = a
or 1
1
4
 = cos a
or a =

cos
1
3
4
Option (c) is correct.
3. Here     v v
2 1
® ®
= = v say
Net momentum of the two elements as
shown in figure = =
®
dp   dp
=2vdmsinq
=
æ
è
ç
ö
ø
÷ 2v
M
R
Rd
p
q q ( )sin
=
2Mv
d
p
q q sin
p
Mv
d =
ò
2
0
2
p
q q
p /
sin
= 
2
0
2
Mv
p
q
p
[cos ]
/
=  
2
0 1
Mv
p
[ ] =
2Mv
p
Option (b) is correct.
4.
1
2
2
1
2
2
0
2
( ) m v kx =
or v
k
m
x =
2
0
or 2 2
0
mv mk x =
or F t mkx
av
´ = D 2
0
or F
mk
t
x
av
=
2
0
D
Option (b) is correct.
190  Mechanics1
l
v
1
m
1
m
Rest
u = Ö5gl
v /3
1
m
H
l
m
a
2m
m
v
m m
Rest
u = Ö2gl
R
q
q
d q
d q
2q
dm
v
2
v
1
dp
dm
v
1
®
®
®
®
dl
kx
0
kx
0
5. As the collisions of the striker and the
walls of the carrom are perfectly elastic the
striker will follow the path OPQROP ...
Change in KE = work done against friction
1
2
0
2
mv mgs  = m
(m = mass of striker, s = displacement of
the striker)
Þ s
v
g
=
2
2m
=
´ ´
( ) 2
2 10
2
0.2
=1 m
PQ OP OA = = 2 =
æ
è
ç
ö
ø
÷
=
1
2 2
2
1
2
m
OP PQ + = + =
1
2
1
2
1 m
\ Striker will stop at point Q where
coordinates are
1
2 2
1
2
,
æ
è
ç
ö
ø
÷
.
Option (a) is correct.
6. As no force is acting on the system along
horizontal, the CM of the system will not
shift horizontally.
4 1 1 p p =  ( )
Þ p =
1
5
m
Displacement ( ) x of bar when pendulum
becomes vertical
x
p
=sinq
x p = sinq
= °
1
5
30 sin =
1
10
m = 0.1 m
When the ball reaches the other extreme
end the bar will further shift to the left by
distance x and as such the net
displacement of the bar will be 2x i.e.,
0.2 m.
Option (b) is correct.
7.
Momentum imparted to the floor in
1st collision =   = + p ep p e ( ) ( ) 1
2nd collision =   = + ep e p ep e ( ) ( )
2
1
3rd collision =   = + e p e p e p e
2 3 2
1 ( ) ( )
As theoretically there will be infinite
collision, total momentum imparted to
floor
= + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1
2
= + + + + ¥ p e e e ( )[ ] 1 1
2
K
= +

p e
e
( ) 1
1
1
=
+

æ
è
ç
ö
ø
÷ p
e
e
1
1
Option (d) is correct.
8. Let F be the frictional force applied by
plate when bullet enters into it
\
1
2
2
mu Fh = …(i)
If plate was free to move
mu M m u + = + ¢ 0 ( )
\ u
m
M m
u ¢ =
+
New KE of bullet =  + ¢
1
2
1
2
2 2
mu M m u ( )
(entering plate)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  191
x
q
p
4 kg
4 kg
x
1 kg
CM
1 – p
x
CM
ep p
During
first
collision
Floor
2
ep ep
3
ep
2
ep
Floor
During
Second
collision
During
third
collision
45°
P R
Q
O
A
45°
Page 3
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m
v
1 1 1
1
0 5
3
+ ´ = +
m
v
m g l
1
1
2
3
5 × =
\ v
m
m
g l
1
1
3
2
5 =
Option (b) is correct.
2. mu mv = 2
Þ v
u
=
2
0 2
2 2
= +  v g H ( )
Þ H
v
g
=
2
2
=
( / ) u
g
2
2
2
=
u
g
2
8
or H
gl
g
=
2
8
or l
l
( cos ) 1
4
 = a
or 1
1
4
 = cos a
or a =

cos
1
3
4
Option (c) is correct.
3. Here     v v
2 1
® ®
= = v say
Net momentum of the two elements as
shown in figure = =
®
dp   dp
=2vdmsinq
=
æ
è
ç
ö
ø
÷ 2v
M
R
Rd
p
q q ( )sin
=
2Mv
d
p
q q sin
p
Mv
d =
ò
2
0
2
p
q q
p /
sin
= 
2
0
2
Mv
p
q
p
[cos ]
/
=  
2
0 1
Mv
p
[ ] =
2Mv
p
Option (b) is correct.
4.
1
2
2
1
2
2
0
2
( ) m v kx =
or v
k
m
x =
2
0
or 2 2
0
mv mk x =
or F t mkx
av
´ = D 2
0
or F
mk
t
x
av
=
2
0
D
Option (b) is correct.
190  Mechanics1
l
v
1
m
1
m
Rest
u = Ö5gl
v /3
1
m
H
l
m
a
2m
m
v
m m
Rest
u = Ö2gl
R
q
q
d q
d q
2q
dm
v
2
v
1
dp
dm
v
1
®
®
®
®
dl
kx
0
kx
0
5. As the collisions of the striker and the
walls of the carrom are perfectly elastic the
striker will follow the path OPQROP ...
Change in KE = work done against friction
1
2
0
2
mv mgs  = m
(m = mass of striker, s = displacement of
the striker)
Þ s
v
g
=
2
2m
=
´ ´
( ) 2
2 10
2
0.2
=1 m
PQ OP OA = = 2 =
æ
è
ç
ö
ø
÷
=
1
2 2
2
1
2
m
OP PQ + = + =
1
2
1
2
1 m
\ Striker will stop at point Q where
coordinates are
1
2 2
1
2
,
æ
è
ç
ö
ø
÷
.
Option (a) is correct.
6. As no force is acting on the system along
horizontal, the CM of the system will not
shift horizontally.
4 1 1 p p =  ( )
Þ p =
1
5
m
Displacement ( ) x of bar when pendulum
becomes vertical
x
p
=sinq
x p = sinq
= °
1
5
30 sin =
1
10
m = 0.1 m
When the ball reaches the other extreme
end the bar will further shift to the left by
distance x and as such the net
displacement of the bar will be 2x i.e.,
0.2 m.
Option (b) is correct.
7.
Momentum imparted to the floor in
1st collision =   = + p ep p e ( ) ( ) 1
2nd collision =   = + ep e p ep e ( ) ( )
2
1
3rd collision =   = + e p e p e p e
2 3 2
1 ( ) ( )
As theoretically there will be infinite
collision, total momentum imparted to
floor
= + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1
2
= + + + + ¥ p e e e ( )[ ] 1 1
2
K
= +

p e
e
( ) 1
1
1
=
+

æ
è
ç
ö
ø
÷ p
e
e
1
1
Option (d) is correct.
8. Let F be the frictional force applied by
plate when bullet enters into it
\
1
2
2
mu Fh = …(i)
If plate was free to move
mu M m u + = + ¢ 0 ( )
\ u
m
M m
u ¢ =
+
New KE of bullet =  + ¢
1
2
1
2
2 2
mu M m u ( )
(entering plate)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  191
x
q
p
4 kg
4 kg
x
1 kg
CM
1 – p
x
CM
ep p
During
first
collision
Floor
2
ep ep
3
ep
2
ep
Floor
During
Second
collision
During
third
collision
45°
P R
Q
O
A
45°
=  +
+
æ
è
ç
ö
ø
÷
1
2
1
2
2
2
mu M m
m
M m
u ( )
= 
+
é
ë
ê
ù
û
ú
1
2
1
2
mu
m
M m
=
+
æ
è
ç
ö
ø
÷
1
2
2
mu
M
M m
= ¢ Fh …(ii)
Dividing Eq. (ii) by Eq. (i),
h
h
M
M m
¢
=
+
i.e., h
M
M m
h ¢ =
+
æ
è
ç
ö
ø
÷
Option (a) is correct.
9.
Now, v
e
v
1 2
1
2
¢ =
+
(entering plase)
\ 10
1
2
16 =
+
×
e
i.e., e=
1
4
Option (b) is correct.
10. y
l b
l
L b l
L
l b L b
CM
=
× + ×
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
× + ×
æ
è
ç
ö
ø
÷
( )
( )
2
1
2 3
1
2
l
bl bL l L b
bl
bL
=
+ +
+
2 2
2 2 6
2
y l
CM
according
toquestion
= é
ë
ê
ê
ê
ù
û
ú
ú
ú
or l l
L l L l L
+
é
ë
ê
ù
û
ú
= + +
2 2 2 6
2 2
or
l L
2 2
2 6
=
or l
L
=
3
11.
mv m v mv
1 1 2 2 1 2
=  …(i)
e =1
\ 2
2 1
v v =
\ m v m v m v
1 2 2 2 1 2
2 ( ) = 
\ 3
1 2
m m =
or
m
m
1
2
1
3
=
12. As the resultant of the velocities of 1st
and 2nd are just opposite to that of 3rd,
the 4th particle will travel in the line in
which 3rd is travelling.
Let the velocity of 4th particle is u as
shown in figure.
\ u v v cos cos 45 45 ° + ° =
i.e., u v =  ( ) 2 1
Total energy released
= + + + 
1
2
1
2
1
2
1
2
2 1
2 2 2 2
mv mv mv m v [ ( )]
= + 
1
2
3 2 1
2 2
mv [ ( ) ] =  mv
2
3 2 [ ]
Option (a) is correct.
13.
192  Mechanics1
1st
2nd
45°
4th
u
3rd
v
v
v
m m
A B A B
Rest
–1
v = 16 ms
2
2
–1
v = 2gs = 10 ms
1 ' v
2
mu mv = 2
1
i.e., v
u
1
2
=
C will increase
the tension in
the string.
mu mv = 2
2
i.e., v
u
2
2
=
C will also increase the
weight of B when collision
takes place. Thus,
mu m mv = + ( ) 2
3
i.e., v
u
3
3
=
A
m
1
v
1
B
m
1
Þ A
v
2 B
v
2
A B
C
u
2 mg mg
A B
C
u
2 mg mg
A B
C u
mg
mg
mg
mg mg
mg
2mg
2mg 2mg
2mg
Page 4
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m
v
1 1 1
1
0 5
3
+ ´ = +
m
v
m g l
1
1
2
3
5 × =
\ v
m
m
g l
1
1
3
2
5 =
Option (b) is correct.
2. mu mv = 2
Þ v
u
=
2
0 2
2 2
= +  v g H ( )
Þ H
v
g
=
2
2
=
( / ) u
g
2
2
2
=
u
g
2
8
or H
gl
g
=
2
8
or l
l
( cos ) 1
4
 = a
or 1
1
4
 = cos a
or a =

cos
1
3
4
Option (c) is correct.
3. Here     v v
2 1
® ®
= = v say
Net momentum of the two elements as
shown in figure = =
®
dp   dp
=2vdmsinq
=
æ
è
ç
ö
ø
÷ 2v
M
R
Rd
p
q q ( )sin
=
2Mv
d
p
q q sin
p
Mv
d =
ò
2
0
2
p
q q
p /
sin
= 
2
0
2
Mv
p
q
p
[cos ]
/
=  
2
0 1
Mv
p
[ ] =
2Mv
p
Option (b) is correct.
4.
1
2
2
1
2
2
0
2
( ) m v kx =
or v
k
m
x =
2
0
or 2 2
0
mv mk x =
or F t mkx
av
´ = D 2
0
or F
mk
t
x
av
=
2
0
D
Option (b) is correct.
190  Mechanics1
l
v
1
m
1
m
Rest
u = Ö5gl
v /3
1
m
H
l
m
a
2m
m
v
m m
Rest
u = Ö2gl
R
q
q
d q
d q
2q
dm
v
2
v
1
dp
dm
v
1
®
®
®
®
dl
kx
0
kx
0
5. As the collisions of the striker and the
walls of the carrom are perfectly elastic the
striker will follow the path OPQROP ...
Change in KE = work done against friction
1
2
0
2
mv mgs  = m
(m = mass of striker, s = displacement of
the striker)
Þ s
v
g
=
2
2m
=
´ ´
( ) 2
2 10
2
0.2
=1 m
PQ OP OA = = 2 =
æ
è
ç
ö
ø
÷
=
1
2 2
2
1
2
m
OP PQ + = + =
1
2
1
2
1 m
\ Striker will stop at point Q where
coordinates are
1
2 2
1
2
,
æ
è
ç
ö
ø
÷
.
Option (a) is correct.
6. As no force is acting on the system along
horizontal, the CM of the system will not
shift horizontally.
4 1 1 p p =  ( )
Þ p =
1
5
m
Displacement ( ) x of bar when pendulum
becomes vertical
x
p
=sinq
x p = sinq
= °
1
5
30 sin =
1
10
m = 0.1 m
When the ball reaches the other extreme
end the bar will further shift to the left by
distance x and as such the net
displacement of the bar will be 2x i.e.,
0.2 m.
Option (b) is correct.
7.
Momentum imparted to the floor in
1st collision =   = + p ep p e ( ) ( ) 1
2nd collision =   = + ep e p ep e ( ) ( )
2
1
3rd collision =   = + e p e p e p e
2 3 2
1 ( ) ( )
As theoretically there will be infinite
collision, total momentum imparted to
floor
= + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1
2
= + + + + ¥ p e e e ( )[ ] 1 1
2
K
= +

p e
e
( ) 1
1
1
=
+

æ
è
ç
ö
ø
÷ p
e
e
1
1
Option (d) is correct.
8. Let F be the frictional force applied by
plate when bullet enters into it
\
1
2
2
mu Fh = …(i)
If plate was free to move
mu M m u + = + ¢ 0 ( )
\ u
m
M m
u ¢ =
+
New KE of bullet =  + ¢
1
2
1
2
2 2
mu M m u ( )
(entering plate)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  191
x
q
p
4 kg
4 kg
x
1 kg
CM
1 – p
x
CM
ep p
During
first
collision
Floor
2
ep ep
3
ep
2
ep
Floor
During
Second
collision
During
third
collision
45°
P R
Q
O
A
45°
=  +
+
æ
è
ç
ö
ø
÷
1
2
1
2
2
2
mu M m
m
M m
u ( )
= 
+
é
ë
ê
ù
û
ú
1
2
1
2
mu
m
M m
=
+
æ
è
ç
ö
ø
÷
1
2
2
mu
M
M m
= ¢ Fh …(ii)
Dividing Eq. (ii) by Eq. (i),
h
h
M
M m
¢
=
+
i.e., h
M
M m
h ¢ =
+
æ
è
ç
ö
ø
÷
Option (a) is correct.
9.
Now, v
e
v
1 2
1
2
¢ =
+
(entering plase)
\ 10
1
2
16 =
+
×
e
i.e., e=
1
4
Option (b) is correct.
10. y
l b
l
L b l
L
l b L b
CM
=
× + ×
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
× + ×
æ
è
ç
ö
ø
÷
( )
( )
2
1
2 3
1
2
l
bl bL l L b
bl
bL
=
+ +
+
2 2
2 2 6
2
y l
CM
according
toquestion
= é
ë
ê
ê
ê
ù
û
ú
ú
ú
or l l
L l L l L
+
é
ë
ê
ù
û
ú
= + +
2 2 2 6
2 2
or
l L
2 2
2 6
=
or l
L
=
3
11.
mv m v mv
1 1 2 2 1 2
=  …(i)
e =1
\ 2
2 1
v v =
\ m v m v m v
1 2 2 2 1 2
2 ( ) = 
\ 3
1 2
m m =
or
m
m
1
2
1
3
=
12. As the resultant of the velocities of 1st
and 2nd are just opposite to that of 3rd,
the 4th particle will travel in the line in
which 3rd is travelling.
Let the velocity of 4th particle is u as
shown in figure.
\ u v v cos cos 45 45 ° + ° =
i.e., u v =  ( ) 2 1
Total energy released
= + + + 
1
2
1
2
1
2
1
2
2 1
2 2 2 2
mv mv mv m v [ ( )]
= + 
1
2
3 2 1
2 2
mv [ ( ) ] =  mv
2
3 2 [ ]
Option (a) is correct.
13.
192  Mechanics1
1st
2nd
45°
4th
u
3rd
v
v
v
m m
A B A B
Rest
–1
v = 16 ms
2
2
–1
v = 2gs = 10 ms
1 ' v
2
mu mv = 2
1
i.e., v
u
1
2
=
C will increase
the tension in
the string.
mu mv = 2
2
i.e., v
u
2
2
=
C will also increase the
weight of B when collision
takes place. Thus,
mu m mv = + ( ) 2
3
i.e., v
u
3
3
=
A
m
1
v
1
B
m
1
Þ A
v
2 B
v
2
A B
C
u
2 mg mg
A B
C
u
2 mg mg
A B
C u
mg
mg
mg
mg mg
mg
2mg
2mg 2mg
2mg
\ v v v
1 2 3
: :
=
u u u
2 2 3
: : = 3 3 2 : :
Option (b) is correct.
14.
v v
CM
= ° cos 30
= v
3
2
Option (a) is correct.
15. r i
®
=
CM
^
r
1
®
(position vector of lighter piece)
= +  3 2 4 i j k
^ ^ ^
r
r r ®
® ®
=
+
+
CM
m m
m m
1 1 2 2
1 2
r
r r
2
1 2 1 1
2
®
® ®
=
+  ( ) m m m
m
CM
=
 +  2
2
3
3 2 4
4
3
i i j k
^ ^ ^ ^
( )
=  + 
1
4
6 2 3 2 4 [ ( )]
^ ^ ^ ^
i i j k
=  +
1
4
4 8 [ ]
^ ^
j k
=  + j k
^ ^
2
\ The heavier part will be at ( , , ) 0 1 2  .
Option (d) is correct.
16. Motion of A :
h
gt
2
1
2
2
= , v
A
(at time t) = = g t gh
i.e., t
h
g
=
Motion of B :
h
v
h
g
g
h
g 2
1
2
= 
i.e., h v
h
g
=
\ v gh =
v gh g
h
g
B
=  × =0
Collision of A and B at time t :
m gh m = 3 w
\ w =
gh
3
Velocity of the combined mass when it
reach ground
v gh ¢ = +
2 2
2 w
= +
gh
g
gh 2
i.e., v
gh
¢ =
19
3
Option (d) is correct.
17. u = velocity of man w.r.t. cart
Let v = velocity of cart w.r.t. ground
\ Velocity of man w.r.t. ground = + u v
m u v mv m ( ) + + = × 2 3 0
\ v
u
= 
3
Work done = KE gained by man and cart
= + +
1
2
1
2
2
2 2
m u v mv ( )
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  193
30°
v
v cos 30°
CM
h/2 h/2
h
Place of
collision
w
3 m
m A
time = t (say)
v
2m
B
g
A B A B
m + 2m 2m m
gh
Rest
w
Page 5
Objective Questions (Level 2)
Single Correct Option
1. m v m m g l m
v
1 1 1
1
0 5
3
+ ´ = +
m
v
m g l
1
1
2
3
5 × =
\ v
m
m
g l
1
1
3
2
5 =
Option (b) is correct.
2. mu mv = 2
Þ v
u
=
2
0 2
2 2
= +  v g H ( )
Þ H
v
g
=
2
2
=
( / ) u
g
2
2
2
=
u
g
2
8
or H
gl
g
=
2
8
or l
l
( cos ) 1
4
 = a
or 1
1
4
 = cos a
or a =

cos
1
3
4
Option (c) is correct.
3. Here     v v
2 1
® ®
= = v say
Net momentum of the two elements as
shown in figure = =
®
dp   dp
=2vdmsinq
=
æ
è
ç
ö
ø
÷ 2v
M
R
Rd
p
q q ( )sin
=
2Mv
d
p
q q sin
p
Mv
d =
ò
2
0
2
p
q q
p /
sin
= 
2
0
2
Mv
p
q
p
[cos ]
/
=  
2
0 1
Mv
p
[ ] =
2Mv
p
Option (b) is correct.
4.
1
2
2
1
2
2
0
2
( ) m v kx =
or v
k
m
x =
2
0
or 2 2
0
mv mk x =
or F t mkx
av
´ = D 2
0
or F
mk
t
x
av
=
2
0
D
Option (b) is correct.
190  Mechanics1
l
v
1
m
1
m
Rest
u = Ö5gl
v /3
1
m
H
l
m
a
2m
m
v
m m
Rest
u = Ö2gl
R
q
q
d q
d q
2q
dm
v
2
v
1
dp
dm
v
1
®
®
®
®
dl
kx
0
kx
0
5. As the collisions of the striker and the
walls of the carrom are perfectly elastic the
striker will follow the path OPQROP ...
Change in KE = work done against friction
1
2
0
2
mv mgs  = m
(m = mass of striker, s = displacement of
the striker)
Þ s
v
g
=
2
2m
=
´ ´
( ) 2
2 10
2
0.2
=1 m
PQ OP OA = = 2 =
æ
è
ç
ö
ø
÷
=
1
2 2
2
1
2
m
OP PQ + = + =
1
2
1
2
1 m
\ Striker will stop at point Q where
coordinates are
1
2 2
1
2
,
æ
è
ç
ö
ø
÷
.
Option (a) is correct.
6. As no force is acting on the system along
horizontal, the CM of the system will not
shift horizontally.
4 1 1 p p =  ( )
Þ p =
1
5
m
Displacement ( ) x of bar when pendulum
becomes vertical
x
p
=sinq
x p = sinq
= °
1
5
30 sin =
1
10
m = 0.1 m
When the ball reaches the other extreme
end the bar will further shift to the left by
distance x and as such the net
displacement of the bar will be 2x i.e.,
0.2 m.
Option (b) is correct.
7.
Momentum imparted to the floor in
1st collision =   = + p ep p e ( ) ( ) 1
2nd collision =   = + ep e p ep e ( ) ( )
2
1
3rd collision =   = + e p e p e p e
2 3 2
1 ( ) ( )
As theoretically there will be infinite
collision, total momentum imparted to
floor
= + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1
2
= + + + + ¥ p e e e ( )[ ] 1 1
2
K
= +

p e
e
( ) 1
1
1
=
+

æ
è
ç
ö
ø
÷ p
e
e
1
1
Option (d) is correct.
8. Let F be the frictional force applied by
plate when bullet enters into it
\
1
2
2
mu Fh = …(i)
If plate was free to move
mu M m u + = + ¢ 0 ( )
\ u
m
M m
u ¢ =
+
New KE of bullet =  + ¢
1
2
1
2
2 2
mu M m u ( )
(entering plate)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  191
x
q
p
4 kg
4 kg
x
1 kg
CM
1 – p
x
CM
ep p
During
first
collision
Floor
2
ep ep
3
ep
2
ep
Floor
During
Second
collision
During
third
collision
45°
P R
Q
O
A
45°
=  +
+
æ
è
ç
ö
ø
÷
1
2
1
2
2
2
mu M m
m
M m
u ( )
= 
+
é
ë
ê
ù
û
ú
1
2
1
2
mu
m
M m
=
+
æ
è
ç
ö
ø
÷
1
2
2
mu
M
M m
= ¢ Fh …(ii)
Dividing Eq. (ii) by Eq. (i),
h
h
M
M m
¢
=
+
i.e., h
M
M m
h ¢ =
+
æ
è
ç
ö
ø
÷
Option (a) is correct.
9.
Now, v
e
v
1 2
1
2
¢ =
+
(entering plase)
\ 10
1
2
16 =
+
×
e
i.e., e=
1
4
Option (b) is correct.
10. y
l b
l
L b l
L
l b L b
CM
=
× + ×
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
× + ×
æ
è
ç
ö
ø
÷
( )
( )
2
1
2 3
1
2
l
bl bL l L b
bl
bL
=
+ +
+
2 2
2 2 6
2
y l
CM
according
toquestion
= é
ë
ê
ê
ê
ù
û
ú
ú
ú
or l l
L l L l L
+
é
ë
ê
ù
û
ú
= + +
2 2 2 6
2 2
or
l L
2 2
2 6
=
or l
L
=
3
11.
mv m v mv
1 1 2 2 1 2
=  …(i)
e =1
\ 2
2 1
v v =
\ m v m v m v
1 2 2 2 1 2
2 ( ) = 
\ 3
1 2
m m =
or
m
m
1
2
1
3
=
12. As the resultant of the velocities of 1st
and 2nd are just opposite to that of 3rd,
the 4th particle will travel in the line in
which 3rd is travelling.
Let the velocity of 4th particle is u as
shown in figure.
\ u v v cos cos 45 45 ° + ° =
i.e., u v =  ( ) 2 1
Total energy released
= + + + 
1
2
1
2
1
2
1
2
2 1
2 2 2 2
mv mv mv m v [ ( )]
= + 
1
2
3 2 1
2 2
mv [ ( ) ] =  mv
2
3 2 [ ]
Option (a) is correct.
13.
192  Mechanics1
1st
2nd
45°
4th
u
3rd
v
v
v
m m
A B A B
Rest
–1
v = 16 ms
2
2
–1
v = 2gs = 10 ms
1 ' v
2
mu mv = 2
1
i.e., v
u
1
2
=
C will increase
the tension in
the string.
mu mv = 2
2
i.e., v
u
2
2
=
C will also increase the
weight of B when collision
takes place. Thus,
mu m mv = + ( ) 2
3
i.e., v
u
3
3
=
A
m
1
v
1
B
m
1
Þ A
v
2 B
v
2
A B
C
u
2 mg mg
A B
C
u
2 mg mg
A B
C u
mg
mg
mg
mg mg
mg
2mg
2mg 2mg
2mg
\ v v v
1 2 3
: :
=
u u u
2 2 3
: : = 3 3 2 : :
Option (b) is correct.
14.
v v
CM
= ° cos 30
= v
3
2
Option (a) is correct.
15. r i
®
=
CM
^
r
1
®
(position vector of lighter piece)
= +  3 2 4 i j k
^ ^ ^
r
r r ®
® ®
=
+
+
CM
m m
m m
1 1 2 2
1 2
r
r r
2
1 2 1 1
2
®
® ®
=
+  ( ) m m m
m
CM
=
 +  2
2
3
3 2 4
4
3
i i j k
^ ^ ^ ^
( )
=  + 
1
4
6 2 3 2 4 [ ( )]
^ ^ ^ ^
i i j k
=  +
1
4
4 8 [ ]
^ ^
j k
=  + j k
^ ^
2
\ The heavier part will be at ( , , ) 0 1 2  .
Option (d) is correct.
16. Motion of A :
h
gt
2
1
2
2
= , v
A
(at time t) = = g t gh
i.e., t
h
g
=
Motion of B :
h
v
h
g
g
h
g 2
1
2
= 
i.e., h v
h
g
=
\ v gh =
v gh g
h
g
B
=  × =0
Collision of A and B at time t :
m gh m = 3 w
\ w =
gh
3
Velocity of the combined mass when it
reach ground
v gh ¢ = +
2 2
2 w
= +
gh
g
gh 2
i.e., v
gh
¢ =
19
3
Option (d) is correct.
17. u = velocity of man w.r.t. cart
Let v = velocity of cart w.r.t. ground
\ Velocity of man w.r.t. ground = + u v
m u v mv m ( ) + + = × 2 3 0
\ v
u
= 
3
Work done = KE gained by man and cart
= + +
1
2
1
2
2
2 2
m u v mv ( )
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  193
30°
v
v cos 30°
CM
h/2 h/2
h
Place of
collision
w
3 m
m A
time = t (say)
v
2m
B
g
A B A B
m + 2m 2m m
gh
Rest
w
= 
æ
è
ç
ö
ø
÷ + 
æ
è
ç
ö
ø
÷
1
2 3
1
2
2
3
2 2
m u
u
m
u
= × +
1
2
4
3
1
2
2
9
2 2
m
u
m
u
= +
2
3 9
2 2
mu mu
=
7
9
2
mu
Option (d) is correct.
18. v
m m
m m
CM
=
+
+
30 50
= 40 m/s upwards.
If the velocity of CM becomes zero at
displacements
0 40 2 10
2 2
= +  ( )s
Þ s =80 m
\ Maximum height attained by CM
= + 20 80 m m
=100 m
Option (c) is correct.
19. As the masses are equal and the collision
is elastic, the particles will exchange their
velocities as shown in figure.
Gain in KE of 1st particle
=  
1
2
2
1
2
2 2
m v mu ( )
=  2
1
2
2 2
mv mv
=
3
2
2
mv
=
3
2
2
p
m
Option (c) is correct.
20. According to question
( ) ( )
( ) ( )
4 4
2
4 4
4
4
m x m
a
m m
m x m a
m m
+
+
=
+
+
or
x a x a
2 4
4
5
+ =
+
i.e., x
a
=
6
Option (b) is correct.
21. x
CM
of Fig. 1 will as that of Fig. 2.
x
a a
CM
=
× + ×
+
40
2
50
3
2
40 50
=
+ 20 75
90
a a
= =
95
90
19
18
a a
194  Mechanics1
m
40 m
20 m
m
50 m/s
Initial position of CM
30 m/s
v 2v
B A
In terms of velocity
v 2v
B A
After collision Before collision
4m
P
CM
4m
x
a
CM
4m
x
a
R
m
O
20 10
40
20
a
a
a a a
x
y
a
O
50
40
a
a
x
y
a
Fig. 1
Fig. 2
p – 2 p
B A
In terms of momentum
Before collision
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