Page 1
Ob jec tive Ques tions (Level 1)
Single Correct Option
1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to
sand while KE does not remain conserved
as it gets used up in doing work against
friction.
2. F M a
ext
net
CM
= ´
\ If F
ext
net
= 0, a
CM
= 0
i.e.,
d
dt
v
CM
= 0
or v
CM
= constant
Option (a) is correct.
3. The forces acting on the blocks would be
equal and opposite as per Newton's 3rd
law of motion. Acceleration of the blocks
will depend upon their masses as per
Newton’s 2nd law of motion. Accelerations
being different velocities will be unequal.
Option (c) is correct.
4. While colliding the balls will apply equal
and opposite impulsive force on each
other. Impulsive forces will change the
momentum of the balls but the total
momentum of the system of 2 balls will
remain conserved impulsive forces being
internal ones. Change in momentum of
the system will definitely be due to
external gravitational forces on the balls
but as the time of impact shall be very less
the impulsive force will over shadow the
weak gravitational force.
5. External force acting on the cannon shell
before explosion is the gravitational force.
Now, as no extra net external force would
be act on the shell during collision the
momentum of the system shall remain
conserved and the CM of the system (now
broken in pieces) will also keep on
following the path which the shell would
have followed had the explosion not taken
place. Further, as the explosion would
never be superelastic, the KE of the
system can’t increase after explosion.
Option (d) is correct.
6.
Velocity of separation
Velocity of approach
= e
As in an elastic collision e < 1
Velocity of separation < velocity of
approach
(when e = 0, the velocity of separation in
zero and the colliding bodies do not
separate from each other.)
Further, whether the collision in elastic or
inelastic the law of conservation of
momentum always hold gord.
\ Option (d) is correct.
7. p v
® ®
= M
CM
, Option (a) is correct.
p p p p
3
® ® ® ®
= + + +
1 2
....,
Option (b) is correct.
Further, we define momentum for every
type of motion.
\ Option (d) is correct.
8. Let us consider a system of two masses as
shown in figure.
Momentum of system about CM
=  + 
® ® ® ®
m m
1 1 2 2
( ) ( ) v v v v
CM CM
= +  +
® ® ®
m m m m
1 1 2 2 1 2
v v v ( )
CM
= +  +
® ®
( ) ( ) m m m m
1 2 1 2
v v
CM CM
=
®
0
Option (c) is correct.
9. Option (a) If collision is inelastic.
Option (b) If collision is perfectly inelastic
Option (c) If the dimensions of the
particles ®0
\ Option (d) would be the answer.
180  Mechanics1
M
1
M
2
F F
v
CM
®
v
1
®
v
2
®
m
1
m
2
CM
m
A
v
v
v
m m
B C
Page 2
Ob jec tive Ques tions (Level 1)
Single Correct Option
1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to
sand while KE does not remain conserved
as it gets used up in doing work against
friction.
2. F M a
ext
net
CM
= ´
\ If F
ext
net
= 0, a
CM
= 0
i.e.,
d
dt
v
CM
= 0
or v
CM
= constant
Option (a) is correct.
3. The forces acting on the blocks would be
equal and opposite as per Newton's 3rd
law of motion. Acceleration of the blocks
will depend upon their masses as per
Newton’s 2nd law of motion. Accelerations
being different velocities will be unequal.
Option (c) is correct.
4. While colliding the balls will apply equal
and opposite impulsive force on each
other. Impulsive forces will change the
momentum of the balls but the total
momentum of the system of 2 balls will
remain conserved impulsive forces being
internal ones. Change in momentum of
the system will definitely be due to
external gravitational forces on the balls
but as the time of impact shall be very less
the impulsive force will over shadow the
weak gravitational force.
5. External force acting on the cannon shell
before explosion is the gravitational force.
Now, as no extra net external force would
be act on the shell during collision the
momentum of the system shall remain
conserved and the CM of the system (now
broken in pieces) will also keep on
following the path which the shell would
have followed had the explosion not taken
place. Further, as the explosion would
never be superelastic, the KE of the
system can’t increase after explosion.
Option (d) is correct.
6.
Velocity of separation
Velocity of approach
= e
As in an elastic collision e < 1
Velocity of separation < velocity of
approach
(when e = 0, the velocity of separation in
zero and the colliding bodies do not
separate from each other.)
Further, whether the collision in elastic or
inelastic the law of conservation of
momentum always hold gord.
\ Option (d) is correct.
7. p v
® ®
= M
CM
, Option (a) is correct.
p p p p
3
® ® ® ®
= + + +
1 2
....,
Option (b) is correct.
Further, we define momentum for every
type of motion.
\ Option (d) is correct.
8. Let us consider a system of two masses as
shown in figure.
Momentum of system about CM
=  + 
® ® ® ®
m m
1 1 2 2
( ) ( ) v v v v
CM CM
= +  +
® ® ®
m m m m
1 1 2 2 1 2
v v v ( )
CM
= +  +
® ®
( ) ( ) m m m m
1 2 1 2
v v
CM CM
=
®
0
Option (c) is correct.
9. Option (a) If collision is inelastic.
Option (b) If collision is perfectly inelastic
Option (c) If the dimensions of the
particles ®0
\ Option (d) would be the answer.
180  Mechanics1
M
1
M
2
F F
v
CM
®
v
1
®
v
2
®
m
1
m
2
CM
m
A
v
v
v
m m
B C
10. F t m v D D =
Þ F
m v
t
=
D
D
=
´  5 65 15
2
( )
=125N
11. m d m d d
c o
´ = ´ 
1 1
( )
Þ d
m
m m
d
o
o c
1
=
+
=
+
8
8 6
d
=
4
7
d
= ´ ´

4
7
10
10
(1.2 m)
= ´

0.64 m 10
10
12. M mv M m V ´ = +  0 ( )
Þ V
m
m m
v = 
 ( )
= 

´
3
9 3
16
= 

8
1
ms
\ KE of 6 kg mass = ´ ´ 
1
2
6 8
2
( )
=192J
13. v
m
m m
v
1
2
1 2
2
2
¢ =
+
=2
2
v (as m m
1 2
< < )
Option (b) is correct.
14. Horizontal velocity of the leaving coal :
v i
1
®
= + v
^
Horizontal velocity of the system
v i
2
®
= + v
^
\ U v v
® ® ® ®
=  =
rel 1 2 1
0
\ F U
th
rel
® ®
®
= =
dm
dt
0
As, the leaving coal does not exert any
thrust force on the wagon, the speed of the
wagon won’t change.
Option (a) is correct.
15. If n be the number of bullet shots per
second
n´ [change in momentum per second] £ F
i.e., n
40
100
1200 0 144 ´ 
é
ë
ê
ù
û
ú
£ ( )
or n £
144
48
or n £ 3
\ Option (a) is correct.
16. Change in momentum along xaxis
= m (v v cos cos q q  ) = 0
\ Net change in momentum
= Change in momentum along yaxis
= +   m v v [( sin ) ( sin )] q q
=2mvsinq
= mv 2 (as q = ° 45 )
Option (a) is correct.
17. Velocity of ball before first impact i.e.,
when it reaches point Q of the horizontal
plane
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181
d
m
o
m
c
d
1
O
C
CM
M – m m M
Rest
V v
3 kg 6 kg 9 kg
v
2
m
2
m
1
v'
2
m
2
v'
1
m
1
v = 0
1
After collision Before collision
x
O
v
i
^
system
v cos q
v sin q
v cos q
q
v
y
P
+
–
q
v cos q
v
Q
x
Page 3
Ob jec tive Ques tions (Level 1)
Single Correct Option
1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to
sand while KE does not remain conserved
as it gets used up in doing work against
friction.
2. F M a
ext
net
CM
= ´
\ If F
ext
net
= 0, a
CM
= 0
i.e.,
d
dt
v
CM
= 0
or v
CM
= constant
Option (a) is correct.
3. The forces acting on the blocks would be
equal and opposite as per Newton's 3rd
law of motion. Acceleration of the blocks
will depend upon their masses as per
Newton’s 2nd law of motion. Accelerations
being different velocities will be unequal.
Option (c) is correct.
4. While colliding the balls will apply equal
and opposite impulsive force on each
other. Impulsive forces will change the
momentum of the balls but the total
momentum of the system of 2 balls will
remain conserved impulsive forces being
internal ones. Change in momentum of
the system will definitely be due to
external gravitational forces on the balls
but as the time of impact shall be very less
the impulsive force will over shadow the
weak gravitational force.
5. External force acting on the cannon shell
before explosion is the gravitational force.
Now, as no extra net external force would
be act on the shell during collision the
momentum of the system shall remain
conserved and the CM of the system (now
broken in pieces) will also keep on
following the path which the shell would
have followed had the explosion not taken
place. Further, as the explosion would
never be superelastic, the KE of the
system can’t increase after explosion.
Option (d) is correct.
6.
Velocity of separation
Velocity of approach
= e
As in an elastic collision e < 1
Velocity of separation < velocity of
approach
(when e = 0, the velocity of separation in
zero and the colliding bodies do not
separate from each other.)
Further, whether the collision in elastic or
inelastic the law of conservation of
momentum always hold gord.
\ Option (d) is correct.
7. p v
® ®
= M
CM
, Option (a) is correct.
p p p p
3
® ® ® ®
= + + +
1 2
....,
Option (b) is correct.
Further, we define momentum for every
type of motion.
\ Option (d) is correct.
8. Let us consider a system of two masses as
shown in figure.
Momentum of system about CM
=  + 
® ® ® ®
m m
1 1 2 2
( ) ( ) v v v v
CM CM
= +  +
® ® ®
m m m m
1 1 2 2 1 2
v v v ( )
CM
= +  +
® ®
( ) ( ) m m m m
1 2 1 2
v v
CM CM
=
®
0
Option (c) is correct.
9. Option (a) If collision is inelastic.
Option (b) If collision is perfectly inelastic
Option (c) If the dimensions of the
particles ®0
\ Option (d) would be the answer.
180  Mechanics1
M
1
M
2
F F
v
CM
®
v
1
®
v
2
®
m
1
m
2
CM
m
A
v
v
v
m m
B C
10. F t m v D D =
Þ F
m v
t
=
D
D
=
´  5 65 15
2
( )
=125N
11. m d m d d
c o
´ = ´ 
1 1
( )
Þ d
m
m m
d
o
o c
1
=
+
=
+
8
8 6
d
=
4
7
d
= ´ ´

4
7
10
10
(1.2 m)
= ´

0.64 m 10
10
12. M mv M m V ´ = +  0 ( )
Þ V
m
m m
v = 
 ( )
= 

´
3
9 3
16
= 

8
1
ms
\ KE of 6 kg mass = ´ ´ 
1
2
6 8
2
( )
=192J
13. v
m
m m
v
1
2
1 2
2
2
¢ =
+
=2
2
v (as m m
1 2
< < )
Option (b) is correct.
14. Horizontal velocity of the leaving coal :
v i
1
®
= + v
^
Horizontal velocity of the system
v i
2
®
= + v
^
\ U v v
® ® ® ®
=  =
rel 1 2 1
0
\ F U
th
rel
® ®
®
= =
dm
dt
0
As, the leaving coal does not exert any
thrust force on the wagon, the speed of the
wagon won’t change.
Option (a) is correct.
15. If n be the number of bullet shots per
second
n´ [change in momentum per second] £ F
i.e., n
40
100
1200 0 144 ´ 
é
ë
ê
ù
û
ú
£ ( )
or n £
144
48
or n £ 3
\ Option (a) is correct.
16. Change in momentum along xaxis
= m (v v cos cos q q  ) = 0
\ Net change in momentum
= Change in momentum along yaxis
= +   m v v [( sin ) ( sin )] q q
=2mvsinq
= mv 2 (as q = ° 45 )
Option (a) is correct.
17. Velocity of ball before first impact i.e.,
when it reaches point Q of the horizontal
plane
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181
d
m
o
m
c
d
1
O
C
CM
M – m m M
Rest
V v
3 kg 6 kg 9 kg
v
2
m
2
m
1
v'
2
m
2
v'
1
m
1
v = 0
1
After collision Before collision
x
O
v
i
^
system
v cos q
v sin q
v cos q
q
v
y
P
+
–
q
v cos q
v
Q
x
v g = + × 0 1 ( ) = g
\ Velocity of ball after 1st impact
= = ev eg
time elapsed between 1st and 2nd impact
with the horizontal plane
( ) ( ) ( ) + = + +  e g e g g t
Þ t e =2
=
4
3
s (as, e =
2
3
L)
Option (c) is correct.
18. x
x dm
dm
x m dx
m dx
CM
= =
ò
ò
ò
ò
(where m = mass per unit length)
=
ò
ò
x
Ax
L
dx
Ax
L
dx
2
2
Q m
Ax
L
=
æ
è
ç
ö
ø
÷
2
=
ò
ò
x dx
x dx
L
L
3
0
2
0
= ´
L
L
4
3
4
3
=
3
4
L
Option (a) is correct.
19. As there is no net external force acting on
the system in the horizontal direction, the
CM of the system shall not shift along
xaxis.
\
50 10 450 5
50 450
50 450 5
50 450
´ + ´
+
=
´ + ´ +
+
( ) ( ) x x
(Initially) (Finally)
Þ x = 1 m
Option (b) is correct.
20. As discussed in the answer to previous
question no. 19.
Mx
M
x
L
M
M
ML
M L
M
M
+ +
æ
è
ç
ö
ø
÷
+
=
+
æ
è
ç
ö
ø
÷
+
3 2
3
3 2
3
(Initially) (Finally)
Þ x M
M
ML +
é
ë
ê
ù
û
ú
=
3
x L =
3
4
i.e., the distance that plank moves =
3
4
L
\ The distance that the man moves
=  L L
3
4
=
L
4
Option (b) is correct.
21.
mv m m mu + × = × + 3 0 0 3
182  Mechanics1
dx
x
L
0
x
450 kg
CM (boat)
10 m
O
50 kg
50 kg
x
450 kg
CM (boat)
X
X
M/3
L Smooth
X
M/3
L Smooth
X
x
O
M
M
O
m 3m m 3m
Rest Rest
v u
Before collision After collision
P
Q
+ eg
2nd
– eg
1st
u = 0
Page 4
Ob jec tive Ques tions (Level 1)
Single Correct Option
1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to
sand while KE does not remain conserved
as it gets used up in doing work against
friction.
2. F M a
ext
net
CM
= ´
\ If F
ext
net
= 0, a
CM
= 0
i.e.,
d
dt
v
CM
= 0
or v
CM
= constant
Option (a) is correct.
3. The forces acting on the blocks would be
equal and opposite as per Newton's 3rd
law of motion. Acceleration of the blocks
will depend upon their masses as per
Newton’s 2nd law of motion. Accelerations
being different velocities will be unequal.
Option (c) is correct.
4. While colliding the balls will apply equal
and opposite impulsive force on each
other. Impulsive forces will change the
momentum of the balls but the total
momentum of the system of 2 balls will
remain conserved impulsive forces being
internal ones. Change in momentum of
the system will definitely be due to
external gravitational forces on the balls
but as the time of impact shall be very less
the impulsive force will over shadow the
weak gravitational force.
5. External force acting on the cannon shell
before explosion is the gravitational force.
Now, as no extra net external force would
be act on the shell during collision the
momentum of the system shall remain
conserved and the CM of the system (now
broken in pieces) will also keep on
following the path which the shell would
have followed had the explosion not taken
place. Further, as the explosion would
never be superelastic, the KE of the
system can’t increase after explosion.
Option (d) is correct.
6.
Velocity of separation
Velocity of approach
= e
As in an elastic collision e < 1
Velocity of separation < velocity of
approach
(when e = 0, the velocity of separation in
zero and the colliding bodies do not
separate from each other.)
Further, whether the collision in elastic or
inelastic the law of conservation of
momentum always hold gord.
\ Option (d) is correct.
7. p v
® ®
= M
CM
, Option (a) is correct.
p p p p
3
® ® ® ®
= + + +
1 2
....,
Option (b) is correct.
Further, we define momentum for every
type of motion.
\ Option (d) is correct.
8. Let us consider a system of two masses as
shown in figure.
Momentum of system about CM
=  + 
® ® ® ®
m m
1 1 2 2
( ) ( ) v v v v
CM CM
= +  +
® ® ®
m m m m
1 1 2 2 1 2
v v v ( )
CM
= +  +
® ®
( ) ( ) m m m m
1 2 1 2
v v
CM CM
=
®
0
Option (c) is correct.
9. Option (a) If collision is inelastic.
Option (b) If collision is perfectly inelastic
Option (c) If the dimensions of the
particles ®0
\ Option (d) would be the answer.
180  Mechanics1
M
1
M
2
F F
v
CM
®
v
1
®
v
2
®
m
1
m
2
CM
m
A
v
v
v
m m
B C
10. F t m v D D =
Þ F
m v
t
=
D
D
=
´  5 65 15
2
( )
=125N
11. m d m d d
c o
´ = ´ 
1 1
( )
Þ d
m
m m
d
o
o c
1
=
+
=
+
8
8 6
d
=
4
7
d
= ´ ´

4
7
10
10
(1.2 m)
= ´

0.64 m 10
10
12. M mv M m V ´ = +  0 ( )
Þ V
m
m m
v = 
 ( )
= 

´
3
9 3
16
= 

8
1
ms
\ KE of 6 kg mass = ´ ´ 
1
2
6 8
2
( )
=192J
13. v
m
m m
v
1
2
1 2
2
2
¢ =
+
=2
2
v (as m m
1 2
< < )
Option (b) is correct.
14. Horizontal velocity of the leaving coal :
v i
1
®
= + v
^
Horizontal velocity of the system
v i
2
®
= + v
^
\ U v v
® ® ® ®
=  =
rel 1 2 1
0
\ F U
th
rel
® ®
®
= =
dm
dt
0
As, the leaving coal does not exert any
thrust force on the wagon, the speed of the
wagon won’t change.
Option (a) is correct.
15. If n be the number of bullet shots per
second
n´ [change in momentum per second] £ F
i.e., n
40
100
1200 0 144 ´ 
é
ë
ê
ù
û
ú
£ ( )
or n £
144
48
or n £ 3
\ Option (a) is correct.
16. Change in momentum along xaxis
= m (v v cos cos q q  ) = 0
\ Net change in momentum
= Change in momentum along yaxis
= +   m v v [( sin ) ( sin )] q q
=2mvsinq
= mv 2 (as q = ° 45 )
Option (a) is correct.
17. Velocity of ball before first impact i.e.,
when it reaches point Q of the horizontal
plane
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181
d
m
o
m
c
d
1
O
C
CM
M – m m M
Rest
V v
3 kg 6 kg 9 kg
v
2
m
2
m
1
v'
2
m
2
v'
1
m
1
v = 0
1
After collision Before collision
x
O
v
i
^
system
v cos q
v sin q
v cos q
q
v
y
P
+
–
q
v cos q
v
Q
x
v g = + × 0 1 ( ) = g
\ Velocity of ball after 1st impact
= = ev eg
time elapsed between 1st and 2nd impact
with the horizontal plane
( ) ( ) ( ) + = + +  e g e g g t
Þ t e =2
=
4
3
s (as, e =
2
3
L)
Option (c) is correct.
18. x
x dm
dm
x m dx
m dx
CM
= =
ò
ò
ò
ò
(where m = mass per unit length)
=
ò
ò
x
Ax
L
dx
Ax
L
dx
2
2
Q m
Ax
L
=
æ
è
ç
ö
ø
÷
2
=
ò
ò
x dx
x dx
L
L
3
0
2
0
= ´
L
L
4
3
4
3
=
3
4
L
Option (a) is correct.
19. As there is no net external force acting on
the system in the horizontal direction, the
CM of the system shall not shift along
xaxis.
\
50 10 450 5
50 450
50 450 5
50 450
´ + ´
+
=
´ + ´ +
+
( ) ( ) x x
(Initially) (Finally)
Þ x = 1 m
Option (b) is correct.
20. As discussed in the answer to previous
question no. 19.
Mx
M
x
L
M
M
ML
M L
M
M
+ +
æ
è
ç
ö
ø
÷
+
=
+
æ
è
ç
ö
ø
÷
+
3 2
3
3 2
3
(Initially) (Finally)
Þ x M
M
ML +
é
ë
ê
ù
û
ú
=
3
x L =
3
4
i.e., the distance that plank moves =
3
4
L
\ The distance that the man moves
=  L L
3
4
=
L
4
Option (b) is correct.
21.
mv m m mu + × = × + 3 0 0 3
182  Mechanics1
dx
x
L
0
x
450 kg
CM (boat)
10 m
O
50 kg
50 kg
x
450 kg
CM (boat)
X
X
M/3
L Smooth
X
M/3
L Smooth
X
x
O
M
M
O
m 3m m 3m
Rest Rest
v u
Before collision After collision
P
Q
+ eg
2nd
– eg
1st
u = 0
\ u
v
=
3
e =
Velocity of separation
Velocity of approach
=


=
u
v
u
v
0
0
=
1
3
Option (d) is correct.
22. Change in momentum of A or B = mu
(As collision is elastic)
Impulse = Change in momentum
Ft
mu
0
2
=
Þ F
mu
t
=
2
0
Option (b) is correct.
23. Acceleration of block A
a
F
m
0
=
¢
Þ F ma ¢ =
0
Acceleration of block B :
a
F F
m
B
=
 ¢
=
 F ma
m
0
= 
F
m
a
0
24.
Option (c) is correct.
Impulse on ball
= Change in momentum of ball
= 
æ
è
ç
ö
ø
÷  +
æ
è
ç
ö
ø
÷
9
20
4
5
0 0
mv mv
= 
5
4
0
mv
25. If a ball dropped from height h rebounds
to a height h¢, then speed of ball just
before 1st impact, u gh = 2
Just after 1st impact u gh ¢ = ¢ 2
\ e
u
u
h
h
=
¢
=
¢
=
64
100
=0.8
i.e., h e h ¢ =
2
Height attained after nth impact
= e h
n 2
= × ( ) 0.8
2
1
n
(as h=1 m)
=( ) 0.8
2n
Option (d) is correct.
26. Momentum of car (+ block) before
throwing block
= Momentum of car after throwing block
+ Momentum of block
500 1 500 25 25 20 ´ =  +
®
i v k
^ ^
( ) ( )
or 475 500 v i k
®
=  ( )
^ ^
or v i k
®
= 
20
19
( )
^ ^
Option (c) is correct.
27. While force is increasing with time
F kt = (where k is + ive constant)
or ma kt =
or
dv
dt
k
m
t =
\ v
k
m
t
C = +
2
2
or v
k
m
t
= ×
2
2
(If at t = 0, v = 0)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  183
37°
53°
3
5
4
37°
53°
v
0
X
Y
+
v cos 37° = v
00
4
5
3
4
v cos 53°
0
v
0
9
20
v
0
3
4
=
u
m m
Rest
m
u
m
After collision Before collision
Rest
A B
F' F'
m m
F
Page 5
Ob jec tive Ques tions (Level 1)
Single Correct Option
1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to
sand while KE does not remain conserved
as it gets used up in doing work against
friction.
2. F M a
ext
net
CM
= ´
\ If F
ext
net
= 0, a
CM
= 0
i.e.,
d
dt
v
CM
= 0
or v
CM
= constant
Option (a) is correct.
3. The forces acting on the blocks would be
equal and opposite as per Newton's 3rd
law of motion. Acceleration of the blocks
will depend upon their masses as per
Newton’s 2nd law of motion. Accelerations
being different velocities will be unequal.
Option (c) is correct.
4. While colliding the balls will apply equal
and opposite impulsive force on each
other. Impulsive forces will change the
momentum of the balls but the total
momentum of the system of 2 balls will
remain conserved impulsive forces being
internal ones. Change in momentum of
the system will definitely be due to
external gravitational forces on the balls
but as the time of impact shall be very less
the impulsive force will over shadow the
weak gravitational force.
5. External force acting on the cannon shell
before explosion is the gravitational force.
Now, as no extra net external force would
be act on the shell during collision the
momentum of the system shall remain
conserved and the CM of the system (now
broken in pieces) will also keep on
following the path which the shell would
have followed had the explosion not taken
place. Further, as the explosion would
never be superelastic, the KE of the
system can’t increase after explosion.
Option (d) is correct.
6.
Velocity of separation
Velocity of approach
= e
As in an elastic collision e < 1
Velocity of separation < velocity of
approach
(when e = 0, the velocity of separation in
zero and the colliding bodies do not
separate from each other.)
Further, whether the collision in elastic or
inelastic the law of conservation of
momentum always hold gord.
\ Option (d) is correct.
7. p v
® ®
= M
CM
, Option (a) is correct.
p p p p
3
® ® ® ®
= + + +
1 2
....,
Option (b) is correct.
Further, we define momentum for every
type of motion.
\ Option (d) is correct.
8. Let us consider a system of two masses as
shown in figure.
Momentum of system about CM
=  + 
® ® ® ®
m m
1 1 2 2
( ) ( ) v v v v
CM CM
= +  +
® ® ®
m m m m
1 1 2 2 1 2
v v v ( )
CM
= +  +
® ®
( ) ( ) m m m m
1 2 1 2
v v
CM CM
=
®
0
Option (c) is correct.
9. Option (a) If collision is inelastic.
Option (b) If collision is perfectly inelastic
Option (c) If the dimensions of the
particles ®0
\ Option (d) would be the answer.
180  Mechanics1
M
1
M
2
F F
v
CM
®
v
1
®
v
2
®
m
1
m
2
CM
m
A
v
v
v
m m
B C
10. F t m v D D =
Þ F
m v
t
=
D
D
=
´  5 65 15
2
( )
=125N
11. m d m d d
c o
´ = ´ 
1 1
( )
Þ d
m
m m
d
o
o c
1
=
+
=
+
8
8 6
d
=
4
7
d
= ´ ´

4
7
10
10
(1.2 m)
= ´

0.64 m 10
10
12. M mv M m V ´ = +  0 ( )
Þ V
m
m m
v = 
 ( )
= 

´
3
9 3
16
= 

8
1
ms
\ KE of 6 kg mass = ´ ´ 
1
2
6 8
2
( )
=192J
13. v
m
m m
v
1
2
1 2
2
2
¢ =
+
=2
2
v (as m m
1 2
< < )
Option (b) is correct.
14. Horizontal velocity of the leaving coal :
v i
1
®
= + v
^
Horizontal velocity of the system
v i
2
®
= + v
^
\ U v v
® ® ® ®
=  =
rel 1 2 1
0
\ F U
th
rel
® ®
®
= =
dm
dt
0
As, the leaving coal does not exert any
thrust force on the wagon, the speed of the
wagon won’t change.
Option (a) is correct.
15. If n be the number of bullet shots per
second
n´ [change in momentum per second] £ F
i.e., n
40
100
1200 0 144 ´ 
é
ë
ê
ù
û
ú
£ ( )
or n £
144
48
or n £ 3
\ Option (a) is correct.
16. Change in momentum along xaxis
= m (v v cos cos q q  ) = 0
\ Net change in momentum
= Change in momentum along yaxis
= +   m v v [( sin ) ( sin )] q q
=2mvsinq
= mv 2 (as q = ° 45 )
Option (a) is correct.
17. Velocity of ball before first impact i.e.,
when it reaches point Q of the horizontal
plane
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181
d
m
o
m
c
d
1
O
C
CM
M – m m M
Rest
V v
3 kg 6 kg 9 kg
v
2
m
2
m
1
v'
2
m
2
v'
1
m
1
v = 0
1
After collision Before collision
x
O
v
i
^
system
v cos q
v sin q
v cos q
q
v
y
P
+
–
q
v cos q
v
Q
x
v g = + × 0 1 ( ) = g
\ Velocity of ball after 1st impact
= = ev eg
time elapsed between 1st and 2nd impact
with the horizontal plane
( ) ( ) ( ) + = + +  e g e g g t
Þ t e =2
=
4
3
s (as, e =
2
3
L)
Option (c) is correct.
18. x
x dm
dm
x m dx
m dx
CM
= =
ò
ò
ò
ò
(where m = mass per unit length)
=
ò
ò
x
Ax
L
dx
Ax
L
dx
2
2
Q m
Ax
L
=
æ
è
ç
ö
ø
÷
2
=
ò
ò
x dx
x dx
L
L
3
0
2
0
= ´
L
L
4
3
4
3
=
3
4
L
Option (a) is correct.
19. As there is no net external force acting on
the system in the horizontal direction, the
CM of the system shall not shift along
xaxis.
\
50 10 450 5
50 450
50 450 5
50 450
´ + ´
+
=
´ + ´ +
+
( ) ( ) x x
(Initially) (Finally)
Þ x = 1 m
Option (b) is correct.
20. As discussed in the answer to previous
question no. 19.
Mx
M
x
L
M
M
ML
M L
M
M
+ +
æ
è
ç
ö
ø
÷
+
=
+
æ
è
ç
ö
ø
÷
+
3 2
3
3 2
3
(Initially) (Finally)
Þ x M
M
ML +
é
ë
ê
ù
û
ú
=
3
x L =
3
4
i.e., the distance that plank moves =
3
4
L
\ The distance that the man moves
=  L L
3
4
=
L
4
Option (b) is correct.
21.
mv m m mu + × = × + 3 0 0 3
182  Mechanics1
dx
x
L
0
x
450 kg
CM (boat)
10 m
O
50 kg
50 kg
x
450 kg
CM (boat)
X
X
M/3
L Smooth
X
M/3
L Smooth
X
x
O
M
M
O
m 3m m 3m
Rest Rest
v u
Before collision After collision
P
Q
+ eg
2nd
– eg
1st
u = 0
\ u
v
=
3
e =
Velocity of separation
Velocity of approach
=


=
u
v
u
v
0
0
=
1
3
Option (d) is correct.
22. Change in momentum of A or B = mu
(As collision is elastic)
Impulse = Change in momentum
Ft
mu
0
2
=
Þ F
mu
t
=
2
0
Option (b) is correct.
23. Acceleration of block A
a
F
m
0
=
¢
Þ F ma ¢ =
0
Acceleration of block B :
a
F F
m
B
=
 ¢
=
 F ma
m
0
= 
F
m
a
0
24.
Option (c) is correct.
Impulse on ball
= Change in momentum of ball
= 
æ
è
ç
ö
ø
÷  +
æ
è
ç
ö
ø
÷
9
20
4
5
0 0
mv mv
= 
5
4
0
mv
25. If a ball dropped from height h rebounds
to a height h¢, then speed of ball just
before 1st impact, u gh = 2
Just after 1st impact u gh ¢ = ¢ 2
\ e
u
u
h
h
=
¢
=
¢
=
64
100
=0.8
i.e., h e h ¢ =
2
Height attained after nth impact
= e h
n 2
= × ( ) 0.8
2
1
n
(as h=1 m)
=( ) 0.8
2n
Option (d) is correct.
26. Momentum of car (+ block) before
throwing block
= Momentum of car after throwing block
+ Momentum of block
500 1 500 25 25 20 ´ =  +
®
i v k
^ ^
( ) ( )
or 475 500 v i k
®
=  ( )
^ ^
or v i k
®
= 
20
19
( )
^ ^
Option (c) is correct.
27. While force is increasing with time
F kt = (where k is + ive constant)
or ma kt =
or
dv
dt
k
m
t =
\ v
k
m
t
C = +
2
2
or v
k
m
t
= ×
2
2
(If at t = 0, v = 0)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  183
37°
53°
3
5
4
37°
53°
v
0
X
Y
+
v cos 37° = v
00
4
5
3
4
v cos 53°
0
v
0
9
20
v
0
3
4
=
u
m m
Rest
m
u
m
After collision Before collision
Rest
A B
F' F'
m m
F
Thus, graph between v and t would be
While force is decreasing with time.
F kt =  (where k is + ive constant)
Þ v
k
m
t
C =  + ¢
2
2
At t = 0, v v =
0
\ C v
k
m
t
¢ = +
0
0
2
2
Thus, v v
k
m
t t = + 
0 0
2 2
2
( )
Thus, graph between v and t would be
\ Option (a) is correct.
28. x
A x A x
A A
CM
=
+
+
1 1 2 2
1 2
0
3 2
3
2 2
1
2
2 2
=
 + 

[ ( ) ] [ ]
( )
p p p
p p
R R x R R
R R
or 8 2
2
1
2
p p R x R R = or 8 2
1
x R =
or x
R
1
4
=
Option (c) is correct.
29. x
A x A x A x
A A A
CM
=
+ +
+ +
1 1 2 2 3 3
1 2 3
\ 0 4
2 2 2
1
=   [ ( ) ] p p p R R R x
+  +  × [ ] [ ] p p R R R
2 2
3 0
or 14 3
1
x R =
or x R
1
3
14
=
Option (d) is correct.
30. As no external force is acting along the
horizontal direction on the system (wedge
+ block). The CM of the system shall not
change along horizontal when the block
moves over the wedge but would change
along vertical as net force (= gravitational
force) is acting on the block.
Further, as no nonconservative force is
acting on the system, its total energy will
not change.
Option (d) is correct.
31.
mv M m v = + ¢ ( ) …(i)
From final position, v gh ¢ = 2
From Eq. (i),
or
m
M m
v gh
+
= 2
or v
M
m
gh = +
æ
è
ç
ö
ø
÷
1 2
Option (c) is correct.
32. As no net extra external force is acting on
the system the CM of the gun and the
bullet system remains at rest. The force
exerted by the trigger of the gun on the
bullet is an interval one.
33. m g T m a
1 1
 =
184  Mechanics1
T T
T T
mg
1
mg
2
a
v
t
0 t
Parabola
v
0
M Rest
v
m
M
v'
m m
h
Final position
Just after
collision
Before collision
Rest
v
t
0
t
Graph would be
parabolic in
nature
v
0
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