DC Pandey Solutions: Centre of Mass, Conservation of Linear Momentum- 1

# DC Pandey Solutions: Centre of Mass, Conservation of Linear Momentum- 1 | DC Pandey Solutions for NEET Physics PDF Download

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Page 1

Introductory Exercise 8.1
1. If a body is placed in a uniform
gravitational field, the CG of the body
coincides with the CM of the body.
r
r
®
=
®
=
=
CM
S
S
i
n
i i
i
n
i
m
m
1
1

while   r
r
®
=
®
=
=
CM
S
S
i
n
i i i
i
n
i i
m g
m g
1
1
If a body is placed in a uniformly
increasing gravitational field ( ) g
®
in the
upward direction the CG of the body will
be higher level than the CM. And, if the
body is placed in a uniformly decreasing
gravitational field in the upward direction
the CG will be at a lower level the CM.
CG shifts from CM according to the
magnitude and direction of the
gravitational field (by some other agency
eg, earth) in which the body is placed.
In zero gravitational field CG has no
meaning while CM still exists, as usual.
2. The cen tre of mass of a rigid body may lie
in side, on the sur face and even out side the
body. The CM of a solid uni form sphere is
at its cen tre. The CM of a solid ring is at
the cen tre of the ring which lies out side
the mass of the body thus, the state ment
is false. (For fur ther de tails see an swer to
1 As ser tion and Rea son JEE corner).
3. Cen tre of mass al ways lies on the axis of
sym me try of the body, if it ex ists. The
state ment is thus true.
4. State ment is true.
5. As more mass is towards base.
Distance  <
r
4
.
6. If two equal masses are kept at
co-ordinates (R, 0) and ( , ) -R 0 , then their
centre of mass will lie at origin.
7.
X
m X m X
m m
CM
=
+
+
1 1 2 2
1 2
Y
m y m y
m m
CM
=
+
+
1 1 2 2
1 2
r X Y = +
CM CM
2 2
= Distance of centre of mass from A
g
increasing
CG
CM
CG
CM
g
decreasing
Centre of Mass, Conservation
of Linear Momentum, Impulse
and Collision
8
3 kg (1/2,  3/2)
2 kg
(1, 0)
1 kg
(0, 0)
Page 2

Introductory Exercise 8.1
1. If a body is placed in a uniform
gravitational field, the CG of the body
coincides with the CM of the body.
r
r
®
=
®
=
=
CM
S
S
i
n
i i
i
n
i
m
m
1
1

while   r
r
®
=
®
=
=
CM
S
S
i
n
i i i
i
n
i i
m g
m g
1
1
If a body is placed in a uniformly
increasing gravitational field ( ) g
®
in the
upward direction the CG of the body will
be higher level than the CM. And, if the
body is placed in a uniformly decreasing
gravitational field in the upward direction
the CG will be at a lower level the CM.
CG shifts from CM according to the
magnitude and direction of the
gravitational field (by some other agency
eg, earth) in which the body is placed.
In zero gravitational field CG has no
meaning while CM still exists, as usual.
2. The cen tre of mass of a rigid body may lie
in side, on the sur face and even out side the
body. The CM of a solid uni form sphere is
at its cen tre. The CM of a solid ring is at
the cen tre of the ring which lies out side
the mass of the body thus, the state ment
is false. (For fur ther de tails see an swer to
1 As ser tion and Rea son JEE corner).
3. Cen tre of mass al ways lies on the axis of
sym me try of the body, if it ex ists. The
state ment is thus true.
4. State ment is true.
5. As more mass is towards base.
Distance  <
r
4
.
6. If two equal masses are kept at
co-ordinates (R, 0) and ( , ) -R 0 , then their
centre of mass will lie at origin.
7.
X
m X m X
m m
CM
=
+
+
1 1 2 2
1 2
Y
m y m y
m m
CM
=
+
+
1 1 2 2
1 2
r X Y = +
CM CM
2 2
= Distance of centre of mass from A
g
increasing
CG
CM
CG
CM
g
decreasing
Centre of Mass, Conservation
of Linear Momentum, Impulse
and Collision
8
3 kg (1/2,  3/2)
2 kg
(1, 0)
1 kg
(0, 0)
8.
Y
A y A y
A A
CM
=
-
-
1 1 2 2
1 2
9. A a x a y a
1
2
1 1
4 = = = , ,
A a x
a
y
a
2
2
2 2
3
2
3
2
= = = , ,
In tro duc tory Ex er cise 8.2
1. Method 1
1 2 7 y y = - ( )
Þ y =
14
3
cm
Displacement of CM
= Position of CM ( ) f - Position of CM ( ) i
= + - ( ) ( ) y x 2
= +
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
14
3
2
20
3
=0
Method 2
Mv mv m v
CM
= +
1 1 2 2
= + -
- -
1 2 2 1
1 1
( ) ( ) ms ms
=0
As velocity of CM is zero, there will not be
any change in the position of the CM.
2.
mg T ma sin 60° - = …(i)
T mg ma - ° = sin 30 …(ii)
Adding above equations,
mg ma (sin sin ) 60 30 2 ° - ° =
Þ a g =
- æ
è
ç
ö
ø
÷
3 1
4
a a
1 2
® ®
and are at right angle
\  a
a a
CM
®
® ®
=
+
= +
m m
m
a a
1 2
1 2
2
1
2
( )
or | |
( )
a
CM
®
= =
- a g
2
3 1
4 2
162 | Mechanics-1
B
30°
60°
a
T
mg sin 30°
mg sin 60°
A
a
+ ve
T
(10 – x) x
2 m
A
1 kg
At t = 0s
t = 0 s
A
1 kg 2 kg t = 0 s
CM (i)
1x = 2 (10 – x)
Þ x =      cm
20
3
y 7 – y
1 m CM (f)
2 kg
t = 1 s
t = 1 s
O
y
x
Page 3

Introductory Exercise 8.1
1. If a body is placed in a uniform
gravitational field, the CG of the body
coincides with the CM of the body.
r
r
®
=
®
=
=
CM
S
S
i
n
i i
i
n
i
m
m
1
1

while   r
r
®
=
®
=
=
CM
S
S
i
n
i i i
i
n
i i
m g
m g
1
1
If a body is placed in a uniformly
increasing gravitational field ( ) g
®
in the
upward direction the CG of the body will
be higher level than the CM. And, if the
body is placed in a uniformly decreasing
gravitational field in the upward direction
the CG will be at a lower level the CM.
CG shifts from CM according to the
magnitude and direction of the
gravitational field (by some other agency
eg, earth) in which the body is placed.
In zero gravitational field CG has no
meaning while CM still exists, as usual.
2. The cen tre of mass of a rigid body may lie
in side, on the sur face and even out side the
body. The CM of a solid uni form sphere is
at its cen tre. The CM of a solid ring is at
the cen tre of the ring which lies out side
the mass of the body thus, the state ment
is false. (For fur ther de tails see an swer to
1 As ser tion and Rea son JEE corner).
3. Cen tre of mass al ways lies on the axis of
sym me try of the body, if it ex ists. The
state ment is thus true.
4. State ment is true.
5. As more mass is towards base.
Distance  <
r
4
.
6. If two equal masses are kept at
co-ordinates (R, 0) and ( , ) -R 0 , then their
centre of mass will lie at origin.
7.
X
m X m X
m m
CM
=
+
+
1 1 2 2
1 2
Y
m y m y
m m
CM
=
+
+
1 1 2 2
1 2
r X Y = +
CM CM
2 2
= Distance of centre of mass from A
g
increasing
CG
CM
CG
CM
g
decreasing
Centre of Mass, Conservation
of Linear Momentum, Impulse
and Collision
8
3 kg (1/2,  3/2)
2 kg
(1, 0)
1 kg
(0, 0)
8.
Y
A y A y
A A
CM
=
-
-
1 1 2 2
1 2
9. A a x a y a
1
2
1 1
4 = = = , ,
A a x
a
y
a
2
2
2 2
3
2
3
2
= = = , ,
In tro duc tory Ex er cise 8.2
1. Method 1
1 2 7 y y = - ( )
Þ y =
14
3
cm
Displacement of CM
= Position of CM ( ) f - Position of CM ( ) i
= + - ( ) ( ) y x 2
= +
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
14
3
2
20
3
=0
Method 2
Mv mv m v
CM
= +
1 1 2 2
= + -
- -
1 2 2 1
1 1
( ) ( ) ms ms
=0
As velocity of CM is zero, there will not be
any change in the position of the CM.
2.
mg T ma sin 60° - = …(i)
T mg ma - ° = sin 30 …(ii)
Adding above equations,
mg ma (sin sin ) 60 30 2 ° - ° =
Þ a g =
- æ
è
ç
ö
ø
÷
3 1
4
a a
1 2
® ®
and are at right angle
\  a
a a
CM
®
® ®
=
+
= +
m m
m
a a
1 2
1 2
2
1
2
( )
or | |
( )
a
CM
®
= =
- a g
2
3 1
4 2
162 | Mechanics-1
B
30°
60°
a
T
mg sin 30°
mg sin 60°
A
a
+ ve
T
(10 – x) x
2 m
A
1 kg
At t = 0s
t = 0 s
A
1 kg 2 kg t = 0 s
CM (i)
1x = 2 (10 – x)
Þ x =      cm
20
3
y 7 – y
1 m CM (f)
2 kg
t = 1 s
t = 1 s
O
y
x
In tro duc tory Ex er cise 8.3
1. v
v
CM
=
+ +
+
=
20 60 3
20 6 0
0
( )
[As there is no force along horizontal
direction].
Þ Velocity of trolley ( ) v = - 9 ms
-1
Total energy produced by man
= KE of man + KE of trolley
= ´ ´ + ´ ´ -
1
2
60 3
1
2
20 9
2 2
( ) = 1.08 kJ
2. On streching and then releasing the
spring the restoring force on each block at
instant will be same (according to
Newton’s 3rd law of motion). Now, as force
is same momentum p of each block will
also be same (Dt being same) [As according
to Newton’s second law of motion rate of
change of momentum of a body is directly
proportional to the net force applied on
the body.]
Now, as KE =
p
m
2
2
,
KE of blocks at any instant will be
inversely proportional to their respective
masses.
3. As no external force acts on the system of
particles, the velocity of CM shall not
change. Thus,
v
CM
=
´ + ´ + ´
+ +
20 20 30 20 40 20
20 30 40
\$ \$ \$
i i k
v
CM
=
´ + + +
+ +
®
20 0 30 10 20 40
20 30 40
( )
^ ^
i k v
300 600 40 400 600 800 i j k i j k
^ ^ ^ ^ ^ ^
+ + = + +
Þ  40 100 600 200 v i j k
®
= + +
^ ^ ^
i.e.,        v i j k
®
= + + ( )
^ ^ ^
2.5 15 5  cms
-1
4. Velocity of projectile at the highest point
before explosion = 10 2 i
^
ms
-1
As no extra external force would be acting
during explosion, the velocity of CM will
not change
m m
m
2 2
10 2
× +
=
® ®
0 v
i
^
Þ         v i
®
=20 2
^
Range of rest half part = ´
R
2
2
(as the velocity of the projectile has
doubled at the highest point)
= R =
u
g
2
(as a = ° 45 )
= =
( ) 20
10
40
2
m
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163
–1
(10i + 20k) cms
30 g
20 g
z
y
x
k
j
i
Rest
v
®
R/2
20 cos 45°
–1
= 10Ö2 ms
45°
20
x
v = 10 Ö2  i
^
®
–1
20 cms
–1
20 cms
30 g
20 g
–1
20 cms
z
y
x
k
j
i
20 kg
+ ve
v
20 kg
3 m/s
60 kg
60 kg
Page 4

Introductory Exercise 8.1
1. If a body is placed in a uniform
gravitational field, the CG of the body
coincides with the CM of the body.
r
r
®
=
®
=
=
CM
S
S
i
n
i i
i
n
i
m
m
1
1

while   r
r
®
=
®
=
=
CM
S
S
i
n
i i i
i
n
i i
m g
m g
1
1
If a body is placed in a uniformly
increasing gravitational field ( ) g
®
in the
upward direction the CG of the body will
be higher level than the CM. And, if the
body is placed in a uniformly decreasing
gravitational field in the upward direction
the CG will be at a lower level the CM.
CG shifts from CM according to the
magnitude and direction of the
gravitational field (by some other agency
eg, earth) in which the body is placed.
In zero gravitational field CG has no
meaning while CM still exists, as usual.
2. The cen tre of mass of a rigid body may lie
in side, on the sur face and even out side the
body. The CM of a solid uni form sphere is
at its cen tre. The CM of a solid ring is at
the cen tre of the ring which lies out side
the mass of the body thus, the state ment
is false. (For fur ther de tails see an swer to
1 As ser tion and Rea son JEE corner).
3. Cen tre of mass al ways lies on the axis of
sym me try of the body, if it ex ists. The
state ment is thus true.
4. State ment is true.
5. As more mass is towards base.
Distance  <
r
4
.
6. If two equal masses are kept at
co-ordinates (R, 0) and ( , ) -R 0 , then their
centre of mass will lie at origin.
7.
X
m X m X
m m
CM
=
+
+
1 1 2 2
1 2
Y
m y m y
m m
CM
=
+
+
1 1 2 2
1 2
r X Y = +
CM CM
2 2
= Distance of centre of mass from A
g
increasing
CG
CM
CG
CM
g
decreasing
Centre of Mass, Conservation
of Linear Momentum, Impulse
and Collision
8
3 kg (1/2,  3/2)
2 kg
(1, 0)
1 kg
(0, 0)
8.
Y
A y A y
A A
CM
=
-
-
1 1 2 2
1 2
9. A a x a y a
1
2
1 1
4 = = = , ,
A a x
a
y
a
2
2
2 2
3
2
3
2
= = = , ,
In tro duc tory Ex er cise 8.2
1. Method 1
1 2 7 y y = - ( )
Þ y =
14
3
cm
Displacement of CM
= Position of CM ( ) f - Position of CM ( ) i
= + - ( ) ( ) y x 2
= +
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
14
3
2
20
3
=0
Method 2
Mv mv m v
CM
= +
1 1 2 2
= + -
- -
1 2 2 1
1 1
( ) ( ) ms ms
=0
As velocity of CM is zero, there will not be
any change in the position of the CM.
2.
mg T ma sin 60° - = …(i)
T mg ma - ° = sin 30 …(ii)
Adding above equations,
mg ma (sin sin ) 60 30 2 ° - ° =
Þ a g =
- æ
è
ç
ö
ø
÷
3 1
4
a a
1 2
® ®
and are at right angle
\  a
a a
CM
®
® ®
=
+
= +
m m
m
a a
1 2
1 2
2
1
2
( )
or | |
( )
a
CM
®
= =
- a g
2
3 1
4 2
162 | Mechanics-1
B
30°
60°
a
T
mg sin 30°
mg sin 60°
A
a
+ ve
T
(10 – x) x
2 m
A
1 kg
At t = 0s
t = 0 s
A
1 kg 2 kg t = 0 s
CM (i)
1x = 2 (10 – x)
Þ x =      cm
20
3
y 7 – y
1 m CM (f)
2 kg
t = 1 s
t = 1 s
O
y
x
In tro duc tory Ex er cise 8.3
1. v
v
CM
=
+ +
+
=
20 60 3
20 6 0
0
( )
[As there is no force along horizontal
direction].
Þ Velocity of trolley ( ) v = - 9 ms
-1
Total energy produced by man
= KE of man + KE of trolley
= ´ ´ + ´ ´ -
1
2
60 3
1
2
20 9
2 2
( ) = 1.08 kJ
2. On streching and then releasing the
spring the restoring force on each block at
instant will be same (according to
Newton’s 3rd law of motion). Now, as force
is same momentum p of each block will
also be same (Dt being same) [As according
to Newton’s second law of motion rate of
change of momentum of a body is directly
proportional to the net force applied on
the body.]
Now, as KE =
p
m
2
2
,
KE of blocks at any instant will be
inversely proportional to their respective
masses.
3. As no external force acts on the system of
particles, the velocity of CM shall not
change. Thus,
v
CM
=
´ + ´ + ´
+ +
20 20 30 20 40 20
20 30 40
\$ \$ \$
i i k
v
CM
=
´ + + +
+ +
®
20 0 30 10 20 40
20 30 40
( )
^ ^
i k v
300 600 40 400 600 800 i j k i j k
^ ^ ^ ^ ^ ^
+ + = + +
Þ  40 100 600 200 v i j k
®
= + +
^ ^ ^
i.e.,        v i j k
®
= + + ( )
^ ^ ^
2.5 15 5  cms
-1
4. Velocity of projectile at the highest point
before explosion = 10 2 i
^
ms
-1
As no extra external force would be acting
during explosion, the velocity of CM will
not change
m m
m
2 2
10 2
× +
=
® ®
0 v
i
^
Þ         v i
®
=20 2
^
Range of rest half part = ´
R
2
2
(as the velocity of the projectile has
doubled at the highest point)
= R =
u
g
2
(as a = ° 45 )
= =
( ) 20
10
40
2
m
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163
–1
(10i + 20k) cms
30 g
20 g
z
y
x
k
j
i
Rest
v
®
R/2
20 cos 45°
–1
= 10Ö2 ms
45°
20
x
v = 10 Ö2  i
^
®
–1
20 cms
–1
20 cms
30 g
20 g
–1
20 cms
z
y
x
k
j
i
20 kg
+ ve
v
20 kg
3 m/s
60 kg
60 kg
Therefore, the rest half part will land at a
distance of 40
40
2
+
æ
è
ç
ö
ø
÷ m i.e., 60 m from the
point of projection.
5. At point P
Horizontal velocity = Initial horizontal
velocity
= °
-
20 2 45
1
cos ms
=20 ms
-1
Vertical velocity
= ° - × 20 2 45 1 sin g (as t =1s)
=70 ms
-1
\ v
®
(velocity of projectile at point P i.e., at
t = 15 just before explosion)
= + 20 10 i j
^ ^
Now, as the projectile breaks up into two
equal parts and one part comes to rest, the
velocity of other half part after explosion
will be
u v
® ®
=2 = + 40 20 i j
^ ^
ms
-1
Angle of projection ( ) a of 2nd half part
after explosion at point P.
a =
é
ë
ê
ù
û
ú
-
tan
1
20
40
=
é
ë
ê
ù
û
ú
-
tan
1
1
2
sin a =
1
5
Maximum height attained by second half
part
= Height of point P + Extra maximum
height attained
by second half part
= ´ - ´
é
ë
ê
ù
û
ú
+ 20 1
1
2
1
2000
2
2
2 2
g
g
( ) sin a
= +
´
´ 15
2000
2 10
1
5

= + 15 20
=35 m
6. Momentum of platform + boy + stone
along x-axis after throwing stone = before
throwing stone
\ ( ) 60 40 1 10
1
2
0 + +
æ
è
ç
ö
ø
÷
= v
Þ v = -
5 2
100
ms
-1
Time of flight of stone
2 10 45
2
´ °
=
sin
g
s
\ Horizontal displacement of platform
(+ boy)
=vT
= - ´
5 2
100
2 = -
1
10
m = - 10 cm
7. Thrust due to the upward component of
the velocity of the bullet will rotate the
movable end of the barrel and thus the
bullet leaving the barrel will travelling at
an angle greater than 45° when it comes
out of the barrel.
164 | Mechanics-1
10 cm
X
T = Ö2
40 kg
X
45°
1 kg
–1
10 ms
60 kg
s
1 kg
–1
10 ms
45°
15 m
–1
20Ö2 ms
P
–1
20 ms
t = 15
a
u 2 2
= Ö(40) + (20)
–1
= Ö2000 ms
t = 0 s
a
Ö5
1
2
R
u
Ö2 20
i
q = 45° Bullet q' > q
Bullet
Page 5

Introductory Exercise 8.1
1. If a body is placed in a uniform
gravitational field, the CG of the body
coincides with the CM of the body.
r
r
®
=
®
=
=
CM
S
S
i
n
i i
i
n
i
m
m
1
1

while   r
r
®
=
®
=
=
CM
S
S
i
n
i i i
i
n
i i
m g
m g
1
1
If a body is placed in a uniformly
increasing gravitational field ( ) g
®
in the
upward direction the CG of the body will
be higher level than the CM. And, if the
body is placed in a uniformly decreasing
gravitational field in the upward direction
the CG will be at a lower level the CM.
CG shifts from CM according to the
magnitude and direction of the
gravitational field (by some other agency
eg, earth) in which the body is placed.
In zero gravitational field CG has no
meaning while CM still exists, as usual.
2. The cen tre of mass of a rigid body may lie
in side, on the sur face and even out side the
body. The CM of a solid uni form sphere is
at its cen tre. The CM of a solid ring is at
the cen tre of the ring which lies out side
the mass of the body thus, the state ment
is false. (For fur ther de tails see an swer to
1 As ser tion and Rea son JEE corner).
3. Cen tre of mass al ways lies on the axis of
sym me try of the body, if it ex ists. The
state ment is thus true.
4. State ment is true.
5. As more mass is towards base.
Distance  <
r
4
.
6. If two equal masses are kept at
co-ordinates (R, 0) and ( , ) -R 0 , then their
centre of mass will lie at origin.
7.
X
m X m X
m m
CM
=
+
+
1 1 2 2
1 2
Y
m y m y
m m
CM
=
+
+
1 1 2 2
1 2
r X Y = +
CM CM
2 2
= Distance of centre of mass from A
g
increasing
CG
CM
CG
CM
g
decreasing
Centre of Mass, Conservation
of Linear Momentum, Impulse
and Collision
8
3 kg (1/2,  3/2)
2 kg
(1, 0)
1 kg
(0, 0)
8.
Y
A y A y
A A
CM
=
-
-
1 1 2 2
1 2
9. A a x a y a
1
2
1 1
4 = = = , ,
A a x
a
y
a
2
2
2 2
3
2
3
2
= = = , ,
In tro duc tory Ex er cise 8.2
1. Method 1
1 2 7 y y = - ( )
Þ y =
14
3
cm
Displacement of CM
= Position of CM ( ) f - Position of CM ( ) i
= + - ( ) ( ) y x 2
= +
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
14
3
2
20
3
=0
Method 2
Mv mv m v
CM
= +
1 1 2 2
= + -
- -
1 2 2 1
1 1
( ) ( ) ms ms
=0
As velocity of CM is zero, there will not be
any change in the position of the CM.
2.
mg T ma sin 60° - = …(i)
T mg ma - ° = sin 30 …(ii)
Adding above equations,
mg ma (sin sin ) 60 30 2 ° - ° =
Þ a g =
- æ
è
ç
ö
ø
÷
3 1
4
a a
1 2
® ®
and are at right angle
\  a
a a
CM
®
® ®
=
+
= +
m m
m
a a
1 2
1 2
2
1
2
( )
or | |
( )
a
CM
®
= =
- a g
2
3 1
4 2
162 | Mechanics-1
B
30°
60°
a
T
mg sin 30°
mg sin 60°
A
a
+ ve
T
(10 – x) x
2 m
A
1 kg
At t = 0s
t = 0 s
A
1 kg 2 kg t = 0 s
CM (i)
1x = 2 (10 – x)
Þ x =      cm
20
3
y 7 – y
1 m CM (f)
2 kg
t = 1 s
t = 1 s
O
y
x
In tro duc tory Ex er cise 8.3
1. v
v
CM
=
+ +
+
=
20 60 3
20 6 0
0
( )
[As there is no force along horizontal
direction].
Þ Velocity of trolley ( ) v = - 9 ms
-1
Total energy produced by man
= KE of man + KE of trolley
= ´ ´ + ´ ´ -
1
2
60 3
1
2
20 9
2 2
( ) = 1.08 kJ
2. On streching and then releasing the
spring the restoring force on each block at
instant will be same (according to
Newton’s 3rd law of motion). Now, as force
is same momentum p of each block will
also be same (Dt being same) [As according
to Newton’s second law of motion rate of
change of momentum of a body is directly
proportional to the net force applied on
the body.]
Now, as KE =
p
m
2
2
,
KE of blocks at any instant will be
inversely proportional to their respective
masses.
3. As no external force acts on the system of
particles, the velocity of CM shall not
change. Thus,
v
CM
=
´ + ´ + ´
+ +
20 20 30 20 40 20
20 30 40
\$ \$ \$
i i k
v
CM
=
´ + + +
+ +
®
20 0 30 10 20 40
20 30 40
( )
^ ^
i k v
300 600 40 400 600 800 i j k i j k
^ ^ ^ ^ ^ ^
+ + = + +
Þ  40 100 600 200 v i j k
®
= + +
^ ^ ^
i.e.,        v i j k
®
= + + ( )
^ ^ ^
2.5 15 5  cms
-1
4. Velocity of projectile at the highest point
before explosion = 10 2 i
^
ms
-1
As no extra external force would be acting
during explosion, the velocity of CM will
not change
m m
m
2 2
10 2
× +
=
® ®
0 v
i
^
Þ         v i
®
=20 2
^
Range of rest half part = ´
R
2
2
(as the velocity of the projectile has
doubled at the highest point)
= R =
u
g
2
(as a = ° 45 )
= =
( ) 20
10
40
2
m
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163
–1
(10i + 20k) cms
30 g
20 g
z
y
x
k
j
i
Rest
v
®
R/2
20 cos 45°
–1
= 10Ö2 ms
45°
20
x
v = 10 Ö2  i
^
®
–1
20 cms
–1
20 cms
30 g
20 g
–1
20 cms
z
y
x
k
j
i
20 kg
+ ve
v
20 kg
3 m/s
60 kg
60 kg
Therefore, the rest half part will land at a
distance of 40
40
2
+
æ
è
ç
ö
ø
÷ m i.e., 60 m from the
point of projection.
5. At point P
Horizontal velocity = Initial horizontal
velocity
= °
-
20 2 45
1
cos ms
=20 ms
-1
Vertical velocity
= ° - × 20 2 45 1 sin g (as t =1s)
=70 ms
-1
\ v
®
(velocity of projectile at point P i.e., at
t = 15 just before explosion)
= + 20 10 i j
^ ^
Now, as the projectile breaks up into two
equal parts and one part comes to rest, the
velocity of other half part after explosion
will be
u v
® ®
=2 = + 40 20 i j
^ ^
ms
-1
Angle of projection ( ) a of 2nd half part
after explosion at point P.
a =
é
ë
ê
ù
û
ú
-
tan
1
20
40
=
é
ë
ê
ù
û
ú
-
tan
1
1
2
sin a =
1
5
Maximum height attained by second half
part
= Height of point P + Extra maximum
height attained
by second half part
= ´ - ´
é
ë
ê
ù
û
ú
+ 20 1
1
2
1
2000
2
2
2 2
g
g
( ) sin a
= +
´
´ 15
2000
2 10
1
5

= + 15 20
=35 m
6. Momentum of platform + boy + stone
along x-axis after throwing stone = before
throwing stone
\ ( ) 60 40 1 10
1
2
0 + +
æ
è
ç
ö
ø
÷
= v
Þ v = -
5 2
100
ms
-1
Time of flight of stone
2 10 45
2
´ °
=
sin
g
s
\ Horizontal displacement of platform
(+ boy)
=vT
= - ´
5 2
100
2 = -
1
10
m = - 10 cm
7. Thrust due to the upward component of
the velocity of the bullet will rotate the
movable end of the barrel and thus the
bullet leaving the barrel will travelling at
an angle greater than 45° when it comes
out of the barrel.
164 | Mechanics-1
10 cm
X
T = Ö2
40 kg
X
45°
1 kg
–1
10 ms
60 kg
s
1 kg
–1
10 ms
45°
15 m
–1
20Ö2 ms
P
–1
20 ms
t = 15
a
u 2 2
= Ö(40) + (20)
–1
= Ö2000 ms
t = 0 s
a
Ö5
1
2
R
u
Ö2 20
i
q = 45° Bullet q' > q
Bullet
In tro duc tory Ex er cise 8.4
1. To just lift rocket off the launching pad
Thrust force = Weight
or v
dm
dt
mg -
æ
è
ç
ö
ø
÷
=
or -
æ
è
ç
ö
ø
÷
=
dm
dt
mg
v
=
+ ´
´
( ) 20 180
10
3
9.8
1.6
=1.225 kgs
-1
(i) Rate of consumption of fuel =
-
2
1
kgs
\ Time required for the consumption of fuel
t = =
-
180
2
90
1
kg
kgs
s
Ultimate speed gained by rocket
v u gt v
m
m
e
= - +
æ
è
ç
ö
ø
÷ log
0
…(i)
Substituting u = 0, v = 1.6 kms
-1
,
m
0
20 180 = + ( ) kg and m = 20 kg in Eq. (i).
v = - ´ + ´
æ
è
ç
ö
ø
÷ 9.8 1.6 90 10
200
20
3
ln
=2.8 kms
-1
(iii) Rate of consumption of fuel =
-
20
1
kgs
\      t = =
-
180
20
9
1
kg
kgs
s
i.e., v = - ´ + ´ 9.8 1.6 9 10
200
20
3
ln
=
-
3.6kms
1
2. Mass at time t, m m t = -
0
m
\
dm
dt
= -m
ma = thrust force - mg
or ma v
dm
dt
mg = -
æ
è
ç
ö
ø
÷
-
or ma v mg = - m
or ( ) ( ) m t
d x
dt
u m t g
0
2
2
0
- = - - m m m (Q v u = )
3. v u gt v
m
m
= - +
æ
è
ç
ö
ø
÷ ln
0
= - +
-
æ
è
ç
ö
ø
÷
0
1
3
0
0
gt u
m
m
t
ln
= - + g u 1
3
2
ln (at t = 1s)
= - u g ln
3
2

Introductory Exercise 8.5
1. u
®
(at t = 0 s) = +
-
( )
^ ^
10 3 10
1
i j ms
At t = 1 s
Horizontal velocity =
-
10 3
1
i
^
ms
Vertical velocity = - ×
-
( )
^
10 1
1
g j ms
=
-
0
1
j
^
ms
\ v i
®
-
= 10 3
1 ^
ms
Change in velocity, D v v u
® ® ®
= -
= - + ( ) ( )
^ ^ ^
10 3 10 3 10 i i j
= -
-
10
1
j
^
ms
i.e.,    Dv
®
-
=10
1
ms , downwards.
2. Impulse ( ) J
®
imparted
= Change in momentum in the time
interval
t=0 s to t = 2 s
= m [(Velocity at t = 2 s)
- (Velocity at t = 0 s)]
= + -
- -
2 4 4 4
1 1
kg ms ms [( ) ( ) ]
^ ^ ^
i j j
= 8 i
^
Ns                   ( Q 1 1
1
kg ms Ns
-
= )
3. Spring will become taut when the ball
would go down by 2 m.
v u g
2 2
2 2 = + ´ ´
or     v =2 10 ms
-1
D D p m v =
= - 12 10 0 ( ) kgms
-1
= 2 10 kg ms
-1

\ Impulse imparted = 2 10 Ns.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 165
60°
20 m/s
2 m

122 docs

## FAQs on DC Pandey Solutions: Centre of Mass, Conservation of Linear Momentum- 1 - DC Pandey Solutions for NEET Physics

 1. What is the concept of centre of mass in physics?
Ans. The concept of centre of mass in physics refers to the point at which the entire mass of a system can be considered to be concentrated. It is a point that represents the average position of all the particles in a system, and it is often used to simplify calculations involving the motion of objects.
 2. How is the centre of mass calculated for a system of particles?
Ans. The centre of mass for a system of particles can be calculated by taking the weighted average of the positions of all the particles, where the weights are given by the masses of the particles. Mathematically, it is given by the formula: Centre of mass = (m1*r1 + m2*r2 + ... + mn*rn)/(m1 + m2 + ... + mn) Here, m1, m2, ..., mn are the masses of the particles, and r1, r2, ..., rn are their respective positions.
 3. What is the principle of conservation of linear momentum?
Ans. The principle of conservation of linear momentum states that the total momentum of a closed system remains constant if no external forces act on it. In other words, the total momentum before an event is equal to the total momentum after the event, as long as there are no external forces involved.
 4. How is the conservation of linear momentum applied in physics?
Ans. The conservation of linear momentum is applied in physics to analyze and understand the motion of objects in various scenarios. By applying this principle, we can predict the final velocities of objects after a collision, determine the recoil velocity of a gun when a bullet is fired, and analyze the motion of objects in explosive events, among other things.
 5. Can the conservation of linear momentum be violated?
Ans. No, the conservation of linear momentum is a fundamental principle in physics and is universally applicable. It cannot be violated under normal circumstances where no external forces act on a closed system. However, in certain extreme scenarios involving relativistic speeds or quantum effects, the conservation of linear momentum may need to be modified, but these cases are beyond the scope of classical mechanics.

## DC Pandey Solutions for NEET Physics

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