DC Pandey Solutions: Laws of Motion- 4

# DC Pandey Solutions: Laws of Motion- 4 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

53. Minimum force required to start the
motion upward
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
4
3 3
3
2
=
7
6
mg
54. Minimum force required to move the block
up the incline with constant speed
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
2
3 3
3
2
=
5
6
mg
55. S
1
2
2
8
=
×
æ
è
ç
ö
ø
÷
=
(5.22)
9.8
14.3 m
Option (c) is correct.
56.    a
g g
¢ =
- 1200 1000
1200
=
g
6
a
g g
=
- 1350 1200
1200
=
g
8
For accelerated motion
v as
max
2 2
1
0 2 = +
Þ s
v
a
1
2
2
=
max
For retarted motion
0 2
2 2
2
= - ¢ v a s
max
Þ s
v
a
2
2
2
=
¢
max
s s
v
a a
1 2
2
2
1 1
+ = +
¢
é
ë
ê
ù
û
ú
max
25
2
8 6
2
= +
é
ë
ê
ù
û
ú
v
g g
max
v
max
=
´ 50
14
9.8
=
-
5.92ms
1
Option (c) is correct.
57. mg sin q = ´ ´ 170 10
8
15
= 906.67 N
f
1
170 10
15
17
(max) = ´ ´ ´ 0.2 = 300 N
f
2
170 10
15
17
(max) = ´ ´ ´ 0.4
= 600 N
The whole system will accelerate as
mgsinq is greater than both f
1
(max) and
f
2
(max).
Total force of friction
= + f f
1 2
(max) (max)
=900 N
Option (a) is correct.
58. mg T ma sin q - - = 300            …(i)
and mg T ma sinq + = …(ii)
Substituting Eq. (i) by Eq. (ii),
2 300 0 T + =
T = - 150 N
= 150 N, compressive.
Option (a) is correct.
Laws of Motion | 101
a
25 m
T' = 1000g
– a'
1200 g
1200 g
T = 1350g
Speed
= v
max
Stops
q
mg sin q
A
B
m = 0.4
f
2
f
1
T
T
m = 0.2
mg sin q
q
8
17
15
Page 2

53. Minimum force required to start the
motion upward
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
4
3 3
3
2
=
7
6
mg
54. Minimum force required to move the block
up the incline with constant speed
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
2
3 3
3
2
=
5
6
mg
55. S
1
2
2
8
=
×
æ
è
ç
ö
ø
÷
=
(5.22)
9.8
14.3 m
Option (c) is correct.
56.    a
g g
¢ =
- 1200 1000
1200
=
g
6
a
g g
=
- 1350 1200
1200
=
g
8
For accelerated motion
v as
max
2 2
1
0 2 = +
Þ s
v
a
1
2
2
=
max
For retarted motion
0 2
2 2
2
= - ¢ v a s
max
Þ s
v
a
2
2
2
=
¢
max
s s
v
a a
1 2
2
2
1 1
+ = +
¢
é
ë
ê
ù
û
ú
max
25
2
8 6
2
= +
é
ë
ê
ù
û
ú
v
g g
max
v
max
=
´ 50
14
9.8
=
-
5.92ms
1
Option (c) is correct.
57. mg sin q = ´ ´ 170 10
8
15
= 906.67 N
f
1
170 10
15
17
(max) = ´ ´ ´ 0.2 = 300 N
f
2
170 10
15
17
(max) = ´ ´ ´ 0.4
= 600 N
The whole system will accelerate as
mgsinq is greater than both f
1
(max) and
f
2
(max).
Total force of friction
= + f f
1 2
(max) (max)
=900 N
Option (a) is correct.
58. mg T ma sin q - - = 300            …(i)
and mg T ma sinq + = …(ii)
Substituting Eq. (i) by Eq. (ii),
2 300 0 T + =
T = - 150 N
= 150 N, compressive.
Option (a) is correct.
Laws of Motion | 101
a
25 m
T' = 1000g
– a'
1200 g
1200 g
T = 1350g
Speed
= v
max
Stops
q
mg sin q
A
B
m = 0.4
f
2
f
1
T
T
m = 0.2
mg sin q
q
8
17
15
More than One Cor rect Options
1. (a) Normal force between A and B m g =
2
= ´ = 1 10 10 N
\ Force of limiting friction by B on A (or by
A on B)
= ´ = m 10 2 N
Total force opposing applied external force
F T = + 2N
= + 2 2 N N
= 4 N
Thus, if F £ 4 N
The block A will remain stationary and so
block B also. The system will be in
equilibrium.
\ Option (a) is correct.
(b) If F > 4 N
F T - - = 2 1 a
and T a - = 2 1 …(i)
F a - = 4 2 …(ii)
i.e., F a = + 4 2
For F > 4 N
2 4 4 a + >
or a > 0
Þ T - > 2 0
i.e., T > 2 N
\ Option (b) is incorrect.
(c) Block A will move over B only when
F > 4 N and then the frictional force
between the blocks will be 2 N if a is just 0
[as explained in (b)].
Option (c) is correct.
(d) If F = 6 N using Eq. (ii)
2 6 4 a = -
Þ a = 1 m/s
2
\ Using Eq. (i),
T - = 2 1
i.e., T = 3 N
Option (d) is correct.
2. At point A
T T mg
1 2
cos cos a b = + …(i)
and T T
1 2
sin sin a b = …(ii)
At point B
T mg
2
cos b = …(iii)
T F mg
2
sin b = - …(iv)
Using Eq. (iii) in Eq. (i),
T T
1 2
2 cos cos a b = …(v)
Dividing Eq. (ii) by Eq. (v),
2 tan tan a b = …(vi)
Option (a) is correct.
Squaring and adding Eqs. (iii) and (v),
T T T
1
2
2
2 2
2
2 2
4 = + cos sin b b …(vii)
Dividing Eq. (iii) by Eq. (iv)
tan b = 1
\ cos sin b b = =
1
2
Substituting the values of sinb and cosb
in Eq. (vii)
T T T
1
2
2
2
2
2
4
1
2
1
2
=
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
=
5
2
2
2
T
102 | Mechanics-1
T T
T T
2 N
F
A
B
2 N
F = mg
B
T
2
T
2
b
A
T
1
a
T
1
mg
b
1
1
Ö2
Page 3

53. Minimum force required to start the
motion upward
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
4
3 3
3
2
=
7
6
mg
54. Minimum force required to move the block
up the incline with constant speed
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
2
3 3
3
2
=
5
6
mg
55. S
1
2
2
8
=
×
æ
è
ç
ö
ø
÷
=
(5.22)
9.8
14.3 m
Option (c) is correct.
56.    a
g g
¢ =
- 1200 1000
1200
=
g
6
a
g g
=
- 1350 1200
1200
=
g
8
For accelerated motion
v as
max
2 2
1
0 2 = +
Þ s
v
a
1
2
2
=
max
For retarted motion
0 2
2 2
2
= - ¢ v a s
max
Þ s
v
a
2
2
2
=
¢
max
s s
v
a a
1 2
2
2
1 1
+ = +
¢
é
ë
ê
ù
û
ú
max
25
2
8 6
2
= +
é
ë
ê
ù
û
ú
v
g g
max
v
max
=
´ 50
14
9.8
=
-
5.92ms
1
Option (c) is correct.
57. mg sin q = ´ ´ 170 10
8
15
= 906.67 N
f
1
170 10
15
17
(max) = ´ ´ ´ 0.2 = 300 N
f
2
170 10
15
17
(max) = ´ ´ ´ 0.4
= 600 N
The whole system will accelerate as
mgsinq is greater than both f
1
(max) and
f
2
(max).
Total force of friction
= + f f
1 2
(max) (max)
=900 N
Option (a) is correct.
58. mg T ma sin q - - = 300            …(i)
and mg T ma sinq + = …(ii)
Substituting Eq. (i) by Eq. (ii),
2 300 0 T + =
T = - 150 N
= 150 N, compressive.
Option (a) is correct.
Laws of Motion | 101
a
25 m
T' = 1000g
– a'
1200 g
1200 g
T = 1350g
Speed
= v
max
Stops
q
mg sin q
A
B
m = 0.4
f
2
f
1
T
T
m = 0.2
mg sin q
q
8
17
15
More than One Cor rect Options
1. (a) Normal force between A and B m g =
2
= ´ = 1 10 10 N
\ Force of limiting friction by B on A (or by
A on B)
= ´ = m 10 2 N
Total force opposing applied external force
F T = + 2N
= + 2 2 N N
= 4 N
Thus, if F £ 4 N
The block A will remain stationary and so
block B also. The system will be in
equilibrium.
\ Option (a) is correct.
(b) If F > 4 N
F T - - = 2 1 a
and T a - = 2 1 …(i)
F a - = 4 2 …(ii)
i.e., F a = + 4 2
For F > 4 N
2 4 4 a + >
or a > 0
Þ T - > 2 0
i.e., T > 2 N
\ Option (b) is incorrect.
(c) Block A will move over B only when
F > 4 N and then the frictional force
between the blocks will be 2 N if a is just 0
[as explained in (b)].
Option (c) is correct.
(d) If F = 6 N using Eq. (ii)
2 6 4 a = -
Þ a = 1 m/s
2
\ Using Eq. (i),
T - = 2 1
i.e., T = 3 N
Option (d) is correct.
2. At point A
T T mg
1 2
cos cos a b = + …(i)
and T T
1 2
sin sin a b = …(ii)
At point B
T mg
2
cos b = …(iii)
T F mg
2
sin b = - …(iv)
Using Eq. (iii) in Eq. (i),
T T
1 2
2 cos cos a b = …(v)
Dividing Eq. (ii) by Eq. (v),
2 tan tan a b = …(vi)
Option (a) is correct.
Squaring and adding Eqs. (iii) and (v),
T T T
1
2
2
2 2
2
2 2
4 = + cos sin b b …(vii)
Dividing Eq. (iii) by Eq. (iv)
tan b = 1
\ cos sin b b = =
1
2
Substituting the values of sinb and cosb
in Eq. (vii)
T T T
1
2
2
2
2
2
4
1
2
1
2
=
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
=
5
2
2
2
T
102 | Mechanics-1
T T
T T
2 N
F
A
B
2 N
F = mg
B
T
2
T
2
b
A
T
1
a
T
1
mg
b
1
1
Ö2
Þ 2 5
1 2
T T =
Option (c) is correct.
3. Displacement of block in 4 s
S = Area under curve
= 16 m.
DK =Workdone by frictional force
1
2
1 4 1 10 16
2
´ ´ = ´ ´ ´ m
Þ m = 0.1
Option (a) is correct.
Option (b) is incorrect.
Acceleration,   a = f tan
= - tan ( ) p q
= - tan q
= - 1 m/s
2
If half rough retardation = 0.5 m/s
2
\ 16 4
1
2
2
= + - t t ( ) 0.5
i.e., t t
2
16 64 0 - + =
or t = 8 s
Option (d) is correct.
Option (c) is incorrect.
4. Let acceleration of wedge ( ) A a =
N ma mg + = sin cos q q
N mg ma = - cos sin q q
Acceleration of
a
N
M
=
sin q
or Ma mg ma = - ( cos sin ) sin q q q
or a M m mg ( sin ) cos sin + =
2
q q q
i.e.,     a
mg
M m
=
+
cos sin
sin
q q
q
2
=
´ ´ ° °
+ ´ °
0.6
1.7 0.6
g cos sin
( sin )
45 45
45
2
=
+
3
17 3
g
=
3
20
g
Let a
B
= Acceleration of block B
Net force on B (along inclined plane)
ma ma mg
B
= + cos sin q q
Þ a a g
B
= + cos sin q q
Thus, ( ) ( cos sin ) cos a a g
B V
= + q q q
= + a g cos sin cos
2
q q q
= + ( ) a g
1
2
= +
æ
è
ç
ö
ø
÷
3
20
1
2
g
g
=
23
40
g

( ) ( cos sin ) sin a a g a
B H
= + - q q q
= -
23
40
3
20
g g

=
17
40
g

5. f m g
A 1 1
(max.) = m
= ´ ´ 0.3 60 10
= 180 N
F
net
on B = + f T
1
(max.)
= + 180 125
= 305 N
A will remain stationary as
T f <
1
(max.)
\ f
1
125 = N
Force of friction acting between A and
B=125 N
\ Options (c) and (d) are incorrect.
Laws of Motion | 103
O
q
–1
v (ms )
f
4 t (s)
q mg
N
ma (Pseudo
force)
F
N sin q
A M
a F = mg cos q
ma sin q
q
B
f
1
f
2
f
1 A
T(pull) = 125 N
Page 4

53. Minimum force required to start the
motion upward
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
4
3 3
3
2
=
7
6
mg
54. Minimum force required to move the block
up the incline with constant speed
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
2
3 3
3
2
=
5
6
mg
55. S
1
2
2
8
=
×
æ
è
ç
ö
ø
÷
=
(5.22)
9.8
14.3 m
Option (c) is correct.
56.    a
g g
¢ =
- 1200 1000
1200
=
g
6
a
g g
=
- 1350 1200
1200
=
g
8
For accelerated motion
v as
max
2 2
1
0 2 = +
Þ s
v
a
1
2
2
=
max
For retarted motion
0 2
2 2
2
= - ¢ v a s
max
Þ s
v
a
2
2
2
=
¢
max
s s
v
a a
1 2
2
2
1 1
+ = +
¢
é
ë
ê
ù
û
ú
max
25
2
8 6
2
= +
é
ë
ê
ù
û
ú
v
g g
max
v
max
=
´ 50
14
9.8
=
-
5.92ms
1
Option (c) is correct.
57. mg sin q = ´ ´ 170 10
8
15
= 906.67 N
f
1
170 10
15
17
(max) = ´ ´ ´ 0.2 = 300 N
f
2
170 10
15
17
(max) = ´ ´ ´ 0.4
= 600 N
The whole system will accelerate as
mgsinq is greater than both f
1
(max) and
f
2
(max).
Total force of friction
= + f f
1 2
(max) (max)
=900 N
Option (a) is correct.
58. mg T ma sin q - - = 300            …(i)
and mg T ma sinq + = …(ii)
Substituting Eq. (i) by Eq. (ii),
2 300 0 T + =
T = - 150 N
= 150 N, compressive.
Option (a) is correct.
Laws of Motion | 101
a
25 m
T' = 1000g
– a'
1200 g
1200 g
T = 1350g
Speed
= v
max
Stops
q
mg sin q
A
B
m = 0.4
f
2
f
1
T
T
m = 0.2
mg sin q
q
8
17
15
More than One Cor rect Options
1. (a) Normal force between A and B m g =
2
= ´ = 1 10 10 N
\ Force of limiting friction by B on A (or by
A on B)
= ´ = m 10 2 N
Total force opposing applied external force
F T = + 2N
= + 2 2 N N
= 4 N
Thus, if F £ 4 N
The block A will remain stationary and so
block B also. The system will be in
equilibrium.
\ Option (a) is correct.
(b) If F > 4 N
F T - - = 2 1 a
and T a - = 2 1 …(i)
F a - = 4 2 …(ii)
i.e., F a = + 4 2
For F > 4 N
2 4 4 a + >
or a > 0
Þ T - > 2 0
i.e., T > 2 N
\ Option (b) is incorrect.
(c) Block A will move over B only when
F > 4 N and then the frictional force
between the blocks will be 2 N if a is just 0
[as explained in (b)].
Option (c) is correct.
(d) If F = 6 N using Eq. (ii)
2 6 4 a = -
Þ a = 1 m/s
2
\ Using Eq. (i),
T - = 2 1
i.e., T = 3 N
Option (d) is correct.
2. At point A
T T mg
1 2
cos cos a b = + …(i)
and T T
1 2
sin sin a b = …(ii)
At point B
T mg
2
cos b = …(iii)
T F mg
2
sin b = - …(iv)
Using Eq. (iii) in Eq. (i),
T T
1 2
2 cos cos a b = …(v)
Dividing Eq. (ii) by Eq. (v),
2 tan tan a b = …(vi)
Option (a) is correct.
Squaring and adding Eqs. (iii) and (v),
T T T
1
2
2
2 2
2
2 2
4 = + cos sin b b …(vii)
Dividing Eq. (iii) by Eq. (iv)
tan b = 1
\ cos sin b b = =
1
2
Substituting the values of sinb and cosb
in Eq. (vii)
T T T
1
2
2
2
2
2
4
1
2
1
2
=
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
=
5
2
2
2
T
102 | Mechanics-1
T T
T T
2 N
F
A
B
2 N
F = mg
B
T
2
T
2
b
A
T
1
a
T
1
mg
b
1
1
Ö2
Þ 2 5
1 2
T T =
Option (c) is correct.
3. Displacement of block in 4 s
S = Area under curve
= 16 m.
DK =Workdone by frictional force
1
2
1 4 1 10 16
2
´ ´ = ´ ´ ´ m
Þ m = 0.1
Option (a) is correct.
Option (b) is incorrect.
Acceleration,   a = f tan
= - tan ( ) p q
= - tan q
= - 1 m/s
2
If half rough retardation = 0.5 m/s
2
\ 16 4
1
2
2
= + - t t ( ) 0.5
i.e., t t
2
16 64 0 - + =
or t = 8 s
Option (d) is correct.
Option (c) is incorrect.
4. Let acceleration of wedge ( ) A a =
N ma mg + = sin cos q q
N mg ma = - cos sin q q
Acceleration of
a
N
M
=
sin q
or Ma mg ma = - ( cos sin ) sin q q q
or a M m mg ( sin ) cos sin + =
2
q q q
i.e.,     a
mg
M m
=
+
cos sin
sin
q q
q
2
=
´ ´ ° °
+ ´ °
0.6
1.7 0.6
g cos sin
( sin )
45 45
45
2
=
+
3
17 3
g
=
3
20
g
Let a
B
= Acceleration of block B
Net force on B (along inclined plane)
ma ma mg
B
= + cos sin q q
Þ a a g
B
= + cos sin q q
Thus, ( ) ( cos sin ) cos a a g
B V
= + q q q
= + a g cos sin cos
2
q q q
= + ( ) a g
1
2
= +
æ
è
ç
ö
ø
÷
3
20
1
2
g
g
=
23
40
g

( ) ( cos sin ) sin a a g a
B H
= + - q q q
= -
23
40
3
20
g g

=
17
40
g

5. f m g
A 1 1
(max.) = m
= ´ ´ 0.3 60 10
= 180 N
F
net
on B = + f T
1
(max.)
= + 180 125
= 305 N
A will remain stationary as
T f <
1
(max.)
\ f
1
125 = N
Force of friction acting between A and
B=125 N
\ Options (c) and (d) are incorrect.
Laws of Motion | 103
O
q
–1
v (ms )
f
4 t (s)
q mg
N
ma (Pseudo
force)
F
N sin q
A M
a F = mg cos q
ma sin q
q
B
f
1
f
2
f
1 A
T(pull) = 125 N
f m m g
A B 2 2
(max) ( ) = + m
= + 0.2 ( ) 60 40 10
= 200 N
f T
1
125 125 250 + = + = N
As, f T f
1 2
+ > (max. block B /along the A
as A is stationary) will move towards right
with acceleration.
Option (a) is correct.
a
f T f
m m
B
B A
=
+ -
+
( ) (max.)
1 2
=
-
+
250 200
40 60
=0.5 m/s
2

Option (b) is correct.
6. (See solution to Question no. 4).
N mg ma = - cos sin q q
Option (c) is correct and option (d) is
incorrect.
As angle between the directions of a and
gsinq will be less than 90°, acceleration of
block A will be more than g sin q.
Option (a) is correct and option (b) is
incorrect.
7. Maximum value of friction.
f
1
= between A and B
= ´ ´ = 0.25 3 10 7.5 N
f
2
= between B and C
= ´ ´ = 0.25 17.5 7 10 N
f
3
= between C and ground
= ´ ´ = 0.25 7.5 15 10 3  N
(a)
T = + = 17.5 7.5 25 N
F T = + + = 37.5 17.5 80 N
(c) T a - - = 7.5 17.5 4 …(i)
F T a - - - = 37.5 17.5 8 …(ii)
F = 200 N …(iii)
Solving these equations we get,
a = 10 m/s
2
8. Maximum value of friction available to
block is less than the maximum value of
friction available to man.
9. N ma sin q = = ´ = 1 5 5 …(i)
N mg cos q = = ´ = 1 10 10 …(ii)
Solving these two equations we get,
tanq=
1
2
and N = 5 5 N.
10. Let f
1
= friction between 2 kg and 4 kg
f
2
= friction between 4 kg and ground
( )
max
f
s1
2 10 8 = ´ ´ = 0.4 N
( ) f
k1
2 10 4 = ´ ´ = 0.2 N
( ) .
max
f
s2
06 6 10 36 = ´ ´ = N
F
k2
0 4 6 10 24 = ´ ´ = . N
(b) At t = 1 s, F = 2 N < ( )
max
f
s2
\   Both the blocks are at rest.
\  f
1
0 =
(c) At t = 4 s, F = 8 N < ( )
max
f
s2
\ Both the blocks are at rest.
f F
2
8 = = N,
11. a = 0, T
1
10 = N,
T T
2 1
20 30 = + = N,
T
3
20 = N.
12. f
max
= ´ ´ = 0.3 2 10 6 N
(a) At t = 2 s, F N = 2
\   f = 2 N
(b) At t = 8 s, F = 8 N > 6 N
\   f = 6 N
104 | Mechanics-1
A
m
a
ma
q
a
A
a > g sin q
net
T
17.5 N
37.5 N
7.5 N
17.5 N
B F T C
Page 5

53. Minimum force required to start the
motion upward
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
4
3 3
3
2
=
7
6
mg
54. Minimum force required to move the block
up the incline with constant speed
= + mg mg
k
sin cos q m q
= + ×
é
ë
ê
ù
û
ú
mg
1
2
2
3 3
3
2
=
5
6
mg
55. S
1
2
2
8
=
×
æ
è
ç
ö
ø
÷
=
(5.22)
9.8
14.3 m
Option (c) is correct.
56.    a
g g
¢ =
- 1200 1000
1200
=
g
6
a
g g
=
- 1350 1200
1200
=
g
8
For accelerated motion
v as
max
2 2
1
0 2 = +
Þ s
v
a
1
2
2
=
max
For retarted motion
0 2
2 2
2
= - ¢ v a s
max
Þ s
v
a
2
2
2
=
¢
max
s s
v
a a
1 2
2
2
1 1
+ = +
¢
é
ë
ê
ù
û
ú
max
25
2
8 6
2
= +
é
ë
ê
ù
û
ú
v
g g
max
v
max
=
´ 50
14
9.8
=
-
5.92ms
1
Option (c) is correct.
57. mg sin q = ´ ´ 170 10
8
15
= 906.67 N
f
1
170 10
15
17
(max) = ´ ´ ´ 0.2 = 300 N
f
2
170 10
15
17
(max) = ´ ´ ´ 0.4
= 600 N
The whole system will accelerate as
mgsinq is greater than both f
1
(max) and
f
2
(max).
Total force of friction
= + f f
1 2
(max) (max)
=900 N
Option (a) is correct.
58. mg T ma sin q - - = 300            …(i)
and mg T ma sinq + = …(ii)
Substituting Eq. (i) by Eq. (ii),
2 300 0 T + =
T = - 150 N
= 150 N, compressive.
Option (a) is correct.
Laws of Motion | 101
a
25 m
T' = 1000g
– a'
1200 g
1200 g
T = 1350g
Speed
= v
max
Stops
q
mg sin q
A
B
m = 0.4
f
2
f
1
T
T
m = 0.2
mg sin q
q
8
17
15
More than One Cor rect Options
1. (a) Normal force between A and B m g =
2
= ´ = 1 10 10 N
\ Force of limiting friction by B on A (or by
A on B)
= ´ = m 10 2 N
Total force opposing applied external force
F T = + 2N
= + 2 2 N N
= 4 N
Thus, if F £ 4 N
The block A will remain stationary and so
block B also. The system will be in
equilibrium.
\ Option (a) is correct.
(b) If F > 4 N
F T - - = 2 1 a
and T a - = 2 1 …(i)
F a - = 4 2 …(ii)
i.e., F a = + 4 2
For F > 4 N
2 4 4 a + >
or a > 0
Þ T - > 2 0
i.e., T > 2 N
\ Option (b) is incorrect.
(c) Block A will move over B only when
F > 4 N and then the frictional force
between the blocks will be 2 N if a is just 0
[as explained in (b)].
Option (c) is correct.
(d) If F = 6 N using Eq. (ii)
2 6 4 a = -
Þ a = 1 m/s
2
\ Using Eq. (i),
T - = 2 1
i.e., T = 3 N
Option (d) is correct.
2. At point A
T T mg
1 2
cos cos a b = + …(i)
and T T
1 2
sin sin a b = …(ii)
At point B
T mg
2
cos b = …(iii)
T F mg
2
sin b = - …(iv)
Using Eq. (iii) in Eq. (i),
T T
1 2
2 cos cos a b = …(v)
Dividing Eq. (ii) by Eq. (v),
2 tan tan a b = …(vi)
Option (a) is correct.
Squaring and adding Eqs. (iii) and (v),
T T T
1
2
2
2 2
2
2 2
4 = + cos sin b b …(vii)
Dividing Eq. (iii) by Eq. (iv)
tan b = 1
\ cos sin b b = =
1
2
Substituting the values of sinb and cosb
in Eq. (vii)
T T T
1
2
2
2
2
2
4
1
2
1
2
=
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷
=
5
2
2
2
T
102 | Mechanics-1
T T
T T
2 N
F
A
B
2 N
F = mg
B
T
2
T
2
b
A
T
1
a
T
1
mg
b
1
1
Ö2
Þ 2 5
1 2
T T =
Option (c) is correct.
3. Displacement of block in 4 s
S = Area under curve
= 16 m.
DK =Workdone by frictional force
1
2
1 4 1 10 16
2
´ ´ = ´ ´ ´ m
Þ m = 0.1
Option (a) is correct.
Option (b) is incorrect.
Acceleration,   a = f tan
= - tan ( ) p q
= - tan q
= - 1 m/s
2
If half rough retardation = 0.5 m/s
2
\ 16 4
1
2
2
= + - t t ( ) 0.5
i.e., t t
2
16 64 0 - + =
or t = 8 s
Option (d) is correct.
Option (c) is incorrect.
4. Let acceleration of wedge ( ) A a =
N ma mg + = sin cos q q
N mg ma = - cos sin q q
Acceleration of
a
N
M
=
sin q
or Ma mg ma = - ( cos sin ) sin q q q
or a M m mg ( sin ) cos sin + =
2
q q q
i.e.,     a
mg
M m
=
+
cos sin
sin
q q
q
2
=
´ ´ ° °
+ ´ °
0.6
1.7 0.6
g cos sin
( sin )
45 45
45
2
=
+
3
17 3
g
=
3
20
g
Let a
B
= Acceleration of block B
Net force on B (along inclined plane)
ma ma mg
B
= + cos sin q q
Þ a a g
B
= + cos sin q q
Thus, ( ) ( cos sin ) cos a a g
B V
= + q q q
= + a g cos sin cos
2
q q q
= + ( ) a g
1
2
= +
æ
è
ç
ö
ø
÷
3
20
1
2
g
g
=
23
40
g

( ) ( cos sin ) sin a a g a
B H
= + - q q q
= -
23
40
3
20
g g

=
17
40
g

5. f m g
A 1 1
(max.) = m
= ´ ´ 0.3 60 10
= 180 N
F
net
on B = + f T
1
(max.)
= + 180 125
= 305 N
A will remain stationary as
T f <
1
(max.)
\ f
1
125 = N
Force of friction acting between A and
B=125 N
\ Options (c) and (d) are incorrect.
Laws of Motion | 103
O
q
–1
v (ms )
f
4 t (s)
q mg
N
ma (Pseudo
force)
F
N sin q
A M
a F = mg cos q
ma sin q
q
B
f
1
f
2
f
1 A
T(pull) = 125 N
f m m g
A B 2 2
(max) ( ) = + m
= + 0.2 ( ) 60 40 10
= 200 N
f T
1
125 125 250 + = + = N
As, f T f
1 2
+ > (max. block B /along the A
as A is stationary) will move towards right
with acceleration.
Option (a) is correct.
a
f T f
m m
B
B A
=
+ -
+
( ) (max.)
1 2
=
-
+
250 200
40 60
=0.5 m/s
2

Option (b) is correct.
6. (See solution to Question no. 4).
N mg ma = - cos sin q q
Option (c) is correct and option (d) is
incorrect.
As angle between the directions of a and
gsinq will be less than 90°, acceleration of
block A will be more than g sin q.
Option (a) is correct and option (b) is
incorrect.
7. Maximum value of friction.
f
1
= between A and B
= ´ ´ = 0.25 3 10 7.5 N
f
2
= between B and C
= ´ ´ = 0.25 17.5 7 10 N
f
3
= between C and ground
= ´ ´ = 0.25 7.5 15 10 3  N
(a)
T = + = 17.5 7.5 25 N
F T = + + = 37.5 17.5 80 N
(c) T a - - = 7.5 17.5 4 …(i)
F T a - - - = 37.5 17.5 8 …(ii)
F = 200 N …(iii)
Solving these equations we get,
a = 10 m/s
2
8. Maximum value of friction available to
block is less than the maximum value of
friction available to man.
9. N ma sin q = = ´ = 1 5 5 …(i)
N mg cos q = = ´ = 1 10 10 …(ii)
Solving these two equations we get,
tanq=
1
2
and N = 5 5 N.
10. Let f
1
= friction between 2 kg and 4 kg
f
2
= friction between 4 kg and ground
( )
max
f
s1
2 10 8 = ´ ´ = 0.4 N
( ) f
k1
2 10 4 = ´ ´ = 0.2 N
( ) .
max
f
s2
06 6 10 36 = ´ ´ = N
F
k2
0 4 6 10 24 = ´ ´ = . N
(b) At t = 1 s, F = 2 N < ( )
max
f
s2
\   Both the blocks are at rest.
\  f
1
0 =
(c) At t = 4 s, F = 8 N < ( )
max
f
s2
\ Both the blocks are at rest.
f F
2
8 = = N,
11. a = 0, T
1
10 = N,
T T
2 1
20 30 = + = N,
T
3
20 = N.
12. f
max
= ´ ´ = 0.3 2 10 6 N
(a) At t = 2 s, F N = 2
\   f = 2 N
(b) At t = 8 s, F = 8 N > 6 N
\   f = 6 N
104 | Mechanics-1
A
m
a
ma
q
a
A
a > g sin q
net
T
17.5 N
37.5 N
7.5 N
17.5 N
B F T C
(c) At t = 10 s, F = 10 N and f = 6 N
\    a =
-
=
10 6
2
2 m/s
2
(d) Block will start at 6 s. After that,
net impulse
= ´ ´ + + ´ - ´
1
2
4 6 10 2 10 6 6 ( )
= 16 N-s = mv
\                v = =
16
2
8 m/s.
13. f
max
= ´ ´ = 0.4 2 10 8 N
(b) At t = 3 s, F = 6 N
\ Common acceleration
a = =
6
6
1 m/s
2
\ Pseudo force on 2 kg
= ´ = 2 1 2  N (backward)
14. N Mg F = - sin q
F N Mg F cos ( sin ) q = m m q = -
\                F
Mg
=
+
m
q m q cos sin
For F to be minimum,
dF
dq
= 0
Match the Columns
1. Acceleration after t = 4 s
At t = 4 s, F = 8 N
\       F
max
= 8
i.e.,      m
s
mg = 8
Þ m
s
mg
= =
´
=
8 8
2 10
0.4
\ (a) ® (r)
At t = 4 s, a =
-
1
2
ms
t = 4 s, F = 8 N
F N ma
k
- = m
i.e.,   m
k
F ma
N
=
-
=
- F ma
mg
=
- ´
´
8 2 1
2 10
( )
= 0.3
(b) = (q)
At t = 01 . s,  F = 0 2 . N
\ Force of friction (at t = 01 . s) = 0.2 N
\      (c) ® (p)
At t = 8 s, F = 16 N
\ a
F mg
m
k
=
- m
=
- ´ ´ 16 2 10
2
( ) 0.3
= 5
i.e.,
a
10
= 0.5
\ (d) ® (s).
2. At q = ° 0 , dragging force = 0
\     Force of force = 0
\  (a) ® (s)
At q = ° 90
Normal force on block by plane will be
zero.
\     Force of friction = 0
\    (b) ® (s)
At q = ° 30
Angle of repose =
-
tan
1
m
=
-
tan ( )
1
1  = ° 45
As q < angle of repose, the block will not
slip and thus,
force of friction = mg sin q
= ´ ´ ° 2 10 30 sin = 10 N
\  (c) ® (p)
At q = ° 60
As q > angle of repose
Block will accelerate and thus force of
friction =m N
= ´ ´ ´ ° 1 2 10 60 cos
= 10 N
\ (d) ® (p).
3. All contact forces (e.g., force of friction and
normal reaction) are electromagnetic in
nature.
\ (a) ® (q), (r)
(b) ® (q), (r).
Laws of Motion | 105
```

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## DC Pandey Solutions for NEET Physics

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