Laws of Motion- 2 Practice Questions - DPP for NEET

 Page 1


(1) (c) Force on the block
= Mass of the block × acceleration of the system
= M × 
P
Mm +
(2) ( b) Mass of the rope = 8 × 
1
2
 = 4 kg
Total mass = 50 + 4 = 54 kg
\ a = 
F 108
m 54
= = 2 m/s
2
Force utilised in pulling the rope =4×2 = 8 N
Force applied on mass = 108 – 8 = 100 N
(3) ( b) Mass of the rope = 15 × 2  = 30 kg
acceleration = 
F 255
m306
== m/s
2
At the point 7 m away from point of application the
mass of first part of rope = 14 kg
\ Force used in pulling 14 kg  = 14 × 
5
6
 = 11.67 N
The remaining force  = (25 –11.67)N = 13.33 N
(4) ( b) The various forces acting are shown in fig.
The force of 100N has
(i) horizontal component of 100 cos 30º = 503 N
and (ii) A vertical component = 100 sin 30º = 50N
50N
W=10×10N
50 3N
R
30°
100N
Since the block is always in contact with the table, the
net vertical force
R = mg + F sin q = (10 × 10 + 50) N = 150N
When the block moves along the table, work is done
by the horizontal component of the force. Since the
distance moves is 10 m, the work done is
503 × 10 = 500 3 Joule.
If v is the speed acquired by the block, the work done
must be equal to the kinetic energy of the block.
Therefore, we have
500 3 = 
1
2
 × 10 × v
2
 Þ v
2
 = 
100 3
 Þ  v = 13.17 m/sec
(5) ( a) We have acceleration
a = 
Fcos 50 3
m 10
q
= = 53 m /sec
2
The velocity after 2 sec,  v = u + at
Þ v = 0 + 53 × 2 = 103 m/sec
(6) (a) All the forces acting on the two blocks are shown in
fig. As the blocks are rigid under the action of a force F,
both will move together with same acceleration.
R
m
R
1
A
F
M
R
1
mg Mg
a = F/(m+M) = 3/(1+2) = 1 m/s
2
Now as the mass of larger block is m and its acceleration
a so force of contact i.e. action on it.
f = Ma = 
MF 23
M m 21
´
=
++
 = 2N
(7) (a) As the same force is applied to the combined
mass, we have
12
1 11
a aa
=+
or a = 
12
12
aa
aa +
 = 
5 15
5 15
´
+
 = 3.75 m/s
2
(8) (a) As net force on the rod = F
1
 – F
2
 and its mass is M so
acceleration of the rod will be
a = (F
1
 – F
2
)/M ...(i)
Now considering the motion of part AB of the rod,
which has mass (M/L)y ,
Acceleration a given by
(i) Assuming that tension at B is T
F
1
 – T = 
M
L
 y × a (from F = ma)
Þ F
1
 – T = 
12
FF M
y
LM
-
(using eq. (1))
Þ T  = 
12
yy
F1F
LL
æ ö æö
-+
ç ÷ ç÷
è ø èø
(9) ( b) The net acceleration of the system is given by
12
1 11
a aa
=+
12n
1111
......
aaaa
= ++ = 1 + 2 + 3 + ......... + n
= 
n
2
 [2 + (n – 1) 1]  = 
n
2
 [n + 1] = 
2
n(n 1) +
Page 2


(1) (c) Force on the block
= Mass of the block × acceleration of the system
= M × 
P
Mm +
(2) ( b) Mass of the rope = 8 × 
1
2
 = 4 kg
Total mass = 50 + 4 = 54 kg
\ a = 
F 108
m 54
= = 2 m/s
2
Force utilised in pulling the rope =4×2 = 8 N
Force applied on mass = 108 – 8 = 100 N
(3) ( b) Mass of the rope = 15 × 2  = 30 kg
acceleration = 
F 255
m306
== m/s
2
At the point 7 m away from point of application the
mass of first part of rope = 14 kg
\ Force used in pulling 14 kg  = 14 × 
5
6
 = 11.67 N
The remaining force  = (25 –11.67)N = 13.33 N
(4) ( b) The various forces acting are shown in fig.
The force of 100N has
(i) horizontal component of 100 cos 30º = 503 N
and (ii) A vertical component = 100 sin 30º = 50N
50N
W=10×10N
50 3N
R
30°
100N
Since the block is always in contact with the table, the
net vertical force
R = mg + F sin q = (10 × 10 + 50) N = 150N
When the block moves along the table, work is done
by the horizontal component of the force. Since the
distance moves is 10 m, the work done is
503 × 10 = 500 3 Joule.
If v is the speed acquired by the block, the work done
must be equal to the kinetic energy of the block.
Therefore, we have
500 3 = 
1
2
 × 10 × v
2
 Þ v
2
 = 
100 3
 Þ  v = 13.17 m/sec
(5) ( a) We have acceleration
a = 
Fcos 50 3
m 10
q
= = 53 m /sec
2
The velocity after 2 sec,  v = u + at
Þ v = 0 + 53 × 2 = 103 m/sec
(6) (a) All the forces acting on the two blocks are shown in
fig. As the blocks are rigid under the action of a force F,
both will move together with same acceleration.
R
m
R
1
A
F
M
R
1
mg Mg
a = F/(m+M) = 3/(1+2) = 1 m/s
2
Now as the mass of larger block is m and its acceleration
a so force of contact i.e. action on it.
f = Ma = 
MF 23
M m 21
´
=
++
 = 2N
(7) (a) As the same force is applied to the combined
mass, we have
12
1 11
a aa
=+
or a = 
12
12
aa
aa +
 = 
5 15
5 15
´
+
 = 3.75 m/s
2
(8) (a) As net force on the rod = F
1
 – F
2
 and its mass is M so
acceleration of the rod will be
a = (F
1
 – F
2
)/M ...(i)
Now considering the motion of part AB of the rod,
which has mass (M/L)y ,
Acceleration a given by
(i) Assuming that tension at B is T
F
1
 – T = 
M
L
 y × a (from F = ma)
Þ F
1
 – T = 
12
FF M
y
LM
-
(using eq. (1))
Þ T  = 
12
yy
F1F
LL
æ ö æö
-+
ç ÷ ç÷
è ø èø
(9) ( b) The net acceleration of the system is given by
12
1 11
a aa
=+
12n
1111
......
aaaa
= ++ = 1 + 2 + 3 + ......... + n
= 
n
2
 [2 + (n – 1) 1]  = 
n
2
 [n + 1] = 
2
n(n 1) +
DPP/ P 10
29
(10) (a) As the mass of the system is 6 + 4 + 2 = 12 kg and
applied force is 60 N, the acceleration of the system
F 60
a
m 12
== = 5 m/s
2
Now at point A as tension in pulling the rope of mass
2kg and block Q of mass 4kg.
T
A
 = (2 + 4) × 5 = 30N
Similarly for B and C,
T
B
= (1 + 4) × 5 = 25N
andT
C
 = (0 + 4) × 5 = 20N
(11) (a) In case (a), the pulling force = 2mg – mg = mg
and the mass is 2m + m = 3m
so acceleration a = mg/3m = g/3
While in case (b), the pulling force = 2mg – mg = mg
but, the mass in motion = m + 0 = m
Acceleration,  a = mg/m = g
(12) (c) It this problem as the pulling force is 2mg while
opposing force is mg, so net force
F= 2mg – mg = mg,
and as the mass in motion = m + m + m = 3m
So the acceleration = 
force mg g
mass 3m3
==
Now as A is accelerated up while B and C down. so
tension T
1
, is such that  mg < T
1
 < 2mg
Actually for the motion of A,
T
1
 = m (g + a) = m(g + g/3) = 
4
3
mg
Now to calculate tension in the string BC we consider
the downward motion of C,
i.e. T
2
 = m (g – a) = m (g – g/3) = (2/3) mg
(13) (a) As pulley Q is not fixed so if it moves a distance d the
length of string between P and Q will changes by
2d (d from above and d from below) i.e. M will move 2d.
This in turn implies that if a (®2d) is the acceleration of
M, the acceleration of Q and so 2M will be of  (a/2)
Now if we consider the motion of mass M, it is
accelerated down so   T =  M(g – a) ...(1)
And for the motion of Q,
2T – T' = 0 × (a/2) = 0 Þ T' = 2T ...(2)
And for the motion of mass 2M,
T' = 2M (a/2)  Þ T' = Ma ...(3)
From equation (2) and (3) T = 
1
2
 Ma, so eq. (1) reduces
1
2
æö
ç÷
èø
 Ma = M (g – a)  Þ a = 
2
3
g
(14) (a) The tension is same in two segments
For B the equation is
(40  × 9.8 – T) = 40a    ...(1)
For C the equation is
(T – 50 × 9.8 × 
1
2
) = 50a      ...(2)
From equation (1) and (2) a = 1.63 m/s
2
distance of fall
S = 
1
2
at
2
 = 
1
2
 × 1.63 × 4
2
 = 13.04 m
(15) (b) The string is massless and inextensible the tension T is
same. Let mass B move down the inclined plane.
For B the equation of motion  m
1
g sin q – T  = m
1
a
  30 × 9.8 × sin 53º – T = 30a
Þ 235.2 – T = 30 a ...(1)
and for A the equation of motion
T – 20 × 9.8 × sin 37º = 20a
T – 117.6 = 20a ...(2)
From (1) & (2) T = 164.64 N
(16) (c)
a
mg
a ma
m
T
(Force diagram in the frame of the car)
Applying Newton’s law perpendicular to string
mg sin ma cos q=q
a
tan
g
q=
Applying Newton’s law along string
Þ
22
T m g a ma - +=
or
22
T m g a ma = ++
(17) (a) As A moves up and B moves down with acceleration a
for the motion of A ,
T – 11 g = 11 a ... (i)
for the motion of B,
11.5 g – T = 11.5 a  ...(ii)
From (i) & (ii) ,
a = 
12
12
mm
mm
-
+
g = 
(11.5 11)9.8
11.5 11
-
+
 = 0.218 m/sec
2
Assuming that the particles are initially at rest, their
velocity at the end of 4 sec will be
v = u + at = 0 + 0.218 × 4 = 0.872 m/s
(18) (a) The height ascended by A in 4 sec
h = ut + 
1
2
at
2
 = 0 + 
1
2
 (0.218) 4
2
 = 1.744 m
This is also the height descended by B in that time.
(19) (c) At the end of 4 sec the string is cut. Now A and B are no
longer connected bodies but become free ones, falling
under gravity .
V elocity of A, when the string was cut
= 0.872 m/s upwards.
Acceleration a = – g (acting downwards),
displacement from this position in the subsequent 2 sec
Page 3


(1) (c) Force on the block
= Mass of the block × acceleration of the system
= M × 
P
Mm +
(2) ( b) Mass of the rope = 8 × 
1
2
 = 4 kg
Total mass = 50 + 4 = 54 kg
\ a = 
F 108
m 54
= = 2 m/s
2
Force utilised in pulling the rope =4×2 = 8 N
Force applied on mass = 108 – 8 = 100 N
(3) ( b) Mass of the rope = 15 × 2  = 30 kg
acceleration = 
F 255
m306
== m/s
2
At the point 7 m away from point of application the
mass of first part of rope = 14 kg
\ Force used in pulling 14 kg  = 14 × 
5
6
 = 11.67 N
The remaining force  = (25 –11.67)N = 13.33 N
(4) ( b) The various forces acting are shown in fig.
The force of 100N has
(i) horizontal component of 100 cos 30º = 503 N
and (ii) A vertical component = 100 sin 30º = 50N
50N
W=10×10N
50 3N
R
30°
100N
Since the block is always in contact with the table, the
net vertical force
R = mg + F sin q = (10 × 10 + 50) N = 150N
When the block moves along the table, work is done
by the horizontal component of the force. Since the
distance moves is 10 m, the work done is
503 × 10 = 500 3 Joule.
If v is the speed acquired by the block, the work done
must be equal to the kinetic energy of the block.
Therefore, we have
500 3 = 
1
2
 × 10 × v
2
 Þ v
2
 = 
100 3
 Þ  v = 13.17 m/sec
(5) ( a) We have acceleration
a = 
Fcos 50 3
m 10
q
= = 53 m /sec
2
The velocity after 2 sec,  v = u + at
Þ v = 0 + 53 × 2 = 103 m/sec
(6) (a) All the forces acting on the two blocks are shown in
fig. As the blocks are rigid under the action of a force F,
both will move together with same acceleration.
R
m
R
1
A
F
M
R
1
mg Mg
a = F/(m+M) = 3/(1+2) = 1 m/s
2
Now as the mass of larger block is m and its acceleration
a so force of contact i.e. action on it.
f = Ma = 
MF 23
M m 21
´
=
++
 = 2N
(7) (a) As the same force is applied to the combined
mass, we have
12
1 11
a aa
=+
or a = 
12
12
aa
aa +
 = 
5 15
5 15
´
+
 = 3.75 m/s
2
(8) (a) As net force on the rod = F
1
 – F
2
 and its mass is M so
acceleration of the rod will be
a = (F
1
 – F
2
)/M ...(i)
Now considering the motion of part AB of the rod,
which has mass (M/L)y ,
Acceleration a given by
(i) Assuming that tension at B is T
F
1
 – T = 
M
L
 y × a (from F = ma)
Þ F
1
 – T = 
12
FF M
y
LM
-
(using eq. (1))
Þ T  = 
12
yy
F1F
LL
æ ö æö
-+
ç ÷ ç÷
è ø èø
(9) ( b) The net acceleration of the system is given by
12
1 11
a aa
=+
12n
1111
......
aaaa
= ++ = 1 + 2 + 3 + ......... + n
= 
n
2
 [2 + (n – 1) 1]  = 
n
2
 [n + 1] = 
2
n(n 1) +
DPP/ P 10
29
(10) (a) As the mass of the system is 6 + 4 + 2 = 12 kg and
applied force is 60 N, the acceleration of the system
F 60
a
m 12
== = 5 m/s
2
Now at point A as tension in pulling the rope of mass
2kg and block Q of mass 4kg.
T
A
 = (2 + 4) × 5 = 30N
Similarly for B and C,
T
B
= (1 + 4) × 5 = 25N
andT
C
 = (0 + 4) × 5 = 20N
(11) (a) In case (a), the pulling force = 2mg – mg = mg
and the mass is 2m + m = 3m
so acceleration a = mg/3m = g/3
While in case (b), the pulling force = 2mg – mg = mg
but, the mass in motion = m + 0 = m
Acceleration,  a = mg/m = g
(12) (c) It this problem as the pulling force is 2mg while
opposing force is mg, so net force
F= 2mg – mg = mg,
and as the mass in motion = m + m + m = 3m
So the acceleration = 
force mg g
mass 3m3
==
Now as A is accelerated up while B and C down. so
tension T
1
, is such that  mg < T
1
 < 2mg
Actually for the motion of A,
T
1
 = m (g + a) = m(g + g/3) = 
4
3
mg
Now to calculate tension in the string BC we consider
the downward motion of C,
i.e. T
2
 = m (g – a) = m (g – g/3) = (2/3) mg
(13) (a) As pulley Q is not fixed so if it moves a distance d the
length of string between P and Q will changes by
2d (d from above and d from below) i.e. M will move 2d.
This in turn implies that if a (®2d) is the acceleration of
M, the acceleration of Q and so 2M will be of  (a/2)
Now if we consider the motion of mass M, it is
accelerated down so   T =  M(g – a) ...(1)
And for the motion of Q,
2T – T' = 0 × (a/2) = 0 Þ T' = 2T ...(2)
And for the motion of mass 2M,
T' = 2M (a/2)  Þ T' = Ma ...(3)
From equation (2) and (3) T = 
1
2
 Ma, so eq. (1) reduces
1
2
æö
ç÷
èø
 Ma = M (g – a)  Þ a = 
2
3
g
(14) (a) The tension is same in two segments
For B the equation is
(40  × 9.8 – T) = 40a    ...(1)
For C the equation is
(T – 50 × 9.8 × 
1
2
) = 50a      ...(2)
From equation (1) and (2) a = 1.63 m/s
2
distance of fall
S = 
1
2
at
2
 = 
1
2
 × 1.63 × 4
2
 = 13.04 m
(15) (b) The string is massless and inextensible the tension T is
same. Let mass B move down the inclined plane.
For B the equation of motion  m
1
g sin q – T  = m
1
a
  30 × 9.8 × sin 53º – T = 30a
Þ 235.2 – T = 30 a ...(1)
and for A the equation of motion
T – 20 × 9.8 × sin 37º = 20a
T – 117.6 = 20a ...(2)
From (1) & (2) T = 164.64 N
(16) (c)
a
mg
a ma
m
T
(Force diagram in the frame of the car)
Applying Newton’s law perpendicular to string
mg sin ma cos q=q
a
tan
g
q=
Applying Newton’s law along string
Þ
22
T m g a ma - +=
or
22
T m g a ma = ++
(17) (a) As A moves up and B moves down with acceleration a
for the motion of A ,
T – 11 g = 11 a ... (i)
for the motion of B,
11.5 g – T = 11.5 a  ...(ii)
From (i) & (ii) ,
a = 
12
12
mm
mm
-
+
g = 
(11.5 11)9.8
11.5 11
-
+
 = 0.218 m/sec
2
Assuming that the particles are initially at rest, their
velocity at the end of 4 sec will be
v = u + at = 0 + 0.218 × 4 = 0.872 m/s
(18) (a) The height ascended by A in 4 sec
h = ut + 
1
2
at
2
 = 0 + 
1
2
 (0.218) 4
2
 = 1.744 m
This is also the height descended by B in that time.
(19) (c) At the end of 4 sec the string is cut. Now A and B are no
longer connected bodies but become free ones, falling
under gravity .
V elocity of A, when the string was cut
= 0.872 m/s upwards.
Acceleration a = – g (acting downwards),
displacement from this position in the subsequent 2 sec
30
DPP/ P 10
(2) If the downward acceleration of the lift is a = g, then the
body will enjoy weightlessness.
(3) If the downward acceleration of the body is a > g, then
the body will rise up to the ceiling of lift
(25) (d), (26) (d), (27) (a).
Initial elongation = 2R cos 30° = 3R
30°
F
N
60°
mg
Extension in the spring is
x = AB – R = 2R cos 30° – R = ( 3 1)R -
\ Spring force
F = kx = 
( 3 1) mg
( 3 1) R 2mg
R
-
-=
FBD of bead is
N = F (mg cos 30°) = (2mg + mg) 
3 33
mg
22
=
Tangential force  F
1
 = F sin 30° – mg sin 30°
              = (2mg – mg) sin 30° = 
mg
2
\   tangential acceleration = g/2
28. ( d) Here the acceleration of both will be same, but their
masses are different. Hence, the net force acting on
each of them will not be same.
29. (c) The FBD of block A in Figure is
mg
N
The force exerted by B on A is N (normal reaction). The
force acting on A are N (horizontal) and mg (weight
downwards). Hence statement I is false.
30. (d) T – m
1
g = m
1
a – .... (1)
m
2
g – T = m
2
a – .... (2)
Solving (1) and (2), 
12
1
12
2mm
Tg
mm
æö
=
ç÷
+ èø
 
21
12
mm
ag
mm
æö -
=
ç÷
+ èø
h = ut + 
1
2
at
2
 = (0.872) × 2 + 
1
2
 (–9.8) 2
2
   = 1.744 – 4.9 × 4 = –17.856m
A descends down by a distance of 17.856 m from the
position it occupied at the end of 4 sec from its start. B
has a free fall. Its position is given by
So the acceleration of mass M is (2/3)g
while tension in the string PQ will be
 T = M(g – (2/3)g) = (1/3)Mg
The force exerted by clamp on the pulley
=  
22
2
TT
3
+= Mg
(20) (a) Here the system behaves as a rigid system, therefore
every part of the system will move with same accelera-
tion. Thus applying newton’s law
mg – T = ma ......... (i)
2T – mg = ma ......... (ii)
Doubling the first equation and adding
mg = 3ma  or acceleration a = 
1
g
3
(21) (c) Tension in the string
T = m (g – a) = m 
g
g
3
æö
-
ç÷
èø
2
T mg
3
=
(22) (a)
(1) Inertia µ mass
(2) 1 Newton = 10
5
 dyne
(3) Thrust on rocket 
F
r
 = 
M
v Mg
t
D
-
D
r r
(4) Apparent weight of a body in the lift accelerated up is
W = m (g + a).
(23) (b)
(1) If a
1
, a
2
, ... a
n
 be the accelerations produced in n
different bodies on applying the same force, the
acceleration produced in their combination due to the
same force will be 
12n
1111
......
aaaa
= ++
(2) Newton's I
st
 and III
rd
 law can be derived from second
law therefore II
nd
 law is the most fundamental law out
of the three law .
(24) (a)
(1) For equilibrium of a body under the action of concurrent
forces 
® ® ® ®
+ + +
n 3 2 1
F ..... F F F
= 0