Laws of Motion- 1 Practice Questions - DPP for NEET

 Page 1


(1) (a) Force causing the acceleration = 400 – 200 = 200N
mass of the boy =  200/9.8
hence acceleration = F/m = 
200
200
 × 9.8 = 9.8 m/s
2
(2) (a) Acceleration = 
F
m
r
= 
ˆˆ
6i 8j
10
+
 in the direction of force
and displacement
2
1
S ut at
2
=+
r
r r
 = 0 + 
1
2
 
ˆˆ
6i 8j
10
æö
+
ç÷
èø
100  = 30i
ˆ
 + 40 j
ˆ
So the displacement is 50 m along tan
–1 
4
3
 with x-axis
(3) (a) From the law of conservation of momentum
1000 × 5 + 0 = (1000 + 60) v
Þ v = 
1060
5 1000 ´
 = 4.71 m/s
(4) (b) Weight of disc = 
10
1000
kg ,
Let speed of the bullet = v
So rate of change of momentum of the bullets
= 
2 105
1000
´´
v = applied force on the disc
Now 
2 105
1000
´´
× v = 
10g
1000
´
 Þ v = 0.98 m/s
2
 = 98 cm/s
2
(5) (c) Total mass = 80 + 40 = 120 kg
The rope cannot with stand this load so the fire man should
slide down the rope with some acceleration
\ The maximum tension = 100 × 9.8 N
m (g – a) = tension ,
120 (9.8 – a) = 100 × 9.8 Þ a = 1.63 m/s
2
(6) (b) Suppose the velocity of the body at the instant when it
reaches the pile of sand be v. Then
v
2
 = 0 + 2 (9.8) × (5 metre) = 98(Q v
2
 = u
2
 + 2as)
a = – 
98
2 (0.05) ´
 = – 980 m/sec
2
Now , retarding force
F = mass × acceleration= 0.02 kg × (– 980 m/sec
2
) = –19.6 N
(7) (b) Impulse = F . t = Area under F-t curve from 4 ms to 16
ms = Area under BCDFB
= Area of trapizium BCEF + area of DCDE
= 
1
2
(200+800) (2×10
–6
) + 
1
2
×10 × 10
–6
 × 800
= 10 × 10
–4 
+ 40 × 10
–4
 N–s = 50 × 10
–4
= 5.0 × 10
–3
 N-s
(8) (a) (a) The elevator having an initial upward speed of 8 m/
sec is brought to rest within a distance of 16 m
Hence, 0 = (8)
2
 + 2a (16) (Q v
2
 = u
2
 + 2as),
a = 
88
2 16
´
-
´
 = – 2 m/sec
2
Resultant upward force on elevator = T – mg. According to
Newton's law.
    T – mg = ma
or  T = mg + ma = m (g + a) = 1000 (9.8 – 2) = 7800 N
(b) Let P be the upward force exerted on the man by the
elevator floor. If m' be the mass of the man, then, weight of
the man acting downward = m' g ,
Upward force on the man = P – m'g
According to Newton's law . P–m' g = m' a or
P = m' (a + g) = (– 2 + 9.8) = 624 N
(9) (d) As P and Q move down, the length l decreases at the
rate of U m/s
A
B
Q
M
q
q
P
b b
y
From figure, l
2  
= b
2
 + y
2
Differentiating with respect to time
2l 
d
dt
l
 = 2y 
dy
dt
(Q b is constant)
\ 
dyd
.
dt y dt
=
ll
 = 
1d
.
cos dt q
l
 = 
U
cosq
(10) (a) The engine, coach, coupling and resistance are, shown
in figure.
COACH ENGINE
DRIVING
FORCE
T
Driving force = 4500 N
Opposing force (Resistance)  = 
4
(5 4)10
100
+
 = 900 N
Resultant force = 4500 – 900 = 3600 N
Mass of engine and coach = 9 × 10
4 
kg
According to Newton's law , F = ma
\  3600 = 9 × 10
4
 a
or a = (3600) / (9 × 10
4
) = 0.04 m/sec
2
So acceleration of the train = 0.04 m/sec
2
Page 2


(1) (a) Force causing the acceleration = 400 – 200 = 200N
mass of the boy =  200/9.8
hence acceleration = F/m = 
200
200
 × 9.8 = 9.8 m/s
2
(2) (a) Acceleration = 
F
m
r
= 
ˆˆ
6i 8j
10
+
 in the direction of force
and displacement
2
1
S ut at
2
=+
r
r r
 = 0 + 
1
2
 
ˆˆ
6i 8j
10
æö
+
ç÷
èø
100  = 30i
ˆ
 + 40 j
ˆ
So the displacement is 50 m along tan
–1 
4
3
 with x-axis
(3) (a) From the law of conservation of momentum
1000 × 5 + 0 = (1000 + 60) v
Þ v = 
1060
5 1000 ´
 = 4.71 m/s
(4) (b) Weight of disc = 
10
1000
kg ,
Let speed of the bullet = v
So rate of change of momentum of the bullets
= 
2 105
1000
´´
v = applied force on the disc
Now 
2 105
1000
´´
× v = 
10g
1000
´
 Þ v = 0.98 m/s
2
 = 98 cm/s
2
(5) (c) Total mass = 80 + 40 = 120 kg
The rope cannot with stand this load so the fire man should
slide down the rope with some acceleration
\ The maximum tension = 100 × 9.8 N
m (g – a) = tension ,
120 (9.8 – a) = 100 × 9.8 Þ a = 1.63 m/s
2
(6) (b) Suppose the velocity of the body at the instant when it
reaches the pile of sand be v. Then
v
2
 = 0 + 2 (9.8) × (5 metre) = 98(Q v
2
 = u
2
 + 2as)
a = – 
98
2 (0.05) ´
 = – 980 m/sec
2
Now , retarding force
F = mass × acceleration= 0.02 kg × (– 980 m/sec
2
) = –19.6 N
(7) (b) Impulse = F . t = Area under F-t curve from 4 ms to 16
ms = Area under BCDFB
= Area of trapizium BCEF + area of DCDE
= 
1
2
(200+800) (2×10
–6
) + 
1
2
×10 × 10
–6
 × 800
= 10 × 10
–4 
+ 40 × 10
–4
 N–s = 50 × 10
–4
= 5.0 × 10
–3
 N-s
(8) (a) (a) The elevator having an initial upward speed of 8 m/
sec is brought to rest within a distance of 16 m
Hence, 0 = (8)
2
 + 2a (16) (Q v
2
 = u
2
 + 2as),
a = 
88
2 16
´
-
´
 = – 2 m/sec
2
Resultant upward force on elevator = T – mg. According to
Newton's law.
    T – mg = ma
or  T = mg + ma = m (g + a) = 1000 (9.8 – 2) = 7800 N
(b) Let P be the upward force exerted on the man by the
elevator floor. If m' be the mass of the man, then, weight of
the man acting downward = m' g ,
Upward force on the man = P – m'g
According to Newton's law . P–m' g = m' a or
P = m' (a + g) = (– 2 + 9.8) = 624 N
(9) (d) As P and Q move down, the length l decreases at the
rate of U m/s
A
B
Q
M
q
q
P
b b
y
From figure, l
2  
= b
2
 + y
2
Differentiating with respect to time
2l 
d
dt
l
 = 2y 
dy
dt
(Q b is constant)
\ 
dyd
.
dt y dt
=
ll
 = 
1d
.
cos dt q
l
 = 
U
cosq
(10) (a) The engine, coach, coupling and resistance are, shown
in figure.
COACH ENGINE
DRIVING
FORCE
T
Driving force = 4500 N
Opposing force (Resistance)  = 
4
(5 4)10
100
+
 = 900 N
Resultant force = 4500 – 900 = 3600 N
Mass of engine and coach = 9 × 10
4 
kg
According to Newton's law , F = ma
\  3600 = 9 × 10
4
 a
or a = (3600) / (9 × 10
4
) = 0.04 m/sec
2
So acceleration of the train = 0.04 m/sec
2
26
DPP/ P 09
Now considering the equilibrium of the coach only, we
have (T – R) = 4 × 10
4
 × 0.04 (Q F = ma)
or T – 
4
4 10
100
´
= 4 × 10
4
 × 0.04,
T = 4 × 10
4
 × 0.04 + 4 × 10
2
 = 1600 + 400 = 2000 N
(11) (d) Given that  
1
F
®
= 
ˆˆ
(8i 10j) + and  
2
F
®
 = 
ˆˆ
(4i 8j) +
Then the total force 
F
r
 = 
ˆˆ
12i 18j +
So acceleration 
ˆˆ
F 12i 18j
a
m6
+
==
r
r
=
ˆˆ
2i 3j + m/sec
2
Net acceleration
| a
®
| = 
22
23 +
= 
49 +
= 
13
 m/sec
2
(12) (c) From the relation
F = ma  Þ a = 
F
m
 = 
1000
1000
 = 1 m/s
2
As the force is brake force, acceleration is – 1 m/s
2
 using
relation v
2
 = u
2
 + 2as, we obtain
2as = u
2
Þ s  =
2
u
2a
 =
2
5
18
18
2
æö
´
ç÷
èø
 = 12.5 m
(13) (a) The water jet striking the block at the rate of 1 kg/s at a
speed of 5 m/s will exert a force on the block
F = v 
dm
dt
  = 5 × 1 = 5 N
2 kg
a
And under the action of this force of 5 N, the block of mass
2 kg will move with an acceleration given by ,
F  = ma Þ  a = F/m = 5/2 = 2.5 m/s
2
(14) (a) Relative speed of the ball = (v + u)
Speed after rebouncing = – (v + u)
So,  F = m 
v
t
D
D
 = 
m[(v u) { (v u)}]
t
+ --+
        = 
2m(v u)
t
+
(15) (b) F = 
dt
dp
 Þ  F dt = dp = p
2
 – p
1
Þ  F × 1 = mnv – 0
Þ  F = mnv
(Total mass of the bullets fired in 1 sec = mn)
(16) (a) The initial momentum  = 15 × 10 = 150 kgm/s  and
Force = 
change in momentum
time
  = 
0 150
15
-
 = –10 N
A constant force of 10 N must be acting in opposite
direction to the motion of body.
(17) (a) The change in momentum in the final direction is equal
to the impulse = 
2.50
1000
× 28 – 
250
24
1000
æö
-´
ç÷
èø
 = 13 Ns
and  force = 
impulse
time
 = 
13
1/100
 = 1300 N
in the direction of the ball.
(18) (b). We know  
dp
F
dt
=
r
r
 Þ Fdt dp =
r
r
Þ  2 × 2 = dp
r
 Þ   4 = dp
r
Therefore change in momentum  = 4 Ns
(19) (a) We know  
dp
F
dt
=
r
r
Þ  Fdt dp =
r
r
 = 
2 1 21
p p mv mv - =-
rr rr
Þ 4 j
ˆ
. 1 = 2 .
2
v
r
 – 2(2i
ˆ
)
Þ 
2
2v
r
 
2
ˆˆ
4j.1 2.v 2 (2i) =-
r
= 
ˆˆ
4j 4i +
Þ 
2
v
r
  = 
ˆˆ
2i 2j +
Þ  
2
|v|
r
= 22 m/s
(20) (c) Initial  momentum of the ball
= 
150
1000
 × 12 = 1.8 kg.m/sec
Final momentum of the ball =–
150
1000
× 20= – 3.0 kg m/sec
Change in momentum = 4.8 kg m/sec
Average force exerted  = Impulse/ time = 
4.8
.01
 = 480 N
(21) (b) Initial momentum of the body = mu = 20 × 3 = 60
and final momentum of the body = – mu = –20 × 3 = – 60
The change in momentum of body in initial direction
  = – 60 – 60 = –120
The change in momemtum imparted to the body in opposite
direction = 120
\ The impulse imparted to the body = 120 Ns
(22) (a) (1) Since the lift is moving down with an acceleration of
3 m/sec
2
, then the inertial force F = ma, acts upwards on
the body
mg
R
F=ma
a=3 m/s
2
Now , R + F= mg
or  R = mg – F = mg – ma = m (g – a) = 60 (9.8 – 3) = 408 N
(2) When the lift is moving down with constant velocity
a = 0  and hence, R = mg = 60 × 9.8 = 588 N
(3) The lift is now moving down with a retardation of
Page 3


(1) (a) Force causing the acceleration = 400 – 200 = 200N
mass of the boy =  200/9.8
hence acceleration = F/m = 
200
200
 × 9.8 = 9.8 m/s
2
(2) (a) Acceleration = 
F
m
r
= 
ˆˆ
6i 8j
10
+
 in the direction of force
and displacement
2
1
S ut at
2
=+
r
r r
 = 0 + 
1
2
 
ˆˆ
6i 8j
10
æö
+
ç÷
èø
100  = 30i
ˆ
 + 40 j
ˆ
So the displacement is 50 m along tan
–1 
4
3
 with x-axis
(3) (a) From the law of conservation of momentum
1000 × 5 + 0 = (1000 + 60) v
Þ v = 
1060
5 1000 ´
 = 4.71 m/s
(4) (b) Weight of disc = 
10
1000
kg ,
Let speed of the bullet = v
So rate of change of momentum of the bullets
= 
2 105
1000
´´
v = applied force on the disc
Now 
2 105
1000
´´
× v = 
10g
1000
´
 Þ v = 0.98 m/s
2
 = 98 cm/s
2
(5) (c) Total mass = 80 + 40 = 120 kg
The rope cannot with stand this load so the fire man should
slide down the rope with some acceleration
\ The maximum tension = 100 × 9.8 N
m (g – a) = tension ,
120 (9.8 – a) = 100 × 9.8 Þ a = 1.63 m/s
2
(6) (b) Suppose the velocity of the body at the instant when it
reaches the pile of sand be v. Then
v
2
 = 0 + 2 (9.8) × (5 metre) = 98(Q v
2
 = u
2
 + 2as)
a = – 
98
2 (0.05) ´
 = – 980 m/sec
2
Now , retarding force
F = mass × acceleration= 0.02 kg × (– 980 m/sec
2
) = –19.6 N
(7) (b) Impulse = F . t = Area under F-t curve from 4 ms to 16
ms = Area under BCDFB
= Area of trapizium BCEF + area of DCDE
= 
1
2
(200+800) (2×10
–6
) + 
1
2
×10 × 10
–6
 × 800
= 10 × 10
–4 
+ 40 × 10
–4
 N–s = 50 × 10
–4
= 5.0 × 10
–3
 N-s
(8) (a) (a) The elevator having an initial upward speed of 8 m/
sec is brought to rest within a distance of 16 m
Hence, 0 = (8)
2
 + 2a (16) (Q v
2
 = u
2
 + 2as),
a = 
88
2 16
´
-
´
 = – 2 m/sec
2
Resultant upward force on elevator = T – mg. According to
Newton's law.
    T – mg = ma
or  T = mg + ma = m (g + a) = 1000 (9.8 – 2) = 7800 N
(b) Let P be the upward force exerted on the man by the
elevator floor. If m' be the mass of the man, then, weight of
the man acting downward = m' g ,
Upward force on the man = P – m'g
According to Newton's law . P–m' g = m' a or
P = m' (a + g) = (– 2 + 9.8) = 624 N
(9) (d) As P and Q move down, the length l decreases at the
rate of U m/s
A
B
Q
M
q
q
P
b b
y
From figure, l
2  
= b
2
 + y
2
Differentiating with respect to time
2l 
d
dt
l
 = 2y 
dy
dt
(Q b is constant)
\ 
dyd
.
dt y dt
=
ll
 = 
1d
.
cos dt q
l
 = 
U
cosq
(10) (a) The engine, coach, coupling and resistance are, shown
in figure.
COACH ENGINE
DRIVING
FORCE
T
Driving force = 4500 N
Opposing force (Resistance)  = 
4
(5 4)10
100
+
 = 900 N
Resultant force = 4500 – 900 = 3600 N
Mass of engine and coach = 9 × 10
4 
kg
According to Newton's law , F = ma
\  3600 = 9 × 10
4
 a
or a = (3600) / (9 × 10
4
) = 0.04 m/sec
2
So acceleration of the train = 0.04 m/sec
2
26
DPP/ P 09
Now considering the equilibrium of the coach only, we
have (T – R) = 4 × 10
4
 × 0.04 (Q F = ma)
or T – 
4
4 10
100
´
= 4 × 10
4
 × 0.04,
T = 4 × 10
4
 × 0.04 + 4 × 10
2
 = 1600 + 400 = 2000 N
(11) (d) Given that  
1
F
®
= 
ˆˆ
(8i 10j) + and  
2
F
®
 = 
ˆˆ
(4i 8j) +
Then the total force 
F
r
 = 
ˆˆ
12i 18j +
So acceleration 
ˆˆ
F 12i 18j
a
m6
+
==
r
r
=
ˆˆ
2i 3j + m/sec
2
Net acceleration
| a
®
| = 
22
23 +
= 
49 +
= 
13
 m/sec
2
(12) (c) From the relation
F = ma  Þ a = 
F
m
 = 
1000
1000
 = 1 m/s
2
As the force is brake force, acceleration is – 1 m/s
2
 using
relation v
2
 = u
2
 + 2as, we obtain
2as = u
2
Þ s  =
2
u
2a
 =
2
5
18
18
2
æö
´
ç÷
èø
 = 12.5 m
(13) (a) The water jet striking the block at the rate of 1 kg/s at a
speed of 5 m/s will exert a force on the block
F = v 
dm
dt
  = 5 × 1 = 5 N
2 kg
a
And under the action of this force of 5 N, the block of mass
2 kg will move with an acceleration given by ,
F  = ma Þ  a = F/m = 5/2 = 2.5 m/s
2
(14) (a) Relative speed of the ball = (v + u)
Speed after rebouncing = – (v + u)
So,  F = m 
v
t
D
D
 = 
m[(v u) { (v u)}]
t
+ --+
        = 
2m(v u)
t
+
(15) (b) F = 
dt
dp
 Þ  F dt = dp = p
2
 – p
1
Þ  F × 1 = mnv – 0
Þ  F = mnv
(Total mass of the bullets fired in 1 sec = mn)
(16) (a) The initial momentum  = 15 × 10 = 150 kgm/s  and
Force = 
change in momentum
time
  = 
0 150
15
-
 = –10 N
A constant force of 10 N must be acting in opposite
direction to the motion of body.
(17) (a) The change in momentum in the final direction is equal
to the impulse = 
2.50
1000
× 28 – 
250
24
1000
æö
-´
ç÷
èø
 = 13 Ns
and  force = 
impulse
time
 = 
13
1/100
 = 1300 N
in the direction of the ball.
(18) (b). We know  
dp
F
dt
=
r
r
 Þ Fdt dp =
r
r
Þ  2 × 2 = dp
r
 Þ   4 = dp
r
Therefore change in momentum  = 4 Ns
(19) (a) We know  
dp
F
dt
=
r
r
Þ  Fdt dp =
r
r
 = 
2 1 21
p p mv mv - =-
rr rr
Þ 4 j
ˆ
. 1 = 2 .
2
v
r
 – 2(2i
ˆ
)
Þ 
2
2v
r
 
2
ˆˆ
4j.1 2.v 2 (2i) =-
r
= 
ˆˆ
4j 4i +
Þ 
2
v
r
  = 
ˆˆ
2i 2j +
Þ  
2
|v|
r
= 22 m/s
(20) (c) Initial  momentum of the ball
= 
150
1000
 × 12 = 1.8 kg.m/sec
Final momentum of the ball =–
150
1000
× 20= – 3.0 kg m/sec
Change in momentum = 4.8 kg m/sec
Average force exerted  = Impulse/ time = 
4.8
.01
 = 480 N
(21) (b) Initial momentum of the body = mu = 20 × 3 = 60
and final momentum of the body = – mu = –20 × 3 = – 60
The change in momentum of body in initial direction
  = – 60 – 60 = –120
The change in momemtum imparted to the body in opposite
direction = 120
\ The impulse imparted to the body = 120 Ns
(22) (a) (1) Since the lift is moving down with an acceleration of
3 m/sec
2
, then the inertial force F = ma, acts upwards on
the body
mg
R
F=ma
a=3 m/s
2
Now , R + F= mg
or  R = mg – F = mg – ma = m (g – a) = 60 (9.8 – 3) = 408 N
(2) When the lift is moving down with constant velocity
a = 0  and hence, R = mg = 60 × 9.8 = 588 N
(3) The lift is now moving down with a retardation of
DPP/ P 09
27
3 m/sec
2
.
The retardation is 3 m/sec
2
 in the downward direction is
equivalent to an acceleration of 3 m/sec
2
 upwards.
Hence the direction of fictitious force is downwards.
Now ,  R = mg + ma = m (g + a)  = 60 (12.8) = 768 N
(23) (b) When the lift is moving up m (g + a)  = force
The scale reading = 
m(g a)
g
+
 = 
g
10g
3
g
æö
+
ç÷
èø
 = 13.3 kg
When lift is moving down the scale reading
= 
m(g a)
g
-
 = 
g
10g
3
g
æö
-
ç÷
èø
= 6.67 kg
(24) (a)
(1) A reference frame in which Newton’s first law is valid is
called an inertial reference frame.
(2) Frame moving at constant velocity relative to a known
inertial frame is also an inertial frame.
(3) Idealy, no inertial frame exists in the universe for
practical purpose, a frame of reference may be considered
as Inertial if its acceleration is negligible with respect to
the acceleration of the object to be observed.
(4) To measure the acceleration of a falling apple, earth can
be considered as an inertial frame.
(25) (a)
(i) In the case of constant velocity of lift, there is no reaction,
therefore the apparent weight = actual weight. Hence the
reading of machine is 50 kg wt.
(ii) In this case the acceleration is upward the reaction R =
ma acts downward, therefore apparent weight is more than
actual weight .
i.e.   W' = W + R  = m (g + a)
Hence, scale show a reading of
m (g + a) Newton = 
50g
50
a
æö
+
ç÷
èø
 kg wt
(26) (a) Tension = m (g + a), when lift moving up, putting the
values, we get
 175 = 25 (9.8 + a) Þ  a = 2.8 m/s
2
[negative sign shows that lift is moving downward]
(27) (b) Apparent tension, T = 2T
0
So,  T = 2T
0
 = T
0
  
0
a
1
g
æö
+
ç÷
èø
or  2 = 1 + 
0
a
g
 Þ a
0
 = g = 9.8 m/s
2
(28) ( b) Cloth can be pulled out without dislodging the dishes
from the table because of inertia. Therefore,
statement- 1 is true.
This is Newton's third law and hence true. But
statement 2 is not a correct explanation of statement 1.
(29) ( d) According to Newton’s second law
Acceleration = 
Force
Mass
 i.e. if net external force on the
body is zero then acceleration will be zero.
(30) (c)
dp
F=
dt
= Slope of momentum -time graph
i.e. Rate of change of meomentum = Slope of
momentum-time graph = force.