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50 Day's Plan For Physics : (Newton's Law of Motion)

ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A. 4 3 1 3 3 1 1 2 3 1 3 2 2 1 3 4 2 3 2 3 3 2 1 2 1 1 4 3 2 1
Q. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
A. 3 4 3 2 2 2 3 2 4 1 4 2 2 1 3 3 4 4 3 2 1 1 3 3 2 3 3 1 1 2
Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
A. 2 2 3 4 3 4 2 3 2 2 3 2 1 1 2 1 2 1 1 4 3 4 1 2 2 4 1 2 2 1
Q. 91 92 93 94 95 96 97 98 99 100
A. 2 4 4 1 2 3 1 4 2 3

HINT – SHEET

1. Ans ( 4 ) 3. Ans ( 1 )
R 3
sin θ = =
250
2 × 1000 × 10 ℓ+R 2+3
Δp 2mv 3
F= = = = 500N sin θ =
Δt Δt 0.01 5

2. Ans ( 3 )
∣ → ∣ = m ∣V
P →∣
∣ ∣ ∣ ∣
θ = 37°
Tcos θ = ω
P = 10 (
√
22 + 12 + 22 )
Tcos37° = ω
4
T= =ω
5
P = 10 × 3 = 30 kg m/s 5ω
T=
4

ADPLNEPH21547 HS-1/10
4. Ans ( 3 ) 10. Ans ( 1 )
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Fnet F
asystem = =
Msystem 4M
3F
T1 = (3M) × asystem ⇒
4
F
T2 = (2M) × asystem ⇒
2
F
T3 = (M) × asystem ⇒
4 Net driving force on the block is
5. Ans ( 3 )
T1 = 0.5 × 8g; mBg = 2T1 mg 3
= mg – = mg
⇒ mB = 8
4 4
3
⇒ (1 + ms ) = 8
∴ μ × 2mg = mg
4
⇒ ms = 7 kg 3
⇒μ=
6. Ans ( 1 ) 11. Ans ( 3 )
8

FL = µS mg
200 = µs × 500
µs = 0.4
fk = µk mg
150 = µk × 500
µk = 0.3 To keep the block stationary
7. Ans ( 1 ) ma cos θ = mg sin θ
⇒ a = g tan θ = g tan 37°
For equilibrium of system, F1 = √ F22 + F32 . As θ 3g
⇒ a=
4
= 90° 12. Ans ( 2 )
In the absence of force F1, N – Mg = Ma
Net force
Acceleration = 3Mg
Mass ⇒ − Mg = Ma
2
√ F22 + F32 g
F1 ⇒a=
= = 2

8. Ans ( 2 )
m m
13. Ans ( 2 )
By Newton's first law
14. Ans ( 1 )
fr = μ mg geff = g + a = 10 m/s2
= 0.6 × 10 × 10 ∴ TP = (M + m) geff
= 60N 0.2
9. Ans ( 3 ) = (4.9 +
2
) × 10 = 50N

Favg =
Δp 2mv cos 30o
= 15. Ans ( 3 )
Δt t Apparent weight = actual weight
2 × 40 × 10−3 × 20 √3
= × = 80√3 N When acceleration / retardation is zero.
10−2 2

HS-2/10
16. Ans ( 4 ) 22. Ans ( 2 )
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S=
u2
=
u2 → = d→p = m d→v
F where →v = d→r ^ ^ ^
= 8ti + 2j − 6k So
2a 2μg dt dt dt
same for both → = 2 [8i^] = 16i^
F
17. Ans ( 2 ) 23. Ans ( 1 )
F = slope of p – t curve Initial reading N1 = (mb + mc)g
at t = 3, Here mb = mass of bird,
slope = – 1. mc = mass of cage, if bird flying upward with
18. Ans ( 3 ) acceleration, let it is a, then new reading.
(A) Definition of force - Newton's first law N2 = mb(g + a) + mcg
(B) Measure of force - Newton's second law ⇒ N2 > N1
(C) Effect of force - change in momentum =
impulse 24. Ans ( 2 )
2mg − mg
(D) Recoiling of gun ⇒ Newton's third law. a1 = =g
m
19. Ans ( 2 ) a2 =
2mg − mg
=
g
In both F = 3ma 3m 3
2g
a= F So, a1 − a2 = ( )
3
3m
(1) N1 = 2ma (2) N2 = ma
2
25. Ans ( 1 )
N1
= For system, 2T = (50 + 30)g ⇒ T = 400 N
N2 1
20. Ans ( 3 )
F=
Δp
= Mg
26. Ans ( 1 )
Δt According to law of inertia (Newton's first law),
⇒ 2mnv = Mg
1 × 9.8 when cloth is pulled from a table, the cloth come in
⇒ v= = 9.8 m/s
2(0.05)(10)
21. Ans ( 3 ) state of motion but dishes remains stationary due to
N = 100 cos 60° = 50 N inertia. Therefore when we pull the cloth suddenly
∴ (fs)max = 0.5 × 50 = 25 N
fk = 0.3 × 50 = 15 N from table the dishes remains stationary.
27. Ans ( 4 )
The time of action of the force is very short,
therefore one can assume that there is no
appreciable change in the position of the body
during the action of the impulsive force.
Net external force, F11 = 100 × 3 − 60
2
√
28. Ans ( 3 )
= 50 3 − 60 ≈ 26.6 N > 25 N
√
T= m1g
∴ The block will move upward , so kinetic friction 7T = m2g
acts on it downward 7m1g = m2g
m2 = 7m1

HS-3/10
29. Ans ( 2 ) 34. Ans ( 2 )
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N = ma

Tx = Mg +
m
(ℓ − x) = Mg + m (
ℓ−x
)
f ≥ mg
ℓ ℓ
μ N ≥ mg
30. Ans ( 1 )
μ ma ≥ mg
N =mg g
a⩾
μ
10
a⩾ ⇒ a ≥ 20 m/s2
0.5
35. Ans ( 2 )
50 − 30
For 10 kg :- a = = 2m/s2
Net force by earth on block 10
For 4 kg :- T – F = 4a
= N→ + −→
mg
F = T – 4a = 30 – 4 × 2 = 22 N
= mg ^j + mg(−j^) 36. Ans ( 2 )
=0 v = u + at ⇒ 1000 = 0 + a × 10

31. Ans ( 3 ) a = 100 m/s2

F 105
m = = = 103 kg
a 100
37. Ans ( 3 )
Conceptual
N – mg = mg
N = 2mg ↑ ­ 38. Ans ( 2 )
32. Ans ( 4 ) Conceptual
39. Ans ( 4 )
Because four discs are above the bottom disc,
F
a=
m
hence force on it = 4 mg.

33. Ans ( 3 ) m1 =
m
L
(L − x)
Impulse = Area under F-t curve enclosed with time
axis. m F F (L − x)
T = m1 a = (L − x) =
L m L

HS-4/10
40. Ans ( 1 ) 46. Ans ( 3 )
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W1 mg 3 Δp
= = ⇒ 2g = 3g − 3a F= = 2mnv
W2 m (g − a) 2 Δt
⇒ a=
g
3
47. Ans ( 4 )
41. Ans ( 4 ) F=
dp
dt
≡
d
dt
(a + bt2 ) = 2bt ∴ F ∝ t
The system, as a whole, will fall towards ground
under gravity. The spring will neither be
48. Ans ( 4 )
2
m (u2 − v2 ) 30 × 10−3 × (120)
compressed nor stretched regardless of the values F = = = 1800N
2S 2 × 12 × 10−2
of m1 and m2.
49. Ans ( 3 )
42. Ans ( 2 ) Initially due to upward acceleration apparent
(10 + 5)g − 5g weight of the body increases but then it decreases
a=
20 due to decrease in gravity.
10g g
a=
20
=
2 50. Ans ( 2 )
43. Ans ( 2 ) Force on particle at 20 cm away F = kx
F F F = 15 × 0.2 = 3 N
a= =
m+m 2m
As k = 15 N/m
[ ]

Force 3
Acceleration = = = 10 m/s2
Mass 0.3
m F 3F
Tension at mid point = ( m+
2
) ×
2m
=
4 51. Ans ( 1 )
44. Ans ( 1 ) We know that
n mv n 35
|F| = ⇒ 93.3 = ( ) × ×v
t t 1000
n
On solving ( ) = 400/ min ute
t
52. Ans ( 1 )
The water jet striking the block at the rate of 1 kg/s

√3 at a speed of 5 m/s will exert a force on the blank

∵ NA = mg
2

∴ NA =
2
× 50 × 10 F = v dm = 5 × 1 = 5 N
dt
√ 3
1000
∴ NA = N Under the action of this force of 5 N, the block of
√ 3
NA 500
∴ NB = = N mass 2 kg will move with an acceleration given by
2 √3

45. Ans ( 3 ) F 5
a= = = 2.5 m/s2
Action Reaction m 2

HS-5/10
53. Ans ( 3 ) 59. Ans ( 1 )
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∫ dp = pf − −pi = ∫ F dt Given: horizontal force, F = 10 N and coefficient of

friction between block and wall ( μ ) = 0.2.
= Area under the curve
We know that at equilibrium horizontal force
pi = 0
provides the normal reaction to the block against
Net Area = 16 – 2 – 1 = 13 N-s
the wall. Therefore, normal reaction to the block
= Vf = 13 = 6.5i m/s (N) = F = 10 N.
2
[As momentum is positive, particle is moving along We also know that weight of the block
positive x axis.] (W) = Frictional force = μ N = 0.2 × 10 = 2 N.
54. Ans ( 3 ) 60. Ans ( 2 )
F = 600 – 2 × 105 t = 0 After string is cut, free body diagram of block A
t = 300 × 10 – 5 gives:
t
I= ∫ 600 − −2 × 105 t) dt
(
0
I = 600 × 300 × 10 – 5
= 1.8 – 0.9 = 0.9 Ns
2m aΑ = 3mg – 2mg
55. Ans ( 2 ) or aA =
mg
=
g
mg sin θ – T = ma 2m 2
T = ma Free body diagram of block B gives:
g sin θ
∴ a=
2
56. Ans ( 3 ) ∴ m aB = mg
dx or aB = g.
v= = 3t2 − 75
dt 61. Ans ( 2 )
dv Free body diagram of ball is
a= = 6t
dt

F = ma = 6 × 6 × t

at t = 4, F = 36 × 4 = 144 newton
62. Ans ( 2 )
57. Ans ( 3 )
T cos θ = T1 = 10 × g .....(i)
T sin θ = 98 .....(ii) 63. Ans ( 3 )
98 since normal force is zero, thus friction force will
∴ tan θ = = 1 or θ = 45°
10 × 9.8 also be zero.
58. Ans ( 1 ) N+50 sin37° = mg
F – Mg = Ma N + 30 = 30
8000 = 2000 a N = 0 (Normal is zero)
∴ Acceleration is 4 ms – 2 upwards fk = μN = 0N

HS-6/10
64. Ans ( 4 ) 71. Ans ( 3 )
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T – g = 1 × 2a For pendulum if θ is angle made by pendulum

4g – 2T = 4a string with vertical then tan θ = a
g
If θ = 90° – θ ⇒ θ = 45° ⇒ a = g
72. Ans ( 2 )
As the block A slides down along the rough
inclined plane, it will lose energy due to opposing
frictional force. Hence, the block B, which is
falling freely, will reach the ground with higher
velocity.
Ans. = 2a (accleration of 1kg)
65. Ans ( 3 ) 73. Ans ( 1 )
→ | = √(6)2 + (8) + (10)2
|F 50 10
ac = = m/s2
= 10 2 √ 15 3
F = ma freq = 10 × ac = 100 N (for lower block)
3
m = F = 10 2 fmax = μ (5) (10) = 25 N
√

m 2 freq > fmax

m = 5 2 kg
√

so there will be slipping.

66. Ans ( 4 ) a10 =
fmax
= 2.5 m/s2
Area under F - t graph = change in momentum. 10
67. Ans ( 2 ) 74. Ans ( 1 )
Solving from the frame of truck
v dm = mg
dt
mg
⇒ dm = = 500 × 10 = 5 kg/sec
dt v 1000

68. Ans ( 3 ) f ≤ μ mg = 6 ⇒ f = 5N.

a=
10
= 1 m/s2
75. Ans ( 2 )
10
Net force acting on 2 Kg block = 2(1) = 2 N
69. Ans ( 2 )
N1 (2m)a 2
= =
N2 (3m)a 3 N = 500 N
fℓ = μN
70. Ans ( 2 ) f ℓ = 0.2 × 500
36 f ℓ = 100 N
acceleration of system = = 2m/s2
18 5g > f ℓ
∴ f ℓ = 50 N
76. Ans ( 1 )
T1 = 4 × 2 = 8 N 36 – T2 = 6 × 2 900 – 600 = 60 amax
T2 = 24 N ⇒ amax = 5 m/s2

HS-7/10
77. Ans ( 2 ) 82. Ans ( 4 )
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fnet 100
a= = = 10m/s2
Σm 10
T = (1+5) (g+a)
T = 6(10+10) ⇒ 120 N
78. Ans ( 1 ) P = (M + m)a ⇒ a =
P
M +m
By impulse momentum theorem →
I = ΔP
F Δ t = m(v2 – v1)
1 3 −3 500
(10 × 10 ) (20 × 10 ) = (v2 − 0)
2 1000
v2= 200 m/sec PM
T = Ma =
M +m
79. Ans ( 1 ) 83. Ans ( 1 )
Force on B = m2a2 ∴ Force on A = m1a1 Since springs are massless, their reading will be
same.
m 2 a2
Acceleration A =
m1 84. Ans ( 2 )
80. Ans ( 4 ) 6g = Mg sin θ (in equilibrium)
dp θ = 30°
As F =
dt
85. Ans ( 2 )
2g + 2g − 2g g
dp = ∫ F dt Acceleration, a = = ; for ‘C’,
∫
2+2+2 3
g
2g – Tc = 2 ( )
3
Δ p = Area under F-t graph 4g
⇒ Tc =
3

1 86. Ans ( 4 )
Δp = × 8 × 1 – 2 × 0.5 = 3 kg m/s As is clear from figure
2
81. Ans ( 3 )
At just sliding condition limiting friction is acting.

F – 50 = 20a ....(i)
f = 10a ....(ii)
50 = 10a T1 = 3g
T2 – T1 = 2g
∴ a = 5 ms – 2
∴ T2 = 2g + T1 = 2g + 3g = 5g
Hence, F = 50+20×5 = 150 N Also, T3 – T2 = 1g
∴ Fmin = 150 N ∴ T3 = 1g + T2 = 1g + 5g = 6g

HS-8/10
87. Ans ( 1 ) 94. Ans ( 1 )
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Here, m1 = 40 kg
m2 = 30 kg
θ = 30º

m1 g sin 30o − m2 g sin 30o

10 10 a=
N = Fsin45°+ mgcos45° = + = 10√2 m1 + m2
√ 2 √ 2 200 − 150
= 0.7 m/s2
88. Ans ( 2 ) =
70

In this case, one 2 kg wt on the left will act as the

95. Ans ( 2 )
2T = (50 + 30) g ⇒ T = 400
support for the spring balance. Hence its reading is 2 kg. 400 + N = 500

89. Ans ( 2 )
FA tan 30∘ 1
F = slope ⇒ = ∘
=
FB tan 45 √3

90. Ans ( 1 )
Freq = mg sin θ + μ mg cos θ ⇒ N = 100

91. Ans ( 2 ) 96. Ans ( 3 )

Static firction is self adjusting but limiting in nature
T1 = 3(g+a) ⇒ 3×12 = 36N and has value
ƒ ≤ µS N
T2 = 7 × 12 ⇒ 84N 97. Ans ( 1 )
For t<0 and t>4s, the position of the particle is not
T3 = 13 × 12 ⇒ 156N
changing i.e., the particle is at rest. So no force is
92. Ans ( 4 ) acting on the particle at these intervals.
Force is in positive direction where acceleration is For 0<t<4s, the position of the particle is

positive i.e. slope of v-t graph is positive. continuously changing. As the position-time graph
is a straight line, the motion of the particle is
93. Ans ( 4 ) uniform, so acceleration, a = 0. Hence no force act
Both static and kinetic friction are independent on
on the particle during this interval also.
the area of contact. Coefficient of static friction
98. Ans ( 4 )
depends on the surfaces in contact. When the lift is accelerating upwards with
acceleration a, then reading on the scale
µk < µs
R = m(g + a) = 80 (10 + 5) N = 1200 N.

HS-9/10
99. Ans ( 2 ) 100. Ans ( 3 )
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Tension between m2 and m3 is given by

F – 70 = (2 + 5) a = 7 × 2
F = 84 N = T

2m1 m3 2×2×2
T= ×g= × 9.8 = 13 N
m1 + m2 + m3 2+2+2

T' – 50 = 5a = 5 × 2 = 10
T' = 60 N

HS-10/10