4 Laws of Motion

10. (3) For the equilibrium of mass m: Ft

Topic-wise Questions T = mg ...(i) 17. (2) n =
mv
1. (1) Linear momentum is conserved For the equilibrium of mass 2 m: 18. (3)
because the net external force acting on 2T cos θ = 2 mg...(ii) C
the system is zero.
Solving equations (i) and (ii), we get
2. (3) T cos θ = T1 = 10 × g ...(i)
θ = 45° Q
T sin θ = 98 ...(ii) R (Larger Force)
98 11. (3) Friction between block A and block
∴ tan θ = =1 (Resultant)
10 × 9.8 B and between block B and surface will
oppose the F.
⇒ θ = 45° A B
As the block is about to move P
3. (1) T – mg = ma (Smaller Force)
∴ F = FAB + FBS
360 – 30 × 10 = 30 × a in ΔABC
= µ ABm Ag + µBS (mA + mB)g
a = 2 m/s2 ∴ R2 + P2 = Q2
= 0.2 × 100 × 10 + 0.3(100 + 200)10
udm ⇒ (12)2 = Q2 – P2...(i)
4. (2) F = = 200 + 900 = 1100 N.
dt P + Q = 18 N (given) ...(ii)
dp
50
= 20N 12. (2) F =    ∴144 = (Q – P) (Q + P)
= 400 × dt
1000 144
dm or Q – P = = 8 ...(iii)
v 13. (1) F = v 18
5. (3) t = dt
a From eqns. (ii) and (iii), P = 5; Q = 13
F
1 14. (3) Vy = U y + y t
∴t ∝ m ∴ Forces are 5 N and 13 N.
a
  15. (3) Applying Newton’s second law for 19. (1) Net force on the rod
 F = ma  the man:
1   = F 1 – F 2 (⸪ F1 > F2)
t∝ ∝m 1000
a  F = same  T + 450 − 1000 = a As mass of the rod is M, hence acceleration
 1  10 of the rod is:
m ∝  T = 550 + 100a ...(i)
 a  (F − F )
⇒t∝m Applying Newton’s second law for the a= 1 2
M
t m 8 4 chair: For the motion of part AB of the rod
∴ 1 = 1 = =
t 2 m 2 10 5  250  [whose mass is equal to (M/L) y],
T − 450 − 250 =  a
 dv   10  M
6. (4) F = m   (using newton's second T = 700 + 25a ...(ii) F1 – T = y×a
 dt  law of motion) L
From equation (i) and (ii) a = 2m/sec2
 8  where T is the tension in the rod at the
F = 150   = 150 × 40 = 6000 N 16. (2) It follows from the figure that the point B.
 0.2 
equations of motion are: M F −F 
7. (1) ⇒ F1 – T = y× 1 2 
v −v  (3.5 − 2) L  M 
F = m  2 1  = 3× = 1.8 N
 t  2.5  y y
In the same direction of motion. ⇒ T = F1 1 −  + F2   .
200 – T = 20 a1 and T = 10 a2.  L L
8. (4) Mass of the body m = 6 kg 20. (1) Net upwards force on the system is 2T
velocity v = v2 – v1 = 5 – 3 = 2 m/s where a1 and a2 are the accelerations for
20 kg and 10 kg respectively. Net downward force on the system is
∴ Momentum p = mv = 6 × 2 = 12 N-s (50 + 30)g = 80g.
But a2 = 12 ms–2
9. (2) After a jerk is provided, impulsive For equilibrium of the system, 2T = 80 g
tension in the string exceeds the breaking ∴ T = 10 × 12 = 120 N
T = 40g
stress. So the string breaks before the 200 − 120 80
∴ a1 = = = 4 ms–2.
jerk can reach point A. 20 20 21. (2) R = m(g – a)
28 Laws of Motion

22. (4) If a be the common acceleration of 27. (1) ma = mg – T 39. (4) For the block on the surface
the system, then equations of motion are: ma = mg –1/3 mg T = 1 × a for the hanging block 1
5a = 5g – T1,...(i) ma = 2/3 mg g –T = 1a, T = ma
3a = T1 – 3g – T2...(ii) a = 2/3 g = 6.53 m/s2 on solving a = 5 m/s2
and a = T2 – g ...(iii) ⇒ a = 6.53m / s 2 40. (2)        
Solving above equations, we get, 28. (2) Change in linear momentum of one 41. (2) J = ∫ F .dt
ball = 2mu, 42. (3) T1 – 20g = 20a
time taken = 1 sec ⇒ T1 = 240 N
Total change in momentum
a T1 T1 a Favg = 43. (4) Let the tension in the string AP2
time taken and P2P1 be T. Considering the force on
n ( 2mu ) pulley P1, we get;
= = 2mnu T = W1...(i)
3kg 5kg 1
Further, let ∠AP2P1 = 2θ
T2 29. (1) using Newton’s 2nd law,
dp d (mv)  dm  Resolving tensions in horizontal and
F= = = v = vM , vertical directions and considering the
1kg 5g
dt dt  dt 
forces on pulley P2, we get;
as velocity v is constant
2T cos θ = W2...(ii)
⸫ F = Mv Newton
1g ∴ 2W1 cos θ = W (⸪ T = W)
g 30. (3) v = 2 ( g + a ) h and 1
a= ⇒ cos θ = or θ = 60°
9 2
g 10g v′ = 2 ( g − a ) h (⸪ W1 = W2 = 100N)
∴ T2 = a + g = +g= .
9 9 31. (4) F = mg sin a ∴ ∠AP2P1 = 2θ = 120°
23. (2) For the monkey to move up with an Where F = ma cos a and a = g tan a
44. (1) m1u1 + m2u2 = (m1 + m2)v
acceleration a the monkey will pull the 32. (2) DW = 2ma 
rope downwards with a force of 40a.  F   
F 45. (1) a = , v = u + at
Tmax = mg + 40amax 33. (1) a = , T =  m + M a m
M +m  
2  46. (2) According to the graph net impulse
600 = 400 + 40a 34. (3) From the figure, it follows that in the time 0 to 8 sec is zero. Hence,
200 T1 = 3g linear momentum gained in the given
amax = = 5m / sec 2
40 2g + T1 = T2 time interval is also zero.
So, the rope will break if the monkey ⇒ T2 = 2g + 3g 47. (3) F = 2r Av2
climbs up with an acceleration of 6m/sec2 ⸫ T2 = 5g. 48. (3) Area under F – t graph gives the
24. (3) F = mB a impulse of the force F .
m2 g
35. (4) a = 1
25. (1) Minimum additional force needed m1 + m2 Area = (3T + T) F0 = 2T F0
2
so that the net force is along the y 36. (2) Change in momentum = 2 mv cosθ [using formula for area of trapezium]
direction. = 20 3 kgm/s ⇒ Impulse = 2T F0
Using horizontal component of force,
37. (1) Let T be the tension in the string. ⇒ mvf – mvi = 2TF0
∑Fx = – (4) sin30° + 1 cos60°
[⸪Impulse = change in momentum]
 + 2 cos 60°
∑Fx = – 0.5 N mvf – 0 = 2TF0
Hence 0.5 N force in +x required to 2TF0
∴ vf =
balance horizontal force. m
1
26. (4) As the block is at rest, therefore 49. (1) Here, y = ut + gt2
2
tension T = Mg dy
Velocity, v = = u + gt
T Let a be the acceleration of the system. dt
dv
The equations of motion are Acceleration, a = =g
dt
Ma = Mg sin 53° – T ...(i) The force acting on the particle is,
and Ma = T – Mg sin 37° ...(ii) F = ma = mg
mg Adding eqns. (i) and (ii), we get; 50. (1) Net force acting on the gun =
Mg(sin 53° − sin 37°)  momentum of each bullet × no.of bullets
T a=
2M Time
Net force on the pulley  4 0.1× 40 × 600
= = 40 N
Fnet = g 4 3 ∵ sin 53° = 5  60
(T + mg) 2 + T 2 a=  −    51. (2) Tension is given by
25 5  sin 37° = 3 
2 2 2m1m3
= g (M + m) + M  5  2 × 6 × 6 × 10
T= ×g =
Since pulley is in equilibrium, clamp ⸫ a = 0.98 m/s2 m1 + m 2 + m3 6+6+6
will exert the same magnitude of force in F 720
38. (2) a = , T = m + 3 m a = = 40N
opposite direction. m1 + m2  1 2 18
 4 
Laws of Motion 29

52. (1) Use equilibrium of forces mv − ( −mv ) 69. (1) Net acceleration a’ on the bob of
2mv
53. (2) Impulse is equal to the change in 61. (1) F= = the pendulum
t t
momentum. For change in momentum to = a 2 + g 2 + 2ag cos θ
2 × 200 × 10−3 × 30
be minimum =
0.01 As, θ = 90°
d
(20 t2 – 40 t) = 0 ⇒ F = 1200N ma
dt a
62. (1) Mass of the body m = 5 kg
40 t – 40 = 0 Change in velocity g mg
⇒ t = 1s
v –v =[(10 – 2) i + (6 – 6) j ] 2 2
f i ∴a ' = a + g
54. (2) J = F1t1 + F2t2 + F3t3, J = mDv Change in momentum = mΔv = 5[8 i ]
55. (3) If F = 0 find ‘t’, ⇒ T = m a 2 + g2
= 40 i kg ms–1
t
63. (3) Acceleration of the system a 70. (4) After the string is cut,
I = ∫ F .dt Using newton’s 2nd law for block (A)
F
0 =
M+m 3 ma = 4mg – 3mg
56. (1) Momentum of each bullet = Mv (common acceleration)
= 50 × 10–3 × 200 g
aA =
3
= 10 kg m/s
Force on the block (M) = kx = Ma
Total force acting on the gum Using newton’s 2nd law for block (B)
m×F
nP 10 × 10 Force on m, Fm =
F= = = 100 N (time = 1 sec) M+m
t 1 F 

⸪ Force = ma  putting a =  ma = mg
 m+M
100 ⇒ aB = g
∴a = 64. (3) Mg sin θ – T = Ma
2800 T = ma 71. (4) Net pulling force on the system is
⇒ a = 0.035 m/s2 = 3.5 cm/s2 Adding (i) and (ii) Mg + mg – mg = Mg. Total mass being
1 pulled is M + 2m.
57. (3) v2 – u2 = 2as ⇒ 0 – u2 = 2 × –5 × 100 a = g sin θ Hence, acceleration of system is,
2
–u2 = –2 × 500 ⇒ u = 1000 = 31.6 m/s As, the block on the ground is moving Mg
a=
Impulse = change in momentum due to tension in the string. M + 2m
Kx
= m (v – u) = 0.1 (0 – 31.6) Mg sin θ
⇒ T = Ma =
= – 3.16 Ns 2
58. (1) Resultant of three forces F1, F2 and F3 65. (4) T1 = (12 + 3)a and T2 = 3a
a
T 15a 5 M
will be ( 2 − 1)F . ∴ 1 = = .
∴ FR = maR T2 3a 1
Therefore, acceleration of body = a
66. (3) v = u + at , tan θ =
( 2 − 1)F g Mg
= ( 2 − 1)a 67. (1) The forces acting on mass m are as Now, since a < g, there should be an
m
shown in the figure shown below. upward force on M so that its acceleration
59. (2) Net upward force on the 2 kg block Acceleration of the system becomes less than g.
in upward direction F = 40 N P Hence, for any value of M spring will be
40 N a=
M+m elongated.
Let the reaction of m on M be f.
72. (1) flim = 0.6 × 10 × 9.8 = 58.8 N
Pm Therefore, the block will slide over the
Then f =
M+m slab & fk = 39.2 N
According to figure, m will be stationary 39.2
2 kg a
when ∴ a slab = = 0.98 m/s 2
40
f cos β = mg sin β
73. (3) Limiting force of friction,
fs = µsR = µsmg = 0.54 × 2 × 10 = 10.8 N
As F < fs, therefore, the block does not
20 N move
Applying Newton’s 2nd law of motion R
40 − 20
a= = 10 ms–2 F = 2.8 N
2 fs m
60. (2) The acceleration of the system Pm
cos β = mg sin β
(m 2 − m1 )g M+m
(3 − 1)g mg
a= = ∴ P = (M + m)g tan β.
(m1 + m 2 + M) (1 + 3 + 6) Further, as static friction is a self adjusting
2g g 10 68. (1) a = Fnet force, hence the frictional force between
= = = = 2 ms–2 m the block and the floor is 2.8 N.
10 5 5
30 Laws of Motion

74. (1) Friction force = mg sin q  0.50  ⇒ µ = tan θ = tan 30°

∴ = × 100 ⇒  = 33.3% 1
75. (4) As the block tends to slide down,  1 + 0.50  ⇒ µ=
therefore the friction must be upwards 3
F
76. (1) f = µN = 0.3 × 250 = 75 N 83. (3) µ = if, normal reaction force 94. (4) fs = μsR = μs mg
R
77. (2) Block is under limiting friction, so becomes double then force also will be = 0.4 × 2 × 9.8
...(i) double. So, μ remains constant. = 7.8N = 7.8 N
µ = tan θ
g If the applied force is smaller than the
Equation of the surface, 84. (1) a =
2
l −1 force of friction, then the block will
x3 not move, because static friction is self
y=
6 85. (3) t1 = sin θ − µ cos θ = 1 − µ cot θ adjusting so, it will be equal to 2.5 N
Y t2 sin θ 95. (4) fmax = µN
h 1
86. (4) tan θ = = µ ⇒ V = πR 2 1
R 3 = µMg = × 60 × 9.8 = 196 N
3
87. (3) The magnitude of limiting friction
= µsR = µsmg. Now, kinetic friction 96. (3) a = (ms – mk)g
m
y µkR opposes the motion and acts mv 2
97. (1) Centripetal force =
θ toward left. r
X F1 3r1
x ∴ Net force towards right = µsR – µkR ∴ = F
dy x 2 So, acceleration produced F2 r1 ⇒ F2 = 3
Slope, = t...(ii)
dx 2 = (µsR – µkR)/m 98. (3) As car is moving on a circle on a
From eqns (i) and (ii), we get = (µs – µk)mg/m level road, then normal reaction force
x2 x 2
= (µs – µk)g balance the weight of the car.
µ= ⇒ 0.5 = ⇒ x2 = 1
2 2 = (0.4 – 0.2)g = 0.2g.  
1 99. (4) h = r 1 − 1 
⇒ x = 1, so y = 88. (4) The retardation α is given by: µ 2 + 1 
6 
g 1 g
78. (4) Frictional force on the block = µMg α = g sin 45° + µg cos 45° = + ×
2 2 2  1 
It will be causing acceleration a = ∴ h = 0.6 × 1 −  = 0.17 m
g  1  1 + 0.44 
µMg/M = µg =  1 + .
2 2 a maω2
The block slides on the belt its velocity 100. (3) 3T = mr ω2 = m ω2 ⇒ T =
becomes v given by the equation, 89. (2) m = tan q 3 3
v2 = 0 + 2ax or v2 = 2µgx w w
90. (4) N + sin 30° = w + 101. (1) v = µrg = 1.5 × 180 × 9.8
v2 2 2
∴x = = 2646 = 51.4m / s
2µg 5w 5w
N= \ f max = µN = µ. 5
4 4 102. (3) v = 900 × = 250m / s
79. (3) Let θ be the angle made by the rod 18
with the track, The block will remain stationary if
v2 v2
Fro equilibrium tan θ = ⇒r=
mv 2 / r W µ5W or W 3 5W rg g tan θ
tan θ = cos 30° < <µ
mg 2 4 2 2 4 (250) 2
∴r =
v2 3 1
⇒ tan θ = or 3 < 5µ or <µ × 10
rg 5 3
(10) 2 ⇒ r = 6250 × 1.73 = 10.8 × 103 m
⇒ tan θ = =1 \ Block will move if m < 3
10 × 10 5 103. (1) T sin θ = mω2r and r = L sin θ
∴ θ = 45° 91. (4) Block does not move ⇒ T = 16 ml
F 2
80. (4) Maximum tension in the string will a= = = 0.09 104. (1) Centripetal force in a string is due to
be for the maximum speed (as m and l m + M 22
tension.
are fixed). 92. (4) Given Horizontal force, F = 10 N
2
T  and coefficient of friction between block mv 2
\ Tmax = mV ⇒ V 2 = max 105. (1) T =
R
 m and wall (µ) = 0.2. 2
Weight of the block (W) = Frictional 106. (1) R = mg cos θ − mv
25 × 1.96 force = µR = 0.2 × 10 = 2N. r
V= ⇒ V = 14 m/s
0.25
93. (1) As the block is sliding down with
81. (1) f1 = 300 N, f2 = 600 N constant velocity
Total force of friction = 900 N Therefore, m(g – a) sin θ = µm(g – a)
cos θ
82. (1) As we know that the relation
 N
µ= ⇒ µ − µ
1−  When θ decreases, cos θ increases i.e.,
=  ⇒ µ =  + µ R increases.
 µ  θ 107. (4) 108. (1)
⇒= 
1+ µ  mg – ma 109. (3) 110. (3)
Laws of Motion 31

6. (4) Both the springs will show same

Morale Booster Questions Hence, AG = 1
reading of 8 kg as the springs are BG 3
1. (1) Non-Inertial Frame massless, therefore tension in both the
springs will be same. 12. (2) Using Newton’s 2nd law of motion
T T Mg – T = Ma ....(i)
K 7. (3) Impulse imparted from 0 to 4 s
a0
= linear momentum gained in 4 s ma
2T a a 2T − mg = …(ii)
Impulse = area enclosed by graph from 2
mT T
2 m1 m2g m2a0 m1g m1a0 (pseudo) 0 to 4s Adding eq. (i) and (ii)
(pseudo) πr 2 m
Let relative to the centre of pulley, m1
Impulse =
2
(∵ r = 2 ) ( 2M − m ) g = a  2M + 
 2
accelerates downward with a and m2 Impulse = 2 π N-s
accelerates upwards with a. ( 2M − m ) 2g
8. (1) Let α = angle of repose ⇒a=
Applying Newton’s 2nd law. 4M + m
⇒ tan α = µ
m1a + m1a0 – T = m1a...(i) ∴ cot α = 3 13. (1) For a straight line motion,
T – m2g – m2a0 = m2a...(ii) v – u = at
a = cot –1 (3)
 −µmg 
On adding (i) and (ii) we get Here, 0 – v = – (μg) t  As, a = 
 m 
 m − m2  Negative sign of a indicates retardation
a= 1 ( g + a0 ) ...(iii)
 m1 + m2  v
∴ t=
Substituting a in equation (i) µg
We get
 14. (1)
2m1m2 ( g + a0 ) 9. (4) p (t) = A (i cos kt − jsin kt)
T= a=g
m1 + m2
F 2T 4m1m2 ( g + a0 )
∴x= = = Fx Px
k k (m1 + m2 )k
2. (2) Given, m = 5 kg
Applied force F = (−3iˆ + 4 ˆj ) N F Fy p ar ar + a

Initial velocity, u = (6iˆ − 12 ˆj )m /s ar ar – a
Py
 ˆ ˆ
Retardation, a = F =  − 3i + 4i  m /s 2  d 
m  5 5 F = [ p (t)] = Ak (−isin kt − jcos kt) Using Newton’s 2nd law for the two
  dt blocks
As final velocity is along Y-axis only, its F·p = A2k (–cos kt sin kt + sin kt cos
x-component must be zero.
kt) = 0 w1 – T = w1 (ar – a) ...(i)
From v = u + at, g
∴ The momentum and force are
3iˆ perpendicular to each other. T – w2 = w1 (ar + a) ...(ii)
for X-component only, 0 = 6iˆ − t
5 g
5× 6 10. (4) According to the law of conservation
t= = 10 s Solving Eqs. (i) and (ii), we get
3 of momentum, total momentum of
bullets = momentum of the disc 4w1w 2
3. (4) Area under the F –t curve = change T= (⸪ a = g)
in momentum i.e., 2m’vn = mg w1 + w 2
1 mg 100 × 980
or × 2 × (10) + ∴ v= = = 980cms −1 15. (2)
2 2m 'n 2 × 5 × 10 C T B
1
2 × 10 + (10 + 20) × 2 11. (3) T F
2 T1 cos 30° T2 cos 60° Am
1
+ × 4 × 20 = m (v – u) 60°
2 T1 sin 30°
A As the pulley is massless,
or 10 + 20 + 30 + 40 = 2 (v –0) Applying Newton’s 2nd law of motion
or 100 = 2v or v = 50 ms–1 the equation of motion is
or T1 sin 30° = T2 sin 60°
4. (2) a =
( m 2 − m1 ) , g  3− m 
or T1 = 3 T2...(i) F − 2T = 0 ⇒ T =
F
g = g
( m1 + m 2 ) 4  3+ m 
For T1 rotational equilibrium, we take
2
⇒ 3 + m = 12 – 4m torque about the centre of gravity The acceleration of the block is,
5m = 9 ⇒ m = 9/5 = 1.8 kg F 
T1 cos30° × AG = T2 cos60° × BG −f 
T − f  2 
5. (1) Initial velocity of a ball = v a= = .
When a ball strikes the wall normally it 3T1 T m m
× AG = 2 × BG
is reflected back, then final velocity = - v 2 2 16. (2) According to the question,
Change in velocity of the ball ∆v = v – AG T2 mg(sin φ + µ cos φ) = 2mg(sin φ − µ cos φ)
(– v) = 2v ∴ =
BG 3T1 ⇒ tan φ = 3µ
Using Newton’s second law,
T2 1 As, μ = tan θ
m∆v (m2v) 2mv But =  [from eq. (i)]
F = ma = = = T1 3 So, tan φ = 3tan θ
∆t t t
32 Laws of Motion

When 4 kg mass is removed then weight

Tsinq Tsinq
∑ Fy =0
of balloon = 60 N
17. (1) N = 50 sin 45° + 150 cos 45° ...(ii) So, F – 60 = 6 × a
1 where, a = new upward acceleration
Solving eq. (i) and (ii), µ = .
2 ⇒ 300 – 60 = 6a ⇒ 240 = 6a
23. (1) g sinθ = μg cosθ
W W 240
2T sin θ = W ⇒ T = = cosecθ ⇒ μ = tanθ ∴a= = 40 ms–2
2sin θ 2 6
When block is projected up
18. (3) Using Newtons; 2nd law of motion 31. (1) Relative vertical acceleration of A
a = sinθ + μg cosθ = 2g sinθ
for A : T = ma with respect to B = g (sin2 60° – sin2 30°)
T kx v 02 v 02 V02 3 1
⇒a= = d stop = = = = 9.8  −  = 4.9m / s 2
m m 2a 2 × 2g sin θ 4g sin θ 4 4
Using Newtons; 2nd law for B : 24. (2) Initially both the blocks will move Mg − F
32. (1) a M =
F – T = ma' together with same acceleration. But M
after some time block B will start moving mg − F
F − kx
⇒ a'= with constant acceleration due to the a=
m m
kinetic friction.
F F
25. (1) Free body diagrams of the two a rel = a M − a m = −
m M
bodies are as follows: 1
&  = a rel t 2
⇒ The relative acceleration F1 = 2N 2
f
F − 2kx 2Mm
a r = a '− a = F2 = 20N f so, F =
m (M − m) t2
19. (1) Let T be the tension in the string Let acceleration of both the blocks v 2 25
connected to 15 kg block. 33. (2) a = = = 2.5m / s 2
towards left is a. Then 2s 10
Therefore, 2T = 250
f − 2 20 − f
a= =
2 4
or 2f – 4 = 20 – f
or 3f = 24 or f = 8N
Maximum friction between two blocks For 6 kg ⇒ F – 2T = 6a ...(i)
can be:
For 2 kg ⇒ T – 2g = 2 (2a) ...(ii)
fmax = µmg = 0.5 × 2 × 10 = 10 N
Adding equation (i) and (ii)
Now, since f < fmax
F = 75 N
For the block having mass 15 kg Therefore, friction force between the
34. (2) Let the tension in the string T = 1 N,
⇒ T + 0.4F = 150 ...(i) two blocks is 8 N.
therefore, TB = 3N and TA = 2N
For the block having mass 25 kg 26. (1) Frictional force = μR = μ (mg + Q T 3
2T = 0.4 F + 0.4 F + 250  ...(ii) cos θ) ∴ v A = B ·v B = v0  (towards right)
TA 2
Solving eq. (i) and (ii) Horizontal Force In = P + Q sin θ 3v v
F = 31.25 N For equilibrium condition net force must ∴ v AB = 0 − v0 = 0 , towards right.
2 2
be zero
20. (3) Wedge moves to the left due to 35. (1) No net force acts on the system when
the horizontal component of normal μ (mg + Q cos θ) = P + Q sin θ the system of two blocks of masses M
reaction. (P + Q sin θ) and m either remaining at rest or moving
µ=
N N sin θ (mg + Q cos θ) with a constant speed, no net force acts
a= H =
M M on them. In this equilibrium state.
27. (2) Net force will be the resultant.
21. (1) kx = 2T & T = 10g T
2d
22. (1) The string is under tension, hence 28. (2) a = m1 − m 2 g , t = μR M
there is limiting friction between the m1 + m 2 a
T
block and the plane. Drawing free body 4g − g 3g
29. (3) a = =
diagram of the block, 4 +1 5 Mg
∑ Fx = 0 ⸫ srel = 6m
µN + 50 cos 45° = 150 sin 45° ...(i)
6g A
f = μN a rel = = 12m / s 2
N 5
M
2s rel
45° t= = 1s
a rel
T = 50N mg
45° T = µR = µMg and T = mg
30. (1) Let upward force on the balloon is F
45° ⇒ F – mg = ma ∴ µMg = mg
y x
⇒ F – 10 × 10 = 10 × 20 m
Mg = 150N ⇒M=
⇒ F = 300 N µ
Laws of Motion 33

36. (2) Let weight of A is w'. = 2 × 4 + 12 × 5 = 8 + 60 = 68 m/s2 g(m1 – m2) = a (m1 + m2)
For equilibrium of the system, Force, F = ma = 2 × 68 = 136 N Putting the value of a in eq (i) we get,
N 41. (3) In this case, vertical component of the m1 ( m1 − m 2 ) g
m1g − T =
force increases the normal reaction, i.e., ( m1 + m 2 )
R = mg + mg cos θ = mg(1 + cos θ)
f = μN
Hence, the block can be pushed along m1g − T =
(m 2
1
− m1m 2 g )
w
the horizontal surface when, horizontal ( m1 + m 2 )
T cos θ = Nµ = µw ...(i)
component of force ≥ frictional force, 2
∴ m g − m1T + m1m 2 g − m 2 T
1
T sin θ = w' ...(ii)
where, T = tension in the thread lying = m12 g − m1m 2 g
between knot and the support. 2m1m 2
g=T
Solving eq. (i) by eq. (ii) we get, m1 + m 2
T sin θ w ' Thrust on the pulley is 2T
=
T cos θ µw 4m1m 2 g
∴ Thrust is equal to
⇒ w' = µw tan θ m1 + m 2
37. (3) 5. (3)
i.e., mg sin θ ≥ µmg(1 + cos θ)
38. (2) Choosing the positive x y-axis as θ
shown in the figure. The momentum of or 2 sin θ cos θ ≥µ. 2 cos2
2 2 2 u
the bead at A is: θ
u v
  ⇒ tan ≥µ.
pi = + mv 2
The momentum of the bead at B is,
Multi-Concept Questions
i P Q
1. (1) As the system is in equilibrium From constrained condition, along the
T sin 60o = 2Kg string the net velocity should be zero.
o
Tsin60 v cos q = u
60o
T60o u
V=
cos θ
o
 Tcos60 T1

pf = − mv 6. (3)
Therefore, the magnitude of the change 2kg - wt
in momentum between A and B is, 3 R T T
    T× = 2 kg
∆ p = pf − pi = −2mv 2
°
i.e., Δp = 2mv (along + ve x-axis) 4 30 30°
T= kg in 2m
gs
The time interval taken by the bead to 3 m mg cos 30°
mg
reach from A to B is, For the equilibrium of the horizontal string 30°
π(d / 2) πd T1 = T cos 60°
Δt = = 2mg –T = 2ma ...(i)
v 2v 4 kg 2 T – mg sin 30° = ma ...(ii)
Therefore, the average force exerted by ⇒ T1 = = kg
3 2 3 On adding eqns. (i) and (ii), we get
the bead on the wire is,
2. (1) mg
∆p 2mv 4mv 2 2mg – = 3ma
Fav. = = = 2 2
∆t πd / 2v πd 3. (2) Initial velocity = = 1 m/s g
or a = .
39. (2) Direction of resultant will be given 2 2 2
Final velocity = − = –1m/s 7. (1) Dowhward Force = mg sin θ
by tan θ, where θ is the angle which  2
pi = 0.4 N-s 1
resultant makes with x-axis.  = 1 × 10 × =5N
y 1 pf = –0.4 N-s 2
∴ tan θ = =    The maximum force of friction
x 2 I = pf − pi = –4.0 – 0.4 = –0.8 N-s = µmg cos θ
or 2y = x
| I | = 0.8 N-s.
or 2y –x = 0. 3
4. (1) using Newton’s 2nd law of motion = 0.8 × 1 × 10 ×
2
40. (1) ⸪ x(t) = pt + qt2 + rt3
= 4 3  7N
Differentiating both sides T
a T T fr
dx a
v= = p + 2qt + 3rt 2 1 kg
dt m2g
dv m2g
⇒a= = 0 + 2q + 6rt
dt m1g – T = m1a  ...(i) θ = 30°
at t = 2s; T – m2g = m2a  ...(ii) Since downward force is less than friction.
a = 2q + 6 × 2 × r = 2q + 12r Adding eq. (i) and (ii) Hence tension in string will be zero.
34 Laws of Motion

8. (2) f = (M + m) µg
N
9. (1) Partial free body diagram of block is ⇒ (15 + 5)(0.6)(10) = 120 N

15
f
shown in figure. P varies from (mg sing

cm
A f=0
θ – µmg cos θ) to (mg sin θ + µmg cos θ). θ
in
It means mgsin θ part of P is always acting m gs
up and term of P changes its direction. cm
10 f=0
When P = mg sin θ i.e., when 2nd varying B F = 80 N
term gets zero. Let this 2nd term is k. θ mg cos θ
F = 80 N
For rotational equilibrium,
As fmax > F
f
0.15
N×x=f× ∴ Frictional force between B and ground
θ 2
in
For no sliding, f < fL will be 80 N.
gs
m
k or mg sin θ < µ × mg cos θ 17. (4) Let the mass of the block B is m.
tan θ < 3 i.e., θs < tan–1 ( 3) = 60°
For P = mg sin θ – µmg cos θ to mg sin θ for no sliding
i.e, k from µmg cos θ to 0, the free body
For no tippling, x < 5 cm
diagram is shown above.
0.15
From free body diagram f = k and f is up mg cos θ × x = mg sin θ ×
2
the incline so f is + ve 0.15
or x = tan θ ×
For P = mg sin θ. k = 0 = f. 2
2
For P = mg sin θ to mg sin θ + µmg cos which gives θT < tan–1   , for no
toppling, 3
θ the friction force would be down the
incline and hence negative equal to 2nd As θS > θT, so block will remain at rest, In equilibrium, T – mg = 0
i.e., won’t slide for some angle upto ⇒ T = mg  ...(i)
term of P. So (1) is correct option.
θT, and then it will topple.
If blocks do not move, then T = fs
mv 2 14. (2) For equilibrium at of the system,
10. (2) Centripetal force f ≤ µs N = T = μsR = μs 2g ...(ii)
r
acting on the moving car on a circular Thus, from Eqs. (i) and (ii), we have
road. Mg = μs2g ⇒ M = μs2 (μs = 0.2)
M = 0.2 × 2 = 0.4 kg
11. (2) ⇒ (m1 v1 ) 2 + (m 2 v 2 ) 2 = (m3 v3 ) 2
18. (1) As the mass m3 is at rest.
⇒ (m × v) 2 + (m × v) 2 = (2m × v3 ) 2 Therefore, 2T = m3g...(i)
Further if m3 is at rest, then pulley P is
⇒ m 2 v 2 + m 2 v 2 = 4m 2 v32 also at rest. Writing equations of motion.
v T1 = T cos 30° and T sin 30° = W
⇒ 2m 2 v 2 = 4m 2 v32 , ⇒ v3 =
2 As, T1 = μR = 12 N
12 × 2 24
12. (2) Δp = OQ sin 30° – (– OP sin 30°) ∴ 12 = T cos30° ⇒ T = =
3 3
= mv sin 30° – (– mv sin 30°) = 2 mv 2T 2T
∴ T1 sin 30° = W
sin 30°
24 1
⇒W= × = 6.92 m3 P
3 2 T
T a
15. (1) In the absence of friction we have a
a1 = g sin θ = 1 a1t12  ...(i) m1
2 m2
When friction is present, then
for the blocks having masses m1 and m2
a 2 = gsin θ − µ k g cos θ
m1g – T = m1a...(ii)
1 2 T – m2g = m2a...(iii)
Its time rate will appear in the form of a2 = a 2 t 2  ...(ii)
2 Solving eq. (ii) and (iii) we get,
average force acting on the wall. From Eqs (i) and (ii) we get m3 = 1 kg
∴ F × t = 2 mv sin 30° 1 2 1 2
a 1 t1 = a 2 t 2 19. (1) If the same wedge is made
2mvsin 30° 2 2 rough, then time taken by the block to
⇒F=
t ⇒ gsin θt12 = [gsin θ − µ gcos θ] × n 2 t12 come down becomes n times more.
⇒ sin 45° = (sin 45° − µ cos 45°) × n 2 The coefficient of friction is
Given, m = 0.5 kg, v = 12 ms–1, k k
oc oc
t = 0.25 s, θ = 30° 1 1 Bl Bl
⇒ µ = 1− ⇒ 1 − µk = 2
n2 n
Hence, F = 2 × 0.5 × 12 × sin 30° = 24N 1
0.25 or µ k = 1 − 2 d d
n
13. (3) For translational equilibrium,
16. (4) Maximum force of friction between Rough surface Smooth surface
N = mg cos θ and mg sin θ = f ground and B is 45
0
45
0
Laws of Motion 35

 1 m1v1 = m2v2 3
µ = 1 − 2  tan θ 4µ cos θ = 3sin θ ⇒ µ = tan θ
 n  1  1 2
4
Hence, m1  m1v12  = m2  m 2 v 2 
 1 2  2  25. (1) According to above question,
⇒ µ = 1 − 2  tan 45° 1
 2  2
and m1 v1 = µm1gs1 M 5
2 m = 2 − M1 = – 10 = 15 kg
µ 0.2
4 −1 1
µ=
4 m 2 v 22 = µm2gs2 26. (4) Extension in the spring is:
2
3 x = AB – R
⇒µ= Therefore, m1 (µm1 gs1) = m2 (µm2 gs2)
4 = 2 R cos 30° – R = ( 3 – 1)R
20. (1) Limiting value of static friction  m2 
∴ s2 =  12  s1. ∴ Spring force,
force on B = µsmBg  m2 
( 3 + 1 mg )
fL = 0.5 × 7 × 10 = 35 N
23. (3) When a horizontal acceleration is
F = kx =
R
( )
3 − 1 R = 2 mg
As F = 100 N
imparted to the inclined plane, a pseudo
Hence, F > fL acceleration acts on the object.
N

and the two bodies will move in the same B

F
direction, (i.e., of applied force) b u t 30°
with different acceleration. Here, force 30° mg
30° 60°
of friction µkmBg will oppose the
motion of B which will cause the motion
of A. So, the equation of motion of block
B will be,
F – µk mBg = mBaB From figure,
For equilibrium of the body on the inclined N = (F + mg) cos 30°
(F − µ k m B g)
i.e., a B = plane, 3 3mg
mB 3
1 = (2 mg + mg) =
ma cos θ = mg sin θ and sin θ = 2 2
(100 − 0.4 × 7 × 10) 
= 27. (1) Force on the may can be rested into
7 1
⇒ a = g tan θ = g
2 − 1 components according to the diagram
100 − 28 72
= = = 10.3 ms –2
7 7 24. (1) The free body diagram shows the d/2
While for body A, fk = mAaA, F F
various forces acting on the block. When
h
µ k m B g 0.4 × 7 × 10 the plane is frictionless, the acceleration F θ F
i.e., a A = = = 0.8ms –2 .
mA 35 of the block, a = g sin θ.
F Since, body starts from rest, initial
21. (1) ∴ F = 300 N so, = 75 N
4 velocity u = 0.
F
or < 100 N mg
4
2F cos θ = mg
Therefore, aM = 0
mg
F = 300 N ∴F=
2cos θ

d2
mg h 2 +
Let time taken is t1. = 4
F/2 F = 75 N
4 2h
F 1 1
∴ s = ut + at12 ⇒ s = gsin θt12
2 2 2 mg 2 2
= 4h d + 4h
F F 2s
4 4 ⇒ t1 = 28. (2) tanθ = μ,
gsin θ
When plane is made rough, then
acceleration a = g (sin θ – μ cos θ), where
50 N 100 N
μ is coefficient of friction.
acceleration of the block of mass m is 1
2s 2s cos θ =
given by ∴ t2 = =2 µ2 + 1
g(sin θ − µ cos θ) gsin θ
75 − 50 h = r – r cosθ
am = = 5m/s2
5 1 4
=
sin θ − µ cos θ sin θ  1 
22. (3) According to law of conservation of h = r [ 1 – cosθ] = r 1 − 
momentum ⇒ sin θ = 4sin θ − 4µ cos θ  µ 2 + 1 
36 Laws of Motion

29. (2) It follows from the figure that C. Newtons IIIrd Law of motion → B. T
hink of Newton’s third law of
F Action – reaction pair (Nature of motion.
force)
M C. mg < fms or mg < ms R
D. Rate of change of linear momentum is
or mg < msma
T θ T define as force.
or g < msa
33. (2) FBD of each block: g
B A
20N 20N 20N 20N or msa > g or µ s >
(–x, 0) (x, 0)O a
Replace ms by m
F
F = 2T cos θ or T = 1 kg 2 kg 3 kg 4 kg D. 
The jumping away of the man
2cos θ
involves upward acceleration.
Force responsible for motion of the
masses on the x-axis is T sin θ.
10N 20N 30N 40N 37. (4)
20 − 10
F A. a1 = = 10 m s 2 ( UP ) 38. (3) 32 – f = 2a …(i)
Hence, ma = T sin θ = · sin θ 1 f – 20 = a …(ii)
2cos θ 20 − 20
B. a 2 = = 0 ( Stationary )
F x 2 Adding (i) and (ii), 12 = 3a
F F OB 30 − 20 10
= tan θ = · = · 2 C. a 3 = = m s 2 ( Down ) or a = 4m s–2
2 2 OM 2 a − x2
3 3
F x 40 − 20 From (ii), f = (20 + 4) N = 24N
·
D. a 4 = = 5 m s 2 ( Down )
∴a= . 4
2m a 2 − x 2 39. (4)
(Note: Consider 80 N instead of 60 N in
30. (3) Given, u = (3iˆ + 4 ˆj )m /s and question) 40. (4) a = 5 – 1 g = 4g = 2g
v = −(3iˆ + 4 ˆj )m /s 5 +1 6 3
34. (4) Total force acting (upside) = 60 N
Mass of the ball = 150 g = 0.15 kg 2 × 5 ×1 5g
Total force acting (downside) T= g=
5 +1 3
Δp = Change in momentum of the ball 1 1 1
= 30 × + 20 × + 10 × + 18 Spring balance reading
Δp = Final momentum – Initial 2 2 2 2T 2 × 5g 10
momentum = m (v–u) = 48 N = = = kgf
g 3g 3
= (0.15)[−(3iˆ + 4 ˆj ) − (3iˆ + 4 ˆj )] A. Acceleration of 2 kg block
41. (4) (B) F = v dm
60 − 48 12 dt
= (0.15)[−6iˆ − 8 ˆj ] = = = 2 m s2
6 6 dm
Both v and (m) are doubled.
∴ ∆p = −[0.9iˆ + 1.20 ˆj ] B. Net force on 3 kg block: dt
For direction, use Newton’s third law of
31. (1) Fnet = 3a = 2 × 6 = 6 N motion.
(µ.N)
on n C. Normal reaction between 2 kg and 1 kg: (C) Volume/time = av
Normal (N) oti tio
M Fric N – 18 – g sin 30° = a Mass/time = avρ
m N = 18 + 5 + 2 = 25 N Momentum/time = αν2ρ
D.  ormal reaction between 3 kg and
N Initial rate of change of momentum
  along horizontal
sin mg cos 2 kg:
mg mg = av2ρcosq
60 – 3g sin 30° – N´ = 3a
Final value is also av2ρcosq with
From FBD: Normal (N) = mg cosq N´ = 60 – 15 – 6 = 39 N. negative sign.
35. (2) Coefficient of friction (µ) for different Now think of force.
Frictional force (f) = µN = µmg cosq
frictions: 42. (4) T1
Force in downside (F) = mg sinq + µ mg T1 cos θ1
Limiting friction > Static friction > θ1
cosq
Kinetic friction > Rolling friction
A. Net force in vertical direction = F sinq
= mg sin2q + µ mg sinq cosq. 36. (4) T1 sin θ1 T2 sin θ2
B. Net force in horizontal direction = F f
cosq
T2 cos θ2
= mg sinq.cosq + µ mg cos2q
ma mg
C. Net force along the plane = F cos 0° = F R
= mg sinq + µmg cosq T2 cos θ2
T2
D. Net force perpendicular to plane = F
cos 90° = 0 θ2
32. (3) mg
A. Newton’s Ist Law of motion → Law T2 sin θ2
mg
of Inertia.
B. Newton’s IInd Law of motion → Real A. g' = g + a = g + 3g = 4g
law of motion mg' = 4mg = 4 W mg
Laws of Motion 37

T2 cos q2 = mg 55. (1) Rolling friction is always less than

or f = 1 mg (2 sin q cos q)
T2 sin q2 = mg 2 sliding friction as the surface area in contact
is very less in that case which reduces the
Dividing tan q2 = 1 1
or f= mg sin 2θ friction which is opposing the motion so
T22[sin2 q2 + cos2 q2] = 2m2g2 2
providing ball bearing in motor’s shaft
For maximum f, the value of sin 2q is minimises friction and they increases its
T2 = 2 mg
maximum i.e., 1. output power.
T1 sin q1 = T2 sin q2 For this, q = 45° R
or T1 sin q1 = mg
1
Again T1 cos q1 = mg + T2 cos q2 \ Maximum frictional force is mg
2
T1 cos q1 = mg + mg
a
47. (3) Using Snth = u + (2n − 1), We get
T1 cos q1 = 2mg 2
T1 sin θ1 mg 1 a 9a sq
= = 13 = u + (2 × 5 − 1) = u + co q
T1 cos θ1 2mg 2 2 2 g mg sin q
M
Again T12 sin2 q1 + T12 cos2 q1 a 11a
15 = u + (2 × 6 − 1) = u +
2 2
= (mg)2 + (2mg)2 = 5m2g2
Subtracting, 2 = a
= T1 = 5 mg mg
2
Again, 13 = u + ×9
43. (3) 2 56. (1) Here the impulse is sharp so the
or u = 4 m s–1
44. (1) a = m1 – m 2 g brick is broken with a bare hand
m1 + m 2 Again, F = ma = 1 kg × 2 m s–2 = 2 N
57. (1) The tension is greater than weight
v = at 48. (1) 49. (3) of bird because the vertical component
Relative velocity = 2at 50. (1) Net force = (m1 – m2)g of the tension will be equal to the
Total mass = m1 + m2 weight of the bird, i.e., Tcosq = W
m1 – m 2
=2 gt, v = 2ah m − m2 \ cosq < 1, i.e., T > W.
m1 + m 2 Acceleration = 1 g
m1 + m 2 58. (1) Rolling friction is always less than
Relative velocity = 2 2ah sliding friction as the surface area in
Heavier mass (m1 shall move down with
m1 – m 2 this acceleration. The lighter mass (m2) contact is very less in that case which
2 2 gh reduces the friction which is opposing
m1 + m 2 shall move up with this acceleration.
the motion so providing ball bearing in
45. (2) (a) The mere fact that the four forces 51. (2) Note that spring balance measures motor’s shaft minimises friction and
act in the same plane does not lead to weight. The common physical balance they increases its output power.
zero resultant. compares masses. When the elevator
59. (3) When a cat falls from more than
(b) Equal magnitudes not given. In both accelerates upwards, there is an apparent
6 to 8 floors, it gets enough velocity or
(c) and (d), the net force is zero. increase in weight but mass is not
the air drag has enough time to build
changed.
46. (4) Consider prism of mass M. up to counteract the weight force of
52. (1) In rolling friction, surfaces in the cat. Hence, there by reducing the
N = Mg + R cos q
contact get momentarily deformed a injury. Also the drag force is directing
f = R sin q little, and this results in a finite area of proportional to area and square of
the point in contact with the surface and velocity.
hence motion is opposed by component
R sin q 60. (3) Air drag is the resistance offered by the
of contact force parallel to surface.
air to the body in motion which increases
Hence, rolling sphere on horizontal
q with increase in area or vice versa.
surface eventually stops.
R R cos q 61. (2) Both of the statement is false as each
53. (1) It is very difficult to produce
f q particle exerts a force on other particle.
slipping motion than to actual rolling.
In bicycle when brakes are on, the
cycles can only do the slipping action NEET Past 10 Year
N
Mg and not rolling action also μ (sliding) Questions
> μ (rolling).
Let us now consider mass m. 1. (4) N = mg and f = ma
54. (2) Both the assertion and reason are
R = mg cos q correct but the reason doesn’t correctly In the frame of car
a N
mg sin q = ma explain the assertion. The tension on
the rope will be uniform throughout the mamax f
or a = g sin q
rope. Therefore, the force or the pull of
From equation (i), f = mg cos q sin q the teams are same. mg