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Laws of Motion

Equation of Motion and

Newton’s Laws of Motion 4 kg
6 kg

1. A small block slides down on a smooth inclined plane, a. g/2 b. g/5

starting from rest at time t = 0. Let Sn be the distance travelled c. g/10 d. g
by the block in the interval t = n – 1 to t = n. The, the ratio
Sn 5. A block of mass m is placed on a smooth inclined wedge ABC
is: (2021)
Sn +1 of inclination θ as shown in the figure. The wedge is given an
2n − 1 2n + 1 acceleration ‘a’ towards the right. The relation between a and
a. b.
2n + 1 2n − 1 θ for the block to remain stationary on the wedge is:  (2018)
2n 2n − 1
c. d.
2n − 1 2n
2. A particle moving with velocity V is acted by three forces
shown by the vector triangle PQR. The velocity of the particle
will:  (2019)
g
a. a = g cos θ b. a =
sin θ
g
c. a d. a = g tan θ
cosec
6. Two blocks A and B of masses 3 m and m
respectively are connected by a massless and
a. Increase inextensible string. The whole system is
b. Decrease suspended by a massless spring as shown in
c. Remain constant figure. The magnitudes of acceleration of A
d. Change according to the smallest force and B immediately after the string is cut, are
3. A balloon with mass m is descending down with an acceleration respectively: (2017-Delhi)
a (where a < g). How much mass should be removed from it g
so that it starts moving up with an acceleration a? (2014) a. g, b. g, g
3
2ma 2ma g g
a. b.
g−a
c. , d. g ,g
g+a 3 3 3
ma ma
c. d. 7. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg
g+a g−a respectively, are in contact on a friction less surface, as
shown. If a force of 14 N is applied on the 4 kg block, then
Motion of Connected Bodies the contact force between A and B is: (2015)

4. Two bodies of mass 4 kg and 6 kg are tied to the ends of

a massless string. The string passes over a pulley which is
frictionless (see figure). The acceleration of the system in a. 6 N b. 8 N
terms of acceleration due to gravity (g) is: (2020) c. 18 N d. 2 N
12 Chapter & Topicwise NEET PYQ's PW

8. A system consists of three masses m1, m2 and m3 is connected

by a string passing over a pulley P. The mass m1 hangs
freely and m2 and m3 are on a rough horizontal table (the
coefficient of friction = μ). The pulley is frictionless and of
negligible mass. The downward acceleration of mass m1 is:
(Assume m1 = m2 = m3 = m) (2014)

mV mV
a. b.
2 3
c. mv d. 2mv
13. The force F acting on a particle of mass m is indicated by the
g (1 − gµ ) 2gµ force-time graph shown. The change in momentum of the
a. b.
9 g particle over the time interval from zero to 8 s is: (2014)

g (1 − 2µ ) g (1 − 2µ )
c. d.
3 2
9. Three blocks with masses m, 2m and 3m are connected by
strings, as shown in the figure. After an upward force F is
applied on block m, the masses move upward at constant
speed v. What is the net force on the block of mass 2m? a. 24 Ns b. 20 Ns
(g is the acceleration due to gravity) (2013) c. 12 Ns d. 6 Ns

Friction

14. Calculate the acceleration of the block and trolly system

shown in the figure. The coefficient of kinetic friction
between the trolly and the surface is 0.05. (g = 10 m/s2,
a. 6 mg b. Zero mass of the string is negligible and no other friction exists).
c. 2 mg d. 3 mg  (2020-Covid)

Conservation of Momentum and

Impulse Momentum Theory
a. 1.50 m/s2 b. 1.66 m/s2
10. A shell of mass m is at rest initially. It explodes into three
c. 1.00 m/s 2
d. 1.25 m/s2
fragments having mass in the ratio 2 : 2 : 1. If the fragments
having equal mass fly off along mutually perpendicular 15. Which one of the following statements is incorrect? (2018)
directions with speed v, the speed of the third (lighter) a. Frictionless force opposes the relative motion.
fragment is: [RC] (2022) b. Limiting value of static friction is directly proportional
to normal reaction.
a. 3 2 v b. v
c. Rolling friction is smaller than sliding friction.
c. 2v d. 2 2 v d. Coefficient of sliding friction has dimensions of length.
11. A ball of mass 0.15 kg is dropped from a height 10 m, strikes 16. A block A of mass m1 rests on a horizontal table. A light
the ground and rebounds to the same height. The magnitude string connected to it passes over a frictionless pulley at the
of impulse imparted to the ball is (g = 10 m/s2) nearly: edge of table and from its other end another block B of mass
 (2021) m2 is suspended. The coefficient of kinetic friction between
the block and table is µk. When the block A is sliding on the
a. 4.2 kg m/s b. 2.1 kg m/s
table, the tension in the string is: (2015)
c. 1.4 kg m/s d. 0 kg m/s
a. ( 2 k 1 )
m − µ m g b.
m m
1 2 (1 + µ k ) g
12. A rigid ball of mass m strikes a rigid wall at 60° and gets
( m1 + m 2 ) ( m1 + m 2 )
reflected without loss of speed as shown in the figure below.
The value of impulse imparted by the wall on the ball will be: m1m 2 (1 − µ k ) g ( m 2 + µ k m1 ) g
c. d.
 (2016 - II) ( m1 + m 2 ) ( m1 + m 2 )
Laws of Motion 13

17. A plank with a box on it at one end is gradually raised about 10

the other end. As the angle of inclination with the horizontal a. 10 rad / s b. rad / s
2π
reaches 30°, the box starts to slip and slides 4.0 m down the c. 10 rad/s d. 10 π rad/s
plank in 4.0 s. The coefficients of static and kinetic friction
20. One end of string of length l is connected to a particle of mass
between the box and the plank will be, respectively:
‘m’ and the other end is connected to a small peg on a smooth
 (2015 - Re)
horizontal table. If the particle moves in circle with speed ‘v’,
the net force on the particle (directed towards center) will be:
(T represents the tension in the string) (2017-Delhi)
mv 2 mv 2
a. T + b. T −
l l
a. 0.4 and 0.3 b. 0.6 and 0.6 c. Zero d. T
c. 0.6 and 0.5 d. 0.5 and 0.6 21. A car is negotiating a curved road of radius R. The road is
banked at an angle θ. The coefficient of friction between the
18. The upper half of an inclined plane of inclination θ is perfectly
tyres of the car and the road is µs. The maximum safe velocity
smooth while lower half is rough. A block starting from rest
on this road is:  (2016 - I)
at the top of the plane will again come to rest at the bottom,
if the coefficient of friction between the block and lower half µs + tan θ
a. gR 2
of the plane is given by (2013) 1 − µs tan θ
a. μ = 2 tan θ b. μ = tan θ
µs + tan θ
1 2 b. gR
c. µ = d. µ = 1 − µs tan θ
tan θ tan θ
gµs + tan θ
c.
R1 − µs tan θ
Circular Motion, g µs + tan θ
d.
Banking of Road R 2 1 − µs tan θ
22. Two stones of masses m and 2 m are whirled in horizontal
19. A block of mass 10 kg is in contact against the inner wall of circles, the heavier one in a radius r/2 and the lighter one
a hollow cylindrical drum of radius 1 m. The coefficient of in radius r. The tangential speed of lighter stone is n times
that of the value of heavier stone when they experience same
friction between the block and the inner wall of the cylinder
centripetal forces. The value of n is: (2015 Re)
is 0.1. The minimum angular velocity needed for the cylinder
to keep the block stationary when the cylinder is vertical and a. 1 b. 2
rotating about its axis, will be: (g = 10 m/s2) (2019) c. 3 d. 4

Answer Key
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
a c a b d d a c b d a c c d d b c
18 19 20 21 22
a c d b b
14 Chapter & Topicwise NEET PYQ's PW

Explanations
1. (a) Using equation of uniform motion, 5. (d) From wedge reference frame
a ma cos q N
Sn =
u+ ( 2n − 1) where,
2
Sn is the distance covered in nth sec
[from t = (n – 1) to t = n] ma
Sn + 1 is the distance covered in (n + 1) sec
th
mg sin q
[from t = n to t = (n + 1)] mg
Now; u = 0 (FBD of mass m)
a For block to be at rest
Sn
= ( 2n − 1) ...(i)
2 ma cos θ = mg sin θ
a a a = g tan θ
S=
n +1  2 ( n + 1) − 1=
 [ 2n + 2 − 1]
2 2
6. (d)    
a
Sn +1
= [ 2n + 1] ...(ii)
2
Dividing eqn (i) and eqn (ii)
a
Sn   (2n − 1) When the string is cut T = 0
⇒ =2
Sn +1  a  for block A
  (2n + 1) 3 m a = 4 mg – 3 mg
2
g
Sn 2n − 1 a=
⇒ = 3
Sn +1 2n + 1
for block B
2. (c) All these forces are forming closed loop in same order. mg = ma ⇒ a = g
So, net force is zero. Fnet
P 7. (a) Acceleration of system =
M total
14
= = 2ms −2
4 + 2 +1

R Q
Force = ma N1 = Normal force between A and B
 14 – N1 = mAa
dv 
⇒ m = 0 ⇒ v = constant 14 – N1 = 4 × 2
dt
3. (a) For downward motion 14 – N1 = 8 ⇒ N1 = 6 N
mg – Fa = ma Net forcein thedirection of motion
8. (c) Acceleration =
⇒ Fa = mg – ma Total massof system
If some mass Δm is removed, then it starts m1g − µ ( m 2 + m3 ) g g (1 − 2µ )
accelerating upwards
= = [ m=1 m=2 m3 ]
m1 + m 2 + m3 3
Fa – (m – Δm)g = (m – Δm)a 9. (b) As block of mass 2m moves with constant velocity so net
mg – ma – mg + gΔm = ma – a Δ m force on it is zero.
gΔm – ma = ma – aΔm
10. (d)
⇒ Δm [g + a] = 2ma v
2m
2ma 2m
∆m = 5
g+a 5
( m 2 − m1 ) g v
4. (b) a =
m1 + m 2
 m/5
(6 − 4) g
2g g v′
a
= = =
6+4 10 5
Laws of Motion 15

As the shell was initially at rest, therefore initial momentum For the motion of both blocks
is zero. m2g – T = m2a
Using conservation of momentum : ( m 2 − µ k m1 ) g
T − µ k m=
1g m1a ⇒=
a
Initial momentum = Final momentum m1 + m 2
2m ˆ + 2m (− vi) 
ˆ + m v ' ⇒ = For the block of mass ‘m2’
0= (− vi) v′ 2v ˆi + 2v ˆj
5 5 5
 m − µ k m1 
 m 2g − T =
m2  2 g
⇒ v'
= (2v)2 + (2v)2 = 2 2 v  m1 + m 2 
11. (a) Given, mass of ball, m = 0.15 kg  m − µ k m1   m + µ k m1 
T= m 2g −  2  m 2g =
m 2g  1 
Height from which it is dropped, h = 10 m  m1 + m 2   m1 + m 2 
Velocity of ball just before reaching the ground, v = 2gh m m (1 + µ k ) g
⇒ T =1 2
m1 + m 2
⇒v= 2 × 10 × 10 = 200 = 10 2 m / s
According to the question, height remaining the same. So, 17. (c) Coefficient of static friction,
velocity will also, only its direction changes. 1
µ= tan 30=
° = 0.6
( ) ()
s
i.e., vi = 10 2 −ˆj and vf =
10 2 ˆj 3
Required magnitude of Impulse imported to the ball = change a = gsin 30° - µkg cos 30°
in linear momentum 1
  ut + at 2
S= [u =
0]
I = ∆P 2
 1  g µkg 3 
=I mvf − mvi 4
⇒=  −  × 16
2  2 2 
= |mvf – m(–vf)| = |mvf + mvf| = |2 mvf| = 2 × 0.15 × 10 2 ⇒ µ k =0.5

I = 3 2 kg m/s 18. (a)    

I = 4.2 kg m/s
 
12. (c) Impulse = ∆P = m∆v = m ( 2v cos60° ) = mv
13. (c) Change in momentum,
∫Δp = ∫Fdt = Area of F – t graph
1
= × 2 × 6 − 3× 2 + 4 × 3
2
= 12 Ns Let m be mass of the block and L be length of the inclined
14. (d)     plane.
For upper half smooth plane
Acceleration of the block, a = gsin θ
Here, u = 0 ( block starts from rest)
Using, v2 – u2 = 2as, we have
15. (d) Coefficient of sliding friction is dimensionless.
L
F v 2 − 0 = 2 × gsin θ ×
Q F = mN ⇒ m = 2
N =v gLsin θ ......(1)
(F and N have same unit i.e., Newton)
For lower half rough plane
16. (b)   
Acceleration of the block, a' = gsin θ – μgcos θ
where μ is the coefficient of friction between the block and
lower half of the plane
Here, u= v= gLsin θ
v = 0 ( block comes to rest)
L
a = a' = gsin θ – μgcos θ, s =
2
16 Chapter & Topicwise NEET PYQ's PW

Again, using v2 – u2 = 2as, we have

g
L ω≥
( )
2
0− gLsin θ = 2 × ( gsin θ − µg cos θ ) × rµ
2
Therefore,
– gLsin θ = (gsin θ – μg cos θ)L
– sin θ = sin θ – μcos θ g
ωmin =
μ cos θ = 2sin θ rµ
⇒ μ = 2tan θ 10
ω=
min = 10 rad / s
19. (c)       0.1× 1
20. (d) In uniform circular motion, tension provides the necessary
centripetal force required to keep particle in motion.
So, net force on particle = T.

v 2  µs + tan θ 
21. (b) We know that = 
gR  1 − µs tan θ 

 µ + tan θ 
v = gR  s 
 1 − µs tan θ 
For equilibrium of the block limiting friction
22. (b) (FC)heavier = (FC)lighter ⇒ 2mV = m ( nV ) ⇒ n 2 = 4 ⇒ n = 2
2 2
fL ≥ mg ⇒ µN ≥ mg
⇒ µmrω2 ≥ mg ( r/2 ) r