##### Document Text Contents

Page 1

Study Material

Laws of Motion & Friction

Nothing in JEE will beyond this model questions in Newton’s Laws of motion & friction

Note:

1) Spring force:

→→

−= xkF x is displacement of the free end from its natural length or deformation of the spring

where K=spring constant.

2) Spring Property: l×K = constant

= Natural length of spring.

3) If spring is cut into two ratio m:n then spring constant is given by

nm

n

nm

m

+

=

+

=

l

l

l

l 21 ; 2211 lll kkk ==

For series combination of springs ........

111

21

++=

kkkeq

For parallel combination of spring 321 kkkkeq ++ …….

4) Spring Balance: It does not measure the weight. It measures the force exerted by the object at the hook.

5) String Constraint: When two objects are connected through a string and if the string have the following

properties.

a) The length of the string remains constant i.e. inextensible string.

b) Always remains tight, does not slacks.

Then the parameters of the motion of the objects along the length of the string and in the direction of extension

have definite relation between them.

6) Wedge Constraint: Conditions:

i) There is a regular contact between two objects

ii) Objects are rigid.

The relative velocity perpendicular to the contact plane of the two rigid objects is always zero if there is a

regular contact between the objects. Wedge constraint is applied for each contact.

Page 2

Study Material

In other words,

Components of velocity along perpendicular direction to the contact plane of the two objects is always equal if

there is no deformations and they remain in contact.

7) Newton’s Law for a System:

..........332211 +++=

→→→→

amamamF ext

=

→

extF Net external force on the system.

321 ,, mmm are the masses of the objects of the system and

321 ,,

→→→

aaa are the acceleration of the objects respectively

8) Newton’s Law for non inertial frame:

→→→

=+ amFF PseudoalRe

Net sum of real and pseudo force is taken in the resultant force.

→

a = Acceleration of the particle in the non inertial frame

FramePseudo amF

→→

−=

Pseudo force is always directed opposite to the direction of the acceleration of the frame.

Pseudo force is an imaginary force and there is no action-reaction for it. So it has nothing to do with Newton’s

Third Law

Reference Frame:

A frame of reference is basically a coordinate system in which motion of object is analyzed. There are two

types of reference frames.

a) Inertial reference frame: Frame of reference moving with constant velocity.

b) Non-inertial reference frame: A frame of reference moving with non-zero acceleration.

Page 29

Study Material

surface

→

BSfB are

B) A particle is performing SHM about x = 0, it crosses a position Q) Different in magnitude

x = x0 at t = t1 and at t = t2, there is no other crossing at x = x0

between t2 – t1. If 1

→

v and 2

→

v are velocity of particle at t1 and

t2 respectively, then 1

→

v and 2

→

v are

C) Two points A and B are marked on the axis of a uniformly R) Same in direction

charged ring at same distance from centre on two sides as

shown. If 1

→

E and 2

→

E are electric field intensities at A and

B respectively, then 1

→

E and 2

→

E are

D) A single turn carrying current of 1A is considered. If

→

A S) Opposite in direction

is its area vector and

→

M is it corresponding magnetic moment,

then

→

A and

→

M are

81. A block of mass 10kg is in equilibrium as shown in the figure. If the block is displaced in vertical direction by

a small amount, then angular frequency of resulting motion is

Page 30

Study Material

a) 210 b) 510 c) 25 d) 55

82. A 5kg ring attached to a spring slides along a smooth fixed rod which is inclined at an angle of 60

0

with

horizontal, as shown in figure. The spring is in its relaxed state when ring is at A, the natural length of spring is

0.5m. The value of spring constant k, so that ring is having zero velocity at B is

a) 500 N/m b) 323 N/m c) 484 N/m d) 286 N/m

Only One Correct Option Questions:

83. In the figure shown the acceleration of A is,

^^

1515 jia A +=

→

then the acceleration of B is: (A remains in

contact with B)

a) 6

^

i b) -15

^

i c) -10

^

i d) -5

^

i

84. Two blocks A and B of masses m & 2m respectively are held at rest such that the spring is in natural length.

Find out the accelerations of both the blocks just after release.

Page 57

Study Material

Laws of Motion & Equilibrium

Answers

121)

θ

θ

2

cos

cos

Mm

Mg

+

145)

( )( )

g

gawW

3

++

; only for 2W > w

122) (a) 2.5m/s

2

(b) 112.5 N (c) 87.5 N 146)2.52N

123)

( )ag

ma

m

+

=∆

2

147) (a) 9.81s, (b) 19.62s,

(c) 15.19ms

2

, 2.69 m/s

2

.

124) (a) 0, 0 (b) 0, 0 (c) 0.05m/s

2

, 9.9m/s

2

148) (a) 117.7N, (b) 147N, (c) m2slips on M

125)

2/

5

2

sm

g

149)

0120,

2

cos2

>= α

α

F

T

126) 1.28 m/s 150) αcos

2

P

T =

127) 1.66m. 151)

α

α

2

cos

sin2

×=

l

hP

F

128) ( )[ ]R

Rg

a /cos1 l

l

−= 152) F = 1.4 kgf.

129) (a) 3.00N (b) 4.00N (c)5.00N 153) 50

4

==

l

hP

F kgf.

130) MC = 10kg, a = 1.96 m/s

2

131)

( ) ( )

++

+

=

++

+−

=

210

20

210

210 1;

mmm

gmm

Tg

mmm

mmm

a

µµ

132) (i) 5.75 m/s

2

(ii) T1 =17.38N, T2 = 40.5N

133) T = 1.05s,

049=α

134) x

L

g

v =

135)

r

mvmg

N

22

2

−=

136) αtan

2

21

21

+ mm

mm

137) 10N-m.

Page 58

Study Material

138) N26 inclined at 450 to the side AB.

139) (a) Bottom: 2W; side; W (b) 2 W

140) 1923.08N

141) 1≥µ

142)

α

µ

sin

W

143) (a) 196N (b) h=1.25m (c) 0.4m to the right

144) (a) 100N, 700N (b) 2.0m

Study Material

Laws of Motion & Friction

Nothing in JEE will beyond this model questions in Newton’s Laws of motion & friction

Note:

1) Spring force:

→→

−= xkF x is displacement of the free end from its natural length or deformation of the spring

where K=spring constant.

2) Spring Property: l×K = constant

= Natural length of spring.

3) If spring is cut into two ratio m:n then spring constant is given by

nm

n

nm

m

+

=

+

=

l

l

l

l 21 ; 2211 lll kkk ==

For series combination of springs ........

111

21

++=

kkkeq

For parallel combination of spring 321 kkkkeq ++ …….

4) Spring Balance: It does not measure the weight. It measures the force exerted by the object at the hook.

5) String Constraint: When two objects are connected through a string and if the string have the following

properties.

a) The length of the string remains constant i.e. inextensible string.

b) Always remains tight, does not slacks.

Then the parameters of the motion of the objects along the length of the string and in the direction of extension

have definite relation between them.

6) Wedge Constraint: Conditions:

i) There is a regular contact between two objects

ii) Objects are rigid.

The relative velocity perpendicular to the contact plane of the two rigid objects is always zero if there is a

regular contact between the objects. Wedge constraint is applied for each contact.

Page 2

Study Material

In other words,

Components of velocity along perpendicular direction to the contact plane of the two objects is always equal if

there is no deformations and they remain in contact.

7) Newton’s Law for a System:

..........332211 +++=

→→→→

amamamF ext

=

→

extF Net external force on the system.

321 ,, mmm are the masses of the objects of the system and

321 ,,

→→→

aaa are the acceleration of the objects respectively

8) Newton’s Law for non inertial frame:

→→→

=+ amFF PseudoalRe

Net sum of real and pseudo force is taken in the resultant force.

→

a = Acceleration of the particle in the non inertial frame

FramePseudo amF

→→

−=

Pseudo force is always directed opposite to the direction of the acceleration of the frame.

Pseudo force is an imaginary force and there is no action-reaction for it. So it has nothing to do with Newton’s

Third Law

Reference Frame:

A frame of reference is basically a coordinate system in which motion of object is analyzed. There are two

types of reference frames.

a) Inertial reference frame: Frame of reference moving with constant velocity.

b) Non-inertial reference frame: A frame of reference moving with non-zero acceleration.

Page 29

Study Material

surface

→

BSfB are

B) A particle is performing SHM about x = 0, it crosses a position Q) Different in magnitude

x = x0 at t = t1 and at t = t2, there is no other crossing at x = x0

between t2 – t1. If 1

→

v and 2

→

v are velocity of particle at t1 and

t2 respectively, then 1

→

v and 2

→

v are

C) Two points A and B are marked on the axis of a uniformly R) Same in direction

charged ring at same distance from centre on two sides as

shown. If 1

→

E and 2

→

E are electric field intensities at A and

B respectively, then 1

→

E and 2

→

E are

D) A single turn carrying current of 1A is considered. If

→

A S) Opposite in direction

is its area vector and

→

M is it corresponding magnetic moment,

then

→

A and

→

M are

81. A block of mass 10kg is in equilibrium as shown in the figure. If the block is displaced in vertical direction by

a small amount, then angular frequency of resulting motion is

Page 30

Study Material

a) 210 b) 510 c) 25 d) 55

82. A 5kg ring attached to a spring slides along a smooth fixed rod which is inclined at an angle of 60

0

with

horizontal, as shown in figure. The spring is in its relaxed state when ring is at A, the natural length of spring is

0.5m. The value of spring constant k, so that ring is having zero velocity at B is

a) 500 N/m b) 323 N/m c) 484 N/m d) 286 N/m

Only One Correct Option Questions:

83. In the figure shown the acceleration of A is,

^^

1515 jia A +=

→

then the acceleration of B is: (A remains in

contact with B)

a) 6

^

i b) -15

^

i c) -10

^

i d) -5

^

i

84. Two blocks A and B of masses m & 2m respectively are held at rest such that the spring is in natural length.

Find out the accelerations of both the blocks just after release.

Page 57

Study Material

Laws of Motion & Equilibrium

Answers

121)

θ

θ

2

cos

cos

Mm

Mg

+

145)

( )( )

g

gawW

3

++

; only for 2W > w

122) (a) 2.5m/s

2

(b) 112.5 N (c) 87.5 N 146)2.52N

123)

( )ag

ma

m

+

=∆

2

147) (a) 9.81s, (b) 19.62s,

(c) 15.19ms

2

, 2.69 m/s

2

.

124) (a) 0, 0 (b) 0, 0 (c) 0.05m/s

2

, 9.9m/s

2

148) (a) 117.7N, (b) 147N, (c) m2slips on M

125)

2/

5

2

sm

g

149)

0120,

2

cos2

>= α

α

F

T

126) 1.28 m/s 150) αcos

2

P

T =

127) 1.66m. 151)

α

α

2

cos

sin2

×=

l

hP

F

128) ( )[ ]R

Rg

a /cos1 l

l

−= 152) F = 1.4 kgf.

129) (a) 3.00N (b) 4.00N (c)5.00N 153) 50

4

==

l

hP

F kgf.

130) MC = 10kg, a = 1.96 m/s

2

131)

( ) ( )

++

+

=

++

+−

=

210

20

210

210 1;

mmm

gmm

Tg

mmm

mmm

a

µµ

132) (i) 5.75 m/s

2

(ii) T1 =17.38N, T2 = 40.5N

133) T = 1.05s,

049=α

134) x

L

g

v =

135)

r

mvmg

N

22

2

−=

136) αtan

2

21

21

+ mm

mm

137) 10N-m.

Page 58

Study Material

138) N26 inclined at 450 to the side AB.

139) (a) Bottom: 2W; side; W (b) 2 W

140) 1923.08N

141) 1≥µ

142)

α

µ

sin

W

143) (a) 196N (b) h=1.25m (c) 0.4m to the right

144) (a) 100N, 700N (b) 2.0m