# Chapter 8 Law of Motion Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 8 Law of Motion Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes

Problem 33.Â Â Â Â  A man weighs

80kg.

He stands on a weighing scale in a lift which is moving upwards with a uniform

acceleration of 5m / s 2. What would be the reading on the scale. (g = 10m / s 2 )

(a) 400 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 800 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 1200 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) Zero

Solution : (c)Â Â Â Â Â  Reading of weighing scale = m(g + a) = 80 (10 + 5) = 1200N

[CBSE 2003]

Problem 34.Â Â Â Â  A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity â€˜gâ€™, the reading on the spring balance will be

(a) 2 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) (4 Â´ g) kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) (2 Â´ g) kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) Zero

Solution : (d)

R = m(g a)

= (g g) = 0

[because the lift is moving downward with a = g]

Problem 35.Â Â Â Â  In the above problem, if the lift moves up with a constant velocity of 2 m/sec, the reading on the balance will be

(a) 2 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 4 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) ZeroÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 1 kg

Solution : (a)

R = mg

= 2g Newton or

2kg

[because the lift is moving with the zero acceleration]

Problem 36.Â Â Â Â  If the lift in problem, moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will be

(a) 2 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) (2 Â´ g) kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) (4 Â´ g) kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 4 kg

Solution : (d)

R = m(g + a)

= m(g + g)

[because the lift is moving upward with a =g]

= 2mg R = 2 Â´ 2g N = 4 g N or 4 kg

Problem 37.Â Â Â Â  A man is standing on a weighing machine placed in a lift, when stationary, his weight is recorded as 40 kg. If the lift is accelerated upwards with an acceleration of 2 m / s 2 , then the weight recorded in the machine will be

(g = 10 m / s 2 )

(a) 32 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 40 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 42 kgÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 48 kg

[MP PMT 1994]

Solution : (d)

R = m(g + a) = 40(10 + 2)

= 480 N

or 48kg

Problem 38.Â Â Â Â  An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 5ms -2 . Taking g to be

10ms 2 , then the tension in the cable isÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [Manipal MEE 1995]

(a) 6000 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 9000 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 60000 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 90000 N

Solution : (d)

T = m(g + a)

= 6000 (10 + 5)

T = 90,000 N

Problem 39.Â Â Â Â  The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration â€˜aâ€™ is 3 : 2. The value of â€˜aâ€™ is (g– Acceleration due to gravity on the earth)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [MP PET 1997]

(a) 3 g

2

(b)

gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 2Â g

3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  3

(d) g

Â

Solution : (b)

weight of a man in stationary lift weight of a man in downward moving lift

=Â Â Â Â Â Â  mgÂ Â Â Â Â Â  =Â 3

m(g a)Â Â Â Â  2

\Â Â  Â gÂ  Â =Â 3

Ãž 2g = 3g – 3a

or a = g

g aÂ Â Â Â Â  2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  3

Problem 40.Â Â Â Â  A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from 60

kg to 50 kg for a while and then comes back to the original mark. What should we conclude

• The lift was in constant motion upwards

• The lift was in constant motion downwards
• The lift while in constant motion upwards, is stopped suddenly
• The lift while in constant motion downwards, is suddenly stopped

Solution : (c)Â Â Â Â Â  For retarding motion of a lift

R = m(g + a) for downward motion

R = m(g a) for upward motion

Since the weight of the body decrease for a while and then comes back to original value it means the lift was moving upward and stops suddenly.

Note : @ Generally we use R = m(g + a) for upward motion

R = m(ga) for downward motion

here a= acceleration, but for the given problem a= retardation

Problem 41. A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight placed on a spring balance. It records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage with an acceleration of 2m / s 2 . The spring balance will now record a weight ofÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [MP PMT 1999]

(a) 24 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 25 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 26 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 27 N

Solution : (b) Since the cage is closed and we can treat bird cage and air as a closed (Isolated) system. In this condition the force applied by the bird on the cage is an internal force due to this reading of spring balance will not change.

Problem 42. A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be W1 . If the bird flies about inside the cage, the reading of the spring balance is W2 . Which of the following is true

(a)

W1 = W2

(b)

W1Â > W2

(c)

W1 < W2

(d) Nothing definite can be predicted

Solution : (b) In this problem the cage is wire-cage the momentum of the system will not be conserved and due to this the weight of the system will be lesser when the bird is flying as compared to the weight of the same system when bird is resting is W2 < W1 .

## Acceleration of Block on Horizontal Smooth Surface.

### (1)Â  When a pull is horizontal

R = mg

andÂ Â Â Â Â Â Â  F = ma

\Â Â Â Â Â Â Â Â  a = F/m

Â

### (2)Â  When a pull is acting at an angle (q) to the horizontal (upward)

R + F sin q = mg

ÃžÂ Â Â Â Â Â Â Â Â Â Â  R = mg â€“ F sinq

andÂ Â Â Â Â Â Â  F cosq = ma

F cosq

\Â Â Â Â Â Â Â Â Â Â Â Â  a =Â Â Â Â  m

### (3)Â  When a push is acting at an angle (q) to the horizontal (downward)

Â

R = mg + F sinq

andÂ Â Â Â Â Â Â  F cosq = ma

a =Â F cosq

m

m2a

## Acceleration of Block on Smooth Inclined Plane.

### (1)Â  When inclined plane is at rest

#### Normal reaction R = mg cosq Force along a inclined plane

F = mg sinq

ma = mg sinq

\Â Â Â Â Â Â Â Â Â Â Â  a = g sinq

### (2)Â  When a inclined plane given a horizontal acceleration â€˜bâ€™

#### Since the body lies in an accelerating frame, an inertial force (mb) acts on it in the opposite direction.

Normal reaction R = mg cosq + mb sinq

andÂ Â Â Â Â Â Â  ma = mg sin q â€“ mb cos q

\Â Â Â Â Â Â Â Â Â Â Â  a = g sinq â€“ b cosq

Â

#### Note : @The condition for the body to be at rest relative to the inclined plane : a = g sinq â€“ b cosq = 0

\Â Â Â Â Â Â Â Â Â Â Â  b = g tanq

## Motion of Blocks in Contact.

Â

Â

Â

Problem 43.Â Â Â Â  Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter mass will be

• 5 N
• 4 N
• 2 N
• None of the above

Solution : (c)Â Â Â Â  Let m1Â = 6kg, m2 = 4kg

and F = 5 N

(given)

Force on the lighter mass =

m2 Â´ F m1 + m2

=Â 4 Â´ 5 6 + 4

= 2N

Problem 44.Â Â Â Â  In the above problem, if a push of 5 N is applied on the lighter mass, the force exerted by the lighter mass on the heavier mass will be

(a) 5 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 4 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 2 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) None of the above

Solution : (d)Â Â Â Â  Force on the heavier mass =

m1 F m1Â + m2

=Â 6 Â´ 5

6 + 4

= 3N

m2a

Problem 45.Â Â Â Â  In the above problem, the acceleration of the lighter mass will be

(a)

0.5 ms 2

(b)

5Â msÂ -2

4

(c)

5Â msÂ -2

6

• None of the above

Solution : (a)

Acceleration =

Net force on the system Total mass of the system

= 5Â = 0.5 m / s 2

10

Problem 46. Â Two blocks are in contact on a frictionless table one has a mass m and the other 2 m as shown in figure. Force F is applied on mass 2m then system moves towards right. Now the same force F is applied on m. The ratio of force of contact between the two blocks will be in the two cases respectively.

(a) 1 : 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 1 : 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 1 : 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 1 : 4

Solution : (b)Â Â Â Â  When the force is applied on mass 2m contact force

f1 = Â Â Â Â Â mÂ Â Â Â Â Â  g = g

m + 2mÂ Â Â Â Â Â Â Â Â Â  3

When the force is applied on mass m contact force

f2 = Â Â Â Â 2mÂ Â Â  Â g = 2Â g

Ratio of contact forces

f1Â =Â 1

m + 2mÂ Â Â Â Â Â Â Â Â Â  3

f2Â Â Â Â Â Â  2

## Motion of Blocks Connected by Mass Less String.

Â

Â

Â

 Â Â Â Â Â Â AT1m1 Â Â Â Â Â Â Bm2 Â Â Â Â Â Â T2 Â Â Â Â Â CÂ m3 Â Â Â Â Â Â Â F m1a a =Â Â Â Â Â Â Â Â Â Â Â Â  Fm1 + m2 + m3Â Â T Â =Â Â Â Â Â Â Â Â Â Â  m1 F1Â Â Â Â  m1 + m2 + m3Â Â TÂ =Â (m1Â + m2 )F2Â Â Â Â  m1 + m2 + m3 m1 T1 T1 = m1a m2a T1 m2 T2 T2Â – T1Â = m2Â a Â m3a T2 Â m3 F F – T2 = m3 a Â Â Â Â Â Â F Â Â Â Â Â Am1 Â Â Â Â Â T1 Â Â Â Â Â Bm2 Â Â Â Â Â T2 Â Â Â Â Â CÂ m3 m1a a =Â Â Â Â Â Â Â Â Â Â Â Â  Fm1 + m2 + m3Â Â TÂ  Â =Â  Â (m2 + m3 )F1Â Â Â Â  m1 + m2 + m3Â TÂ  Â =Â Â Â Â Â Â Â Â Â Â Â  m3 F2Â Â Â Â  m1 + m2 + m3 F m1 T1 F – T1 = m1a m2a T1 m2 T2 T1Â – T2Â = m2Â a Â Â T2 m3a m3 Â T2 = m3 a

Problem 47.Â Â Â Â  A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey

can climb up along the rope (g = 10m / s 2 )

[CBSE 2003]

(a)

10 m / s 2

(b)

25 m / s 2

(c)

2.5m / s 2

(d)

5m / s 2

Solution : (c)Â Â Â Â  Maximum tension that string can bear (Tmax) = 25 Â´ g N = 250 N

Tension in rope when the monkey climb up T = m(g + a)

For limiting condition T = Tmax Ãž

m(g + a) = 250

Ãž 20 (10 + a)= 250

\ a = 2.5 m / s2

Problem 48.Â Â Â Â  Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10N, then tension T1 =

 Â 10N Â Â 2kg Â T1 Â Â 3kg Â T2 Â Â 5kg

[Orissa JEE 2002]

Â

Â

Â

Â

Â

(a) 1NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 5 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 8 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 10 N

Solution : (c)Â Â Â Â Â  By comparing the above problem with general expression. T1 =

(m2Â + m3 )F m1 + m2 + m3

=Â (3 + 5)10 2 + 3 + 5

= 8 Newton

Â

Problem 49.Â Â Â Â  Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A

force F applied on the upper string produces an acceleration of

2m / s 2

in the upward direction in both the

blocks. If T and T Â¢Â be the tensions in the two parts of the string, thenÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [AMU (Engg.) 2000]

(a)

(b)

(c)

(d)

T = 70.8 N T = 58.8 N T = 70.8 N T = 70.8 N

and T Â¢Â = 47.2N and T Â¢Â = 47.2N and T Â¢Â = 58.8 N and T Â¢Â = 0

Solution : (a)Â Â Â Â  From F.B.D. of mass 4 kg

From F.B.D. of mass 2 kg

For total system upward force

4a = T‘-4 g

2a = T T‘-2g

.â€¦.(i)

.â€¦.(ii)

F = T = (2 + 4)(g + a) = 6 (18 + 2) N = 70.8 N by substituting the value of T in equation (i) and (ii) and solving we get T‘ = 47.2N

Problem 50. Three masses of 15 kg. 10 kg and 5 kg are suspended vertically as shown in the fig. If the string attached to the support breaks and the system falls freely, what will be the tension in the string between 10 kg and 5 kg masses. Take g = 10 ms -2 . It is assumed that the string remains tight during the motion

 15kg 10kg

(a) 300 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 250 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 50 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) Zero

Solution : (d)Â Â Â Â  In the condition of free fall, tension becomes zero.

Problem 51.Â Â Â Â  A sphere is accelerated upwards with the help of a cord whose breaking strength is five times its weight. The maximum acceleration with which the sphere can move up without cord breaking is

• 4g (b) 3gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 2gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) g

Solution : (a)Â Â Â Â  Tension in the cord = m(g + a) and breaking strength = 5 mg

For critical condition m(g + a)= 5mg

Ãž a = 4 g

This is the maximum acceleration with which the sphere can move up with cord breaking.

## Motion of Connected Block Over a Pulley.

 ConditionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Free body diagramÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  EquationÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Tension and acceleration Â T1m1Â Â Â Â Â Â  m1a m1gÂ T1m2Â Â Â Â Â Â  m2a m2gÂ T2Â Â Â T1Â Â Â Â Â  T1Â Â T1m1Â Â Â Â Â Â  m1a m1gÂ T1m2Â Â Â Â Â Â Â Â  m2a m2g + T2Â Â T2m3Â Â Â Â Â Â Â  m3a m3g Â m1a = T1Â – m1Â g T =Â 2m1m2Â  Â g1Â Â Â Â  m1 + m2Â Â T =Â 4m1m2Â  Â g2Â Â Â Â  m1 + m2Â Â a =Â Ã©Â m2 – m1 Ã¹gÃªÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  ÃºÃ« m1 +Â m2 Ã»Â Â Â T =Â 2m1[m2 + m3 ] g1Â Â Â Â  m1 + m2 + m3Â Â TÂ  Â =Â Â Â Â Â Â  2m1m3Â Â Â Â Â Â Â  g2Â Â Â Â  m1 + m2 + m3Â Â T =Â 4m1[m2 + m3 ] g3Â Â Â Â  m1 + m2 + m3Â Â a =Â [(m2 + m3 ) – m1 ]g m1 + m2 + m3 Â T2 P Â m2a = m2 g – T1 T1 T1 a m1 A m2 a Â T2 = 2T1 B Â Â Â m1a = T1Â – m1g T3 p T1 m1 T1 Â m2 a = m2 g + T2 – T1 A m2 T3 BÂ  Â T2 m3C a Â m3a = m3 g – T2 T1 T1 Â Â T3 = 2T1

Â

Â

Â

##### ConditionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Free body diagramÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  EquationÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Tension and acceleration

When pulley have a finiteÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T1

m1a = m1Â gT1

a =Â Â Â Â Â Â  m1 – m2

mass M and radius R then tension in two segments of string are different

m1

m1g

m1a

m1 + m2 + M

 2

Â m2a = T2 – m2 g

T

+Â MÂ Ã¹

2

 T

m2Â Â Â Â Â Â  m2a

2

MÂ Â Â Â Â Â Â Â Â Â Â  RÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2g

T1 =

1 ÃªÃ«Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  2

m1 + m2

2Â ÃºÃ»Â g

• M

2

T2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Torque

= (T1T2 )R = Ia

+Â MÂ Ã¹

m2Â Â Â Â Â Â Â Â Â  T1

aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (T1 – T2 )R = I a

Â

T2 =

2 ÃªÃ«Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  1

2Â ÃºÃ»Â g

M

BÂ Â Â Â Â Â Â Â  m1

R

1Â Â Â Â Â Â  2 a

m1 + m2 + 2

AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (T1 – T2 )R = 2 MRÂ Â  Â R

T2Â Â Â Â Â  T1

T1 – T2

=Â Ma

2

AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1a

T = m1a

m

 2

a =Â mÂ  Â + mÂ  Â g

 m

TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â  P

1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1Â Â Â Â Â Â Â Â Â Â  T

Â

T

Â

 m
 1
 2

m2a = m2 g T

1Â Â Â Â Â Â Â Â  2

m1m2

m2Â Â Â Â Â  a B

T

m2

m2g

m2a

T =Â Â Â Â Â Â Â Â  +Â mÂ Â  Â g

Â

TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1a = T m1 g sinq

a =Â Ã©Â m2 – m1 sinq Ã¹Â g

 Ã«
 Ã»

m1a

m1

PÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1g sinq

TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  q

ÃªÂ Â Â Â  m1 + m2Â Â Â Â Â  Ãº

AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T

m1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2a = m2 g T

 1
 2

m1m2(1 + sinq )

m2

qÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  BÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2

T

Â

Â

m2g

m2a

T =Â Â Â Â Â Â Â  mÂ  Â + mÂ Â Â Â Â Â Â Â Â Â  g

Â

Â

Â

aÂ Â Â Â Â Â Â Â Â Â Â Â Â  TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T

a

m1aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T

m1

m1g sina

T m1 g sin a = m1a

a =Â (m2 sin bm1 sina) g m1 + m2

mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  a

A

T

aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  b

m2

b

Â

m2a

Â

Â

m2g sinb

Â

Â

Â

Â

 Condition Free body diagram Equation Tension and acceleration Â Â Â Â Â Â Â PTam1AqÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2BÂ Â Â Â Â a1Am1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  PTTÂ Â As d 2 (x 2 )Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2Â Â Â Â Â Â Â Â Â  a2dt 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  B=Â 1 d 2 (x1 )2Â Â Â  dt 2\ a2 = a12a1 = acceleration of block Aa 2 = acceleration of block B m1aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Tm1m1g sinqqÂ Â Â Â m2aTm2Â Â Â Â Â Â Â Â m1am1Â Â Â Â Â Â Â Â  TÂ Â Â Â Â 2TÂ m2Â Â Â Â  Â Â Â Â Â  m2(a/2)Â m2g m1g sinq – T = m1aÂ Â Â T = m2aÂ Â Â Â Â Â Â T = m1aÂ Â Â Â m2 a = m2 g – 2T 2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a a =Â m1 g sinqm1 + m2Â Â Â Â T =Â  Â 2m1m2Â Â Â  g4m1 + m2Â Â Â Â Â =Â a =Â Â Â Â Â Â  2m2 g1Â Â Â Â Â Â Â Â Â Â  4m1 + m2Â aÂ Â  Â =Â Â Â Â Â Â  m2 g2Â Â Â Â  4m1 + m2Â T =Â 2m1m2 g4m1 + m2

Â

Â

Â

Problem 52.Â Â Â Â  A light string passing over a smooth light pulley connects two blocks of masses m1 and m2

(vertically). If the

acceleration of the system is g/8 then the ratio of the masses isÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [AIEEE 2002]

(a) 8 : 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 9 : 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 4 : 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 5 : 3

Ã¦Â m2 m1 Ã¶Â Â Â Â Â Â Â Â Â Â Â Â  gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m2

Solution : (b)

a = Ã§

m + m

Ã·Â g =Â Â Â Â  ;Â Â Â Â  by solving

8

= 9 / 7

m

Ã¨Â  Â 1Â Â Â Â Â Â Â Â  2 Ã¸Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  1

Problem 53.Â Â Â Â  A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given g = 10 ms -2 )

[Kerala (Engg.) 2000]

(a)

100 ms 2

3ms 2 10ms 2 30ms 2

Ã¦Â Â Â  Â m2Â Â Â Â Â  Ã¶

Ã¦Â  Â 3Â  Â Ã¶Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  2

Solution : (b)

a = Ã§Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Ã·Â g

m + m

=Â Ã§ 7 +Â 3 Ã· 10 =Â 3m / s

Ã¨Â  Â 1Â Â Â Â Â Â Â Â  2 Ã¸Â Â Â Â Â Â Â Â Â  Ã¨Â Â Â Â Â Â Â Â Â Â  Ã¸

Problem 54.Â Â Â Â  Two masses m1

and m2

are attached to a string which passes over a frictionless smooth pulley. When

m1 = 10 kg,

• 20 m / s 2

m2 = 6 kg, the acceleration of masses is

[Orissa JEE 2002]

• 5m / s 2

(c) 2.5 m / s 2

(d) 10 m / s 2

Solution : (c)

a = m1 Â m2Â g = Ã¦Â 10 –Â 6 Ã¶Â 10 = 2.5 m / s2

m + mÂ Â Â Â Â Â Â Â Â Â Â  Ã§Â 10 + 6 Ã·

1Â Â Â Â Â Â Â  2Â Â Â Â Â Â Â Â  Ã¨Â Â Â Â Â Â Â Â Â Â Â Â Â  Ã¸

Problem 55.Â Â Â Â  Two weights W1

and W2

are suspended from the ends of a light string passing over a smooth fixed pulley. If

the pulley is pulled up with an acceleration g, the tension in the string will be

(a)

4W1W2

W1Â + W2

(b)

2W1W2

W1Â + W2

(c)

W1W2 W1Â + W2

(d)

Â Â Â  Â W1W2Â Â Â Â

2(W1Â + W2Â )

 (Â Â Â Â Â Â Â Â  )

Solution : (a)Â Â Â Â  When the system is at rest tension in string T = 2m1m2Â g

m1 + m2

If the system moves upward with acceleration g then T =

2m1m2

m1Â + m2

(g + g )

=Â 4m1m2 g m1 + m2

or T =Â 4w1w2 Â

w1Â + w2

Problem 56.Â Â Â Â  Two masses M1 and M2 are attached to the ends of a string which passes over a pulley attached to the top of an inclined plane. The angle of inclination of the plane in q. Take g = 10 msâ€“2.

If M1 = 10 kg, M2 = 5 kg, q = 30o, what is the acceleration of mass M2

(a)

(b)

(c)

10ms 2

5ms 2

• ms – 2

3

(d) Zero

Solution : (d)Â Â Â Â  Acceleration = m2Â –Â m1 sinqÂ g = 5 Â 10. sin 30Â g

=Â 5 5 g = 0

m1 + m2

5 + 10

5 + 10

Problem 57.Â Â Â Â  In the above problem, what is the tension in the string

(a) 100 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 50 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 25 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) Zero

Solution : (b)

TÂ =Â m1m2Â (1 + sinq ) gÂ m1 + m2

=Â 10 Â´ 5(1 + sin 30).10

10 + 5

= 50 N

Problem 58.Â Â Â Â  In the above problem, given that M2 = 2M1 and M2 moves vertically downwards with acceleration a. If the position of the masses are reversed the acceleration of M2 down the inclined plane will be

(a) 2 aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) aÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) a/2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) None of the above

Solution : (d)Â Â Â Â  If m2 = 2m1 , then m2Â moves vertically downward with acceleration

a =Â m2 m1 sinq g =Â 2m1 m1 sin 30 g = g/2

m1 + m2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1 + 2m1

If the position of masses are reversed then m2 moves downward with acceleration

a‘ =Â m2 sinq m1Â gÂ =Â 2m1Â sin 30 –Â m1Â . g =Â 0

[As m

 2Â Â Â Â Â Â Â Â Â Â Â  1

= 2m ]

m1 + m2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  m1 + 2m1

i.e. the m2 will not move.

Problem 59.Â Â Â Â  In the above problem, given that M2 = 2M1 and the tension in the string is T. If the positions of the masses are reversed, the tension in the string will be

(a) 4 TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 1 TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) T/2

Solution : (c)Â Â Â Â  Tension in the string T = m1m2 (1 +Â sinqÂ )Â g

m1 + m2

If the position of the masses are reversed then there will be no effect on tension.

Problem 60.Â Â Â Â  In the above problem, given that M1 = M2 and q = 30o . What will be the acceleration of the system

(a)

10ms 2

(b)

5ms 2

(c)

2.5ms 2

(d) Zero

Solution : (c)

a =Â m2 m1 sinq g

=Â 1 sin 30 g

= gÂ = 2.5 m / s 2

[As m

 1Â Â Â Â Â Â Â Â Â  2

= m ]

m1 + m2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  2Â Â Â Â Â Â Â Â Â Â Â Â Â  4

Problem 61.Â Â Â Â  In the above problem, given that M1 = M2 = 5 kg and q = 30o . What is tension in the string (a) 37.5 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 25 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 12.5 NÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) Zero

Solution : (a)

T =Â m1m2 (1 + sinq ) gÂ =Â 5 Â´Â 5 (1 +Â sin 30)Â Â´10 =Â 37.5 N

m1 + m2

5 + 5

Problem 62.Â Â Â Â  Two blocks are attached to the two ends of a string passing over a smooth pulley as shown in the figure. The acceleration of the block will be (in m/s2) (sin 37o = 0.60, sin 53o = 0.80)

(a) 0.33

(b) 0.133

(c) 1

(d) 0.066

Solution : (b)

a =Â m2 sin b m1 sina gÂ =Â 50 sin 53Â°Â –Â 100 sin 37Â°Â gÂ =Â –Â 0.133 m / s 2

m1 + m2

100 + 50

Problem 63. The two pulley arrangements shown in the figure are identical. The mass of the rope is negligible. In (a) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In (b). m is lifted up by pulling the other end of the rope with a constant downward force of 2mg. The

ratio of accelerations in two cases will be

(a) 1 : 1

(b) 1 : 2

(c) 1 : 3

(d) 1 : 4

Solution : (c)Â Â Â Â Â  For first case a

=Â m2Â –Â m1Â g =Â 2m m =Â g

Â

â€¦..(i)

1

For second case

m1 + m2Â Â Â Â Â Â Â Â  m + 2mÂ Â Â Â Â  3

from free body diagram of m

ma2 = T mg ma2 = 2mg mg

Â

[As T= 2mg]

a2 = g

From (i) and (ii)

â€¦â€¦..(ii)

a1Â = g / 3 =1 / 3

a2Â Â Â Â Â Â Â  g

Problem 64.Â Â Â Â  In the adjoining figure m1 = 4m2. The pulleys are smooth and light. At time t = 0, the system is at rest. If the system is released and if the acceleration of mass m1 is a, then the acceleration of m2 will be

• g
• a
• a

2

• 2a

Solution : (d)Â Â Â Â  Since the mass m2 travels double distance in comparison to mass m1 therefore its acceleration will be double i.e. 2a

Â

Problem 65.Â Â Â Â  In the above problem (64), the value of a will be

(a) gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) g

2

(c)

gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) g

4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  8

Solution : (c)Â Â Â Â  By drawing the FBD of m1

m1a = m1g – 2T

m2 (2a) = T m2 g

and m2

â€¦..(i)

â€¦..(ii)

by solving these equation a = g / 4

Problem 66.Â Â Â Â  In the above problem, the tension T in the string will be

• m2g (b)

m2Â g

2

(c)

2

• m2g

(d)

3

2 m2 g

Solution : (d)Â Â Â Â  From the solution (65) by solving equation

T = 3Â m2g

2

Problem 67.Â Â Â Â  In the above problem, the time taken by m1 in coming to rest position will be

(a) 0.2 sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 0.4 sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 0.6 sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 0.8 s

Solution : (b)Â Â Â Â  Time taken by mass m2 to cover the distance 20 cm

Â

t =Â Â Â Â Â Â Â Â Â Â  =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  = 0.4 sec

Problem 68.Â Â Â Â  In the above problem, the distance covered by m2 in 0.4 s will be

(a) 40 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 20 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 10 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 80 cm Solution : (a)Â Â Â Â  Since the m2 mass cover double distance therefore S = 2 Â´ 20 = 40 cm

Problem 69.Â Â Â Â  In the above problem, the velocity acquired by m2 in 0.4 second will be

(a) 100 cm/sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 200 cm/sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 300 cm/sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 400 cm/s Solution : (b)Â Â Â Â  Velocity acquired by mass m2 in 0.4 sec

From v = u + at

[As a = g / 2 = 10 = 5 m / s2 ]

2

v = 0 + 5 Â´ 0.4 = 2m / s = 200 cm / sec.

Problem 70.Â Â Â Â  In the above problem, the additional distance traversed by m2 in coming to rest position will be

(a) 20 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 40 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 60 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 80 cm Solution : (a)Â Â Â Â  When m2 mass acquired velocity 200 cm/sec it will move upward till its velocity becomes zero.

H = u 2

2g

=Â (200)2

2 Â´ 100

= 20 cm

Problem 71.Â Â Â Â  The acceleration of block B in the figure will be

(a)

m2 g

(4m1 + m2 )

2m2 g

(4m1 + m2 )

2m1Â g

(m1 + 4m2 )

2m1Â g

(m1 + m2 )

Solution : (a)Â Â Â Â  When the block m2 moves downward with acceleration a, the acceleration of mass m1

covers double distance in the same time in comparison to m2 . Let T is the tension in the string.

By drawing the free body diagram of A and B

will be

2a because it

T = m1Â 2a

m2g – 2T = m2a

by solving (i) and (ii)

â€¦â€¦..(i)

â€¦â€¦..(ii)

a = (

m2 g

# )

4m1 + m2

## Motion of Massive String.

##### ConditionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Free body diagramÂ Â Â Â Â Â Â Â Â Â Â Â Â Â  EquationÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Tension and acceleration

Â

a

Â

MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T1

Â F = (M + m)a

Â

Â T1 = Ma

Â

MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  FÂ Â Â  Â T1 = force applied by the

string on the block

m/2

MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T2

Â T2 =

Tension at mid point

of the rope

m

Â LÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  F

TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  FÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  a

Â x Â

Â F = ma

Â a = F / m

Â

Â

m = Mass of string

T = Tension in string at a distance x from the end where the force is applied

m [(L â€“ x)/L]

T

a

Â

Â

F2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  L Â

A

F1

Â xÂ Â Â Â Â Â Â Â  B

A (M/L)x B

TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  F1

a

Â

F2

M = Mass of uniform rod

L = Length of rod

MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  F1

a

Â F1 – F2 = Ma

Â

A

Â

 B

Â LÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  TÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  A

T

Â

Â x Â

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  B

FÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  T

Mass of segment BC

Â

Â

Â

Spring Balance and Physical Balance.

#### (1)Â  Spring balance : When its upper end is fixed with rigid support and body of mass m hung from its lower end. Spring is stretched and the weight of the body can be measured by the reading of spring

balance R = W = mg

Â

#### The mechanism of weighing machine is same as that of spring balance.

Effect of frame of reference : In inertial frame of reference the reading of spring balance shows the actual weight of the body but in non-inertial frame of reference reading of spring balance increases or decreases in accordance with the direction of acceleration

#### [for detail refer Article (4.13)]

• Physical balance : In physical balance actually we compare the mass of body in both the pans. Here we does not calculate the absolute weight of the

#### Here X and Y are the mass of the empty pan.

• Perfect physical balance :

#### Weight of the pan should be equal i.e. X = Y

and the needle must in middle of the beam i.e. a = b.

#### Effect of frame of reference : If the physical balance is perfect then there will be no effect of frame of reference (either inertial or non- inertial) on the measurement. It is always errorless.

• False balance : When the masses of the pan are not equal then balance shows the error in measurement. False balance may be of two types

X > Y

and

a < b

#### For rotational equilibrium about point â€˜Oâ€™

Xa = YbÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  â€¦â€¦(i)

W1 in pan Y.

#### For rotational equilibrium about point â€˜Oâ€™

(X + W) a = (Y + W1)b

#### â€¦..(ii)

Now if the pans are changed then to balance the body we have to put a weight W2 in pan X. For rotational equilibrium about point â€˜Oâ€™

(X + W2) a = (Y + W)b

â€¦â€¦(iii)

#### (b)Â  If the beam of physical balance is not horizontal (when the pans are empty) and the arms are equal

i.e.Â Â Â Â Â Â Â Â Â Â Â Â Â Â  X > Y and a = b

For equilibrium

X + W = Y + W1

#### â€¦..(i)

Now if pans are changed then to balance the body we have to put a weight W2 in X Pan.

X + W2 = Y + W

#### â€¦..(ii)

True weight W = W1 +Â W2

2

## Modification of Newtonâ€™s Laws of Motion.

#### According to Newton, direction and time i.e., time and space are absolute. The velocity of observer has no effect on it. But, according to special theory of relativity Newtonâ€™s laws are true, as long as we are dealing with velocities which are small compare to velocity of light. Hence the time and space measured by two observers in relative motion are not same. Some conclusions drawn by the special theory of relativity about mass, time and distance which are as follows :

• Let the length of a rod at rest with respect to an observer is

#### L0. If the rod moves with velocity v w.r.t.

observer and its length is L, then

L = L0

#### where, c is the velocity of light.

Now, as v increases L decreases, hence the length will appear shrinking.

#### Let a clock reads T0 for an observer at rest. If the clock moves with velocity v and clock reads T with respect to observer, then T =

Hence, the clock in motion will appear slow.

#### (3)Â  Let the mass of a body is m0 at rest with respect to an observer. Now, the body moves with velocity v with respect to observer and its mass is m, then m =

m0 is called the rest mass.

#### Hence, the mass increases with the increases of velocity.

Note : @Â Â Â Â Â Â Â Â Â Â Â Â Â  If

#### v << c, i.e., velocity of the body is very small w.r.t. velocity of light, then

m = m0 . i.e., in

#### the practice there will be no change in the mass.

@Â Â Â  If v is comparable to c, then m > m0, i.e., mass will increase.

#### @Â Â Â  If

v = c, then

m =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  or

m = m0

= Â¥.

#### Hence, the mass becomes infinite, which is not

possible, thus the speed cannot be equal to the velocity of light.

#### @Â Â Â  The velocity of particles can be accelerated up to a certain limit. In cyclotron the speed of charged particles cannot be increased beyond a certain limit.

Problem 72.Â Â Â Â  A body weighs 8 gm, when placed in one pan and 18gm, when placed in the other pan of a false balance. If the beam is horizontal (when both the pans are empty), the true weight of the body is

(a) 13 gmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 12 gmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 15.5 gmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 15 gm

Solution : (b)Â Â Â Â  For given condition true weight =Â Â Â Â Â Â Â Â Â Â Â Â Â  =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  =12 gm.

Problem 73.Â Â Â Â  A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a. What will be the angle of inclination with verticalÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [Orissa JEE 2003]

(a)

tan1(a / g)

(b)

tan1(g / a)

cos 1(a / g)

cos 1(g / a)

Solution : (a)Â Â Â Â  From the figure

tanq = a

g

q = tan1 (a / g )

Problem 74.Â Â Â Â  A block of mass

5 kg

is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it

for 4 sec. What will be the distance of the block from the point where the force started actingÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [Pb PMT 2002]

(a) 10 mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 8 mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 6 mÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 2 m

Â

Solution : (a)Â Â Â Â  In the given problem force is working in a direction perpendicular to initial velocity. So the body will move under the effect of constant velocity in horizontal direction and under the effect of force in vertical direction.

Sx = ux Â´ t = 1.5 Â´ 4 = 6m

SyÂ = uy t + 1Â at 2 = 0 + 1Â (F / m)t 2

= 1Â (5 / 5)(4)2

= 8m

2Â Â Â Â Â Â Â Â Â Â Â Â Â Â  2

\ Â S =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  =

2

= 100 =10m

Problem 75.Â Â Â Â  The velocity of a body of rest mass m0

isÂ  3Â c

2

(where c is the velocity of light in vacuum). Then mass of this

body isÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [Orissa JEE 2002]

• (

/ 2)m0

(1 / 2)m0

2 m0

(2 /

• m0

Solution : (c)Â Â Â Â Â  From Einsteinâ€™s formulaÂ Â Â Â Â Â  m =

=Â Â Â Â Â Â Â Â Â Â Â Â  m0Â Â Â Â Â Â Â Â Â  =Â Â Â Â Â Â Â Â  m0

= 2m0

Problem 76.Â Â Â Â  Three weights W, 2W and 3W are connected to identical springs suspended form rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weights from the rod are such that

[Roorkee 1999]

• 3 W will be farthest (b) W will be farthest

(c) All will be at the same distanceÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 2 W will be farthest

Solution : (c)Â Â Â Â Â  For W, 2W, 3W apparent weight will be zero because the system is falling freely. So there will be no extension in any spring i.e. the distances of the weight from the rod will be same.

Problem 77.Â Â Â Â  A bird is sitting on stretched telephone wires. If its weight is W then the additional tension produced by it in the wires will be

(a) T = WÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) T > WÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) T < WÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) T = 0

Solution : (b)Â Â Â Â  For equilibrium

2T sinq = W

T =Â Â  Â W

2 sinq

q lies between 0 to 90Â° i.e. sinq < 1

\ T > W

Problem 78.Â Â Â Â  With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is 2

3

his weightÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  [CPMT 1979]

• 2 g

3

• g (c) 1 g

3

(d) Zero

Solution : (c)Â Â Â Â  When fireman slides down, Tension in the rope T = m(g a)

For critical condition m(g â€“ a) = 2/3 mg Ãž mg ma = 2 mg

3

\ a = g

3

So, this is the minimum acceleration by which a fireman can slides down on a rope.

Problem 79.Â Â Â Â  A car moving at a speed of 30 kilometres per hourâ€™s is brought to a halt in 8 metres by applying brakes. If the same car is moving at 60 km. per hour, it can be brought to a halt with same braking power in

(a) 8 metresÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (b) 16 metresÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (c) 24 metresÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  (d) 32 metres Solution : (d)Â Â Â Â  From v2 = u2 – 2as

0 = u 2 – 2as

s = u 2

2a

Ãž s Âµ u2 (if a = constant)

sÂ Â Â Â Â Â  Ã¦Â uÂ  Â Ã¶Â 2Â Â Â Â  Ã¦Â 60 Ã¶Â 2

Â Â Â 2 =Â Ã§Â 2 Ã·Â =Â Ã§

Ã·Â  Â =Â 4 ÃžÂ s2Â =Â 4s1Â =Â 4 Â´Â 8

= 32 metres.

s1Â Â Â Â  Ã¨ u1 Ã¸

Ã¨Â 30 Ã¸