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

File name : Chapter-8-Law-of-Motion-Part-2.pdf

 

 

 

Problem 33.     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 N                                 (b) 800 N                             (c) 1200 N                              (d) Zero

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

[CBSE 2003]

 

Problem 34.     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 ‘g’, the reading on the spring balance will be

(a) 2 kg                                    (b) (4 ´ g) kg                      (c) (2 ´ g) kg                          (d) Zero

 

Solution : (d)

R = m(g a)

= (g g) = 0

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

 

Problem 35.     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 kg                                    (b) 4 kg                                (c) Zero                          (d) 1 kg

 

Solution : (a)

R = mg

= 2g Newton or

2kg

[because the lift is moving with the zero acceleration]

 

Problem 36.     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 kg                                    (b) (2 ´ g) kg                      (c) (4 ´ g) kg                          (d) 4 kg

 

Solution : (d)

R = m(g + a)

= m(g + g)

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

 

= 2mg R = 2 ´ 2g N = 4 g N or 4 kg

Problem 37.     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 kg                                  (b) 40 kg                              (c) 42 kg                                 (d) 48 kg

[MP PMT 1994]

 

Solution : (d)

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

= 480 N

or 48kg

 

Problem 38.     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 is                                                                                [Manipal MEE 1995]

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

 

Solution : (d)

T = m(g + a)

= 6000 (10 + 5)

T = 90,000 N

 

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

 

(a) 3 g

2

(b)

g                                    (c) 2 g

3                                  3

(d) g

 

 

Solution : (b)

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

=       mg       = 3

m(g a)     2

 

\    g   3

Þ 2g = 3g – 3a

or a = g

 

g a      2                                             3

Problem 40.     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)      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 of                                                                   [MP PMT 1999]

(a) 24 N                                   (b) 25 N                               (c) 26 N                                   (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)

W1 > 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)  When a pull is horizontal

R = mg

and        F = ma

\         a = F/m

 

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

R + F sin q = mg

Þ            R = mg F sinq

and        F cosq = ma

F cosq

\             a =     m

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

 

 

 

R = mg + F sinq

and        F cosq = ma

a F cosq

m

m2a

 

Acceleration of Block on Smooth Inclined Plane.

(1)  When inclined plane is at rest

Normal reaction R = mg cosq Force along a inclined plane

F = mg sinq

ma = mg sinq

\            a = g sinq

(2)  When a inclined plane given a horizontal acceleration ‘b

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

and        ma = mg sin q – mb cos q

\            a = g sinqb cosq

 

Note : @The condition for the body to be at rest relative to the inclined plane : a = g sinqb cosq = 0

\            b = g tanq

Motion of Blocks in Contact.

 

 

 

 

Problem 43.     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)     Let m1 = 6kg, m2 = 4kg

and F = 5 N

(given)

 

 

Force on the lighter mass =

m2 ´ F m1 + m2

= 4 ´ 5 6 + 4

= 2N

 

Problem 44.     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 N                                      (b) 4 N                                  (c) 2 N                                     (d) None of the above

 

 

Solution : (d)     Force on the heavier mass =

m1 F m1 + m2

= 6 ´ 5

6 + 4

= 3N

m2a

 

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

 

 

 

(a)

0.5 ms 2

(b)

ms -2

4

(c)

ms -2

6

  • None of the above

 

 

Solution : (a)

Acceleration =

Net force on the system Total mass of the system

= 5 = 0.5 m / s 2

10

 

Problem 46.  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 : 1                           (b) 1 : 2                        (c) 1 : 3                           (d) 1 : 4

 

 

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

f1 =      m       g = g

 

m + 2m           3

 

When the force is applied on mass m contact force

f2 =     2m     g = 2 g

 

Ratio of contact forces

f1 = 1

 

m + 2m           3

 

f2       2

Motion of Blocks Connected by Mass Less String.

 

 

 

 

  

 

 

 

 

 

A

T1

m1

  

 

 

 

 

 

B

m2

 

 

 

 

 

 

T2

  

 

 

 

 

C

 

m3

 

 

 

 

 

 

 

F

 m1a  a =             F

m1 + m2 + m3

 

 

T  =           m1 F

1     m1 + m2 + m3

 

 

T = (m1 + m2 )F

2     m1 + m2 + m3

 m1T1T1 = m1a
 m2a  
T1m2T2T2 – T1 = ma
  

m3a

  
T2 

m3

FF T2 = m3 a
 

 

 

 

 

 

F

 

 

 

 

 

A

m1

 

 

 

 

 

T1

 

 

 

 

 

B

m2

  

 

 

 

 

T2

 

 

 

 

 

C

 

m3

  m1a  a =             F

m1 + m2 + m3

 

 

T   =   (m2 + m3 )F

1     m1 + m2 + m3

 

T   =            m3 F

2     m1 + m2 + m3

Fm1T1F T1 = m1a
 m2a  
T1m2T2T1 – T2 = ma
 

 

T2

m3a m3  

T2 = m3 a

Problem 47.     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)     Maximum tension that string can bear (Tmax) = 25 ´ g N = 250 N

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

 

For limiting condition T = Tmax Þ

m(g + a) = 250

Þ 20 (10 + a)= 250

\ a = 2.5 m / s2

 

Problem 48.     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 =

 

10N

 

 

2kg

 

T1

 

 

3kg

 

T2

 

 

5kg

 

 

[Orissa JEE 2002]

 

 

 

 

 

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

 

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

(m2 + m3 )F m1 + m2 + m3

(3 + 5)10 2 + 3 + 5

= 8 Newton

 

 

Problem 49.     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 ¢ be the tensions in the two parts of the string, then                                          [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 ¢ = 47.2N and T ¢ = 47.2N and T ¢ = 58.8 N and T ¢ = 0

 

Solution : (a)     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

.….(i)

.….(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 N                                (b) 250 N                             (c) 50 N                                   (d) Zero

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

Problem 51.     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) 3g                                   (c) 2g                                       (d) g

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

 

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

Þ 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.

Condition                    Free body diagram                    Equation                    Tension and acceleration
      

T1

m1       m1a m1g

 

T1

m2       m2a m2g

 

T2

 

 

 

T1      T1

 

 

T1

m1       m1a m1g

 

T1

m2         m2a m2g + T2

 

 

T2

m3        m3a m3g

 m1a = T1 – mgT = 2m1m g

1     m1 + m2

 

 

T = 4m1m g

2     m1 + m2

 

 

a = é m2 m1 ùg

ê                  ú

ë m1 + m2 û

 

 

 

T = 2m1[m2 + m3 ] g

1     m1 + m2 + m3

 

 

T   =       2m1m3        g

2     m1 + m2 + m3

 

 

T = 4m1[m2 + m3 ] g

3     m1 + m2 + m3

 

 

a = [(m2 + m3 ) – m1 ]g m1 + m2 + m3

    

T2

  
  P   m2a = m2 g T1
  T1   
   T1  
 am1   
  Am2a T2 = 2T1
   B  
      

 

 m1a = T1 – m1g

   T3  
  p   
  T1   
  m1T1  m2 a = m2 g + T2 – T1
  Am2  
  T3B   T2  
   m3

C

a m3a = m3 g T2
T1 T1   
      

 T3 = 2T1

 

 

 

 

Condition                    Free body diagram                    Equation                    Tension and acceleration

 

When pulley have a finite                          T1

m1a = mgT1

a =       m1 – m2

 

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

m1

m1g

m1a

m1 + m2 + M

 

2

 m2a = T2 – m2 g

T

m  é2m

M ù

 

2

T

m2       m2a

2

M            R                                                                                                                                                                   m2g

T1 =

1 êë                    2

m1 + m2

2 úû g

  • M

2

 

T2                                                                                                                                               Torque

= (T1T2 )R = Ia

m é2m

M ù

 

 

m2          T1

a                                 (T1 – T2 )R = I a

 

T2 =

2 êë                    1

2 úû g

M

 

B         m1

R

1       2 a

m1 + m2 + 2

 

A                                                                                                                                 (T1 – T2 )R = 2 MR    R

 

T2      T1

T1 – T2

Ma

2

 

A                                                                                                 m1a

T = m1a

m

2

a m   + m   g

 

m

T               P

1                                                                 m1           T

 

T

 

m
1
2

m2a = m2 g T

1         2

 

 

m1m2

 

 

m2      a B

T

m2

m2g

m2a

T =         + m    g

 

 

 

T                                         m1a = T m1 g sinq

a = é m2 – m1 sinq ù g

 

ë
û

m1a

m1

P                                     m1g sinq

T                                                                                        q

ê     m1 + m2      ú

 

A                                               T

m1                                           a                                                                                                                       m2a = m2 g T

1
2

m1m2(1 + sinq )

 

m2

q                        B                                                               m2

T

 

 

m2g

m2a

T =        m   + m           g

 

 

 

 

 

a              T                     T

a

m1a                  T

m1

m1g sina

T m1 g sin a = m1a

a = (m2 sin bm1 sina) g m1 + m2

 

m                                       m                                                      a

A

T

a                           b

m2

 

b

 

m2a

 

 

m2g sinb

 

 

 

 

 

ConditionFree body diagramEquationTension and acceleration
 

 

 

 

 

 

 

P

T

a

m1

A

q                                   m2

B

 

 

 

 

 

a1

A

m1                                 P

T

T

 

 

As d 2 (x 2 )               m2          a2

dt 2                       B

= 1 d 2 (x1 )

2    dt 2

\ a2 = a1

2

a1 = acceleration of block A

a 2 = acceleration of block B

m1a                  T

m1

m1g sinq

q

 

 

 

 

m2a

T

m2

 

 

 

 

 

 

 

 

m1a

m1         T

 

 

 

 

 

2T

 

m2           m2(a/2) m2g

m1g sinqT = m1a

 

 

 

T = m2a

 

 

 

 

 

 

 

T = m1a

 

 

 

 

m2 a = m2 g – 2T 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a

a m1 g sinq

m1 + m2

 

 

 

 

T =   2m1m2    g

4m1 + m2

 

 

 

 

 

a =       2m2 g

1           4m1 + m2

 

a    =       m2 g

2     4m1 + m2

 

T = 2m1m2 g

4m1 + m2

 

 

 

 

 

Problem 52.     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 is                                                            [AIEEE 2002]

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

æ m2 m1 ö             g                                  m2

 

Solution : (b)

a = ç

m + m

÷ g =     ;     by solving

8

= 9 / 7

m

 

è   1         2 ø                                                   1

Problem 53.     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

æ     m2      ö

 

 

 

 

æ   3   ö                       2

 

Solution : (b)

a = ç                 ÷ g

m + m

= ç 7 + 3 ÷ 10 = 3m / s

 

è   1         2 ø          è           ø

 

Problem 54.     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  m2 g = æ 10 – 6 ö 10 = 2.5 m / s2

 

 

m + m            ç 10 + 6 ÷

1        2         è              ø

 

Problem 55.     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

W1 + W2

(b)

2W1W2

W1 + W2

(c)

W1W2 W1 + W2

(d)

     W1W2    

2(W1 + W2 )

 

(         )

Solution : (a)     When the system is at rest tension in string T = 2m1mg

m1 + m2

 

If the system moves upward with acceleration g then T =

2m1m2

m1 + m2

(g + g )

= 4m1m2 g m1 + m2

or T = 4w1w2  

w1 + w2

 

Problem 56.     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 ms–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)     Acceleration = m2 – m1 sinq g = 5  10. sin 30 g

= 5 5 g = 0

 

m1 + m2

5 + 10

5 + 10

 

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

(a) 100 N                                 (b) 50 N                               (c) 25 N                                  (d) Zero

 

 

Solution : (b)

T m1m2 (1 + sinq ) g m1 + m2

10 ´ 5(1 + sin 30).10

10 + 5

= 50 N

 

Problem 58.     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 a                                       (b) a                                      (c) a/2                            (d) None of the above

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

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

m1 + m2                    m1 + 2m1

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

 

a‘ = m2 sinq m1 g = 2m1 sin 30 – m1 . g = 0

[As m

2            1

= 2m ]

 

m1 + m2                     m1 + 2m1

i.e. the m2 will not move.

Problem 59.     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 T                                       (b) 1 T                                  (c) T                                         (d) T/2

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

m1 + m2

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

Problem 60.     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 m2 m1 sinq g

1 sin 30 g

= g = 2.5 m / s 2

[As m

1          2

= m ]

 

m1 + m2                        2              4

 

 

 

 

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

 

Solution : (a)

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

 

m1 + m2

5 + 5

 

Problem 62.     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 m2 sin b m1 sina g = 50 sin 53° – 100 sin 37° g = – 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)      For first case a

m2 – m1 g 2m mg

 

…..(i)

 

1

 

For second case

m1 + m2         m + 2m      3

 

from free body diagram of m

ma2 = T mg ma2 = 2mg mg

 

[As T= 2mg]

 

a2 = g

From (i) and (ii)

……..(ii)

a1 = g / 3 =1 / 3

 

a2        g

Problem 64.     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)     Since the mass m2 travels double distance in comparison to mass m1 therefore its acceleration will be double i.e. 2a

 

 

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

 

(a) g                                          (b) g

2

(c)

g                                        (d) g

4                                     8

 

Solution : (c)     By drawing the FBD of m1

m1a = m1g – 2T

m2 (2a) = T m2 g

and m2

…..(i)

…..(ii)

 

by solving these equation a = g / 4

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

 

  • m2g (b)

mg

2

(c)

2

  • m2g

(d)

3

2 m2 g

 

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

T = 3 m2g

2

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

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

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

 

t =           =                  =                   = 0.4 sec

 

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

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

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

(a) 100 cm/s                            (b) 200 cm/s                       (c) 300 cm/s                           (d) 400 cm/s Solution : (b)     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 ´ 0.4 = 2m / s = 200 cm / sec.

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

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

 

H = u 2

2g

= (200)2

2 ´ 100

= 20 cm

 

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

 

 

(a)

 

 

m2 g

(4m1 + m2 )

2m2 g

(4m1 + m2 )

2mg

(m1 + 4m2 )

 

 

 

 

2mg

(m1 + m2 )

 

Solution : (a)     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 = m1 2a

m2g – 2T = m2a

by solving (i) and (ii)

……..(i)

……..(ii)

 

a = (

m2 g

)

 

4m1 + m2

 

 

Motion of Massive String.

Condition                            Free body diagram               Equation                         Tension and acceleration

 

 

a

 

M                     T1

 F = (M + m)a

 

 T1 = Ma

 

 

M                    m                                    F     T1 = force applied by the

string on the block

 

m/2

M                                       T2

 

 

 

 T2 =

Tension at mid point

 

of the rope

m

 L                                                                                                                                   F

T                                F                                                                 a

 x  

 F = ma

 a = F / m

 

 

 

 

m = Mass of string

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

m [(L x)/L]

T

a

 

 

 

 

F2                            L  

A

F1

 x         B

A (M/L)x B

T                                                F1

a

 

 

 

F2

 

M = Mass of uniform rod

L = Length of rod

M                          F1

a

 F1 – F2 = Ma

 

 

A

 

B

 L                 T                                                                                                                                                                                                    A

T

 

 x  

C                                                                                                                                                                                                                  B

F                                                                                                                                                                                                           T

Mass of segment BC

 

 

 

 

Spring Balance and Physical Balance.

(1)  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

 

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

(a)   If the beam of physical balance is horizontal (when the pans are empty) but the arms are not equal

 

X > Y

and

a < b

 

For rotational equilibrium about point ‘O

Xa = Yb                                               ……(i)

In this physical balance if a body of weight W is placed in pan X then to balance it we have to put a weight

W1 in pan Y.

For rotational equilibrium about point ‘O

 

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

…..(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 ‘O

 

 

 

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

From (i), (ii) and (iii) True weight W =

……(iii)

 

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

i.e.               X > Y and a = b

In this physical balance if a body of weight W is placed in X Pan then to balance it. We have to put a weight W1 in Y Pan

 

For equilibrium

X + W = Y + W1

…..(i)

 

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

 

For equilibrium From (i) and (ii)

X + W2 = Y + W

…..(ii)

 

True weight W = W1 + W2

2

Modification of Newton’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 Newton’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.

(2)

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)  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 : @              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.

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

 

 

@    If

v = c, then

m =                 or

m = m0

0

= ¥.

Hence, the mass becomes infinite, which is not

 

 

 

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

@    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.     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 gm                                 (b) 12 gm                             (c) 15.5 gm                             (d) 15 gm

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

Problem 73.     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 vertical                                                                                            [Orissa JEE 2003]

 

(a)

tan1(a / g)

(b)

tan1(g / a)

cos 1(a / g)

cos 1(g / a)

 

Solution : (a)     From the figure

tanq = a

g

q = tan1 (a / g )

 

 

 

 

Problem 74.     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 acting                   [Pb PMT 2002]

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

 

Solution : (a)     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 ´ t = 1.5 ´ 4 = 6m

 

Sy = uy t + 1 at 2 = 0 + 1 (F / m)t 2

= 1 (5 / 5)(4)2

= 8m

 

2               2

\  S =                       =

2

= 100 =10m

 

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

is  3 c

2

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

 

body is                                                                                                                          [Orissa JEE 2002]

 

  • (

/ 2)m0

(1 / 2)m0

2 m0

(2 /

  • m0

 

 

 

 

Solution : (c)      From Einstein’s formula       m =

=             m0          =         m0

= 2m0

 

 

 

Problem 76.     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 distance                               (d) 2 W will be farthest

Solution : (c)      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.     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 = W                                (b) T > W                            (c) T < W                                (d) T = 0

Solution : (b)     For equilibrium

2T sinq = W

T =    W

2 sinq

q lies between 0 to 90° i.e. sinq < 1

\ T > W

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

3

his weight                                                                                                                              [CPMT 1979]

 

  • 2 g

3

  • g (c) 1 g

3

(d) Zero

 

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

For critical condition m(g a) = 2/3 mg Þ 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.     A car moving at a speed of 30 kilometres per hour’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 metres                             (b) 16 metres                      (c) 24 metres                          (d) 32 metres Solution : (d)     From v2 = u2 – 2as

0 = u 2 – 2as

 

s = u 2

2a

Þ s µ u2 (if a = constant)

 

s       æ u   ö 2     æ 60 ö 2

 

   2 = ç 2 ÷ = ç

÷   = 4 Þ s2 = 4s1 = 4 ´ 8

= 32 metres.

 

s1     è u1 ø

 

è 30 ø

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