Laws of Motion

Question 453
CBSEENPH11020070

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s–1. What is the trajectory of the bob if the string is cut when the bob is

(a) at one of its extreme positions,

(b) at its mean position.

Solution
(a) When bob is at its extreme position:
The trajectory of the bob if the string is cut at one of it's extreme positions will be in vertically downward direction.
At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground. 
(b) When the bob is at its mean position:
When the string is cut at its mean position, the trajectory of the bob will be a parabolic path. 
At the mean position, the velocity of the bob is 1 m/s.
The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only.
Hence, the bob will follow a parabolic path. 
Question 454
CBSEENPH11020071

A man of mass 70 kg stands on a weighing scale in a lift which is moving, 

(a) upwards with a uniform speed of 10 m s–1

(b) downwards with a uniform acceleration of 5 m s–2

(c) upwards with a uniform acceleration of 5 m s–2.
 
What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Solution
(a) 
Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, we can write the equation of motion as, 
                      R – mg = ma

where,
 ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
          R = m
              = 70 × 10
              = 700 N 
Therefore, reading on the weighing scale = 700 over straight g
                                                     = 700 over 10
                                                     = 70 space kg
(b) 
Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 , downward
Using Newton’s second law of motion, we can write the equation of motion as: 
                      R + mg = ma 

                               R = m(g – a
                                  = 70 (10 – 5)
                                  = 70 × 5 
                                          = 350 N 
∴ Reading on the weighing scale = 350 g = 350 over 10= 35 kg
(c) 
Mass of the man, m = 70 kg
Acceleration, a = 5 m/supward 
Using Newton’s second law of motion, we can write the equation of motion as:

                        R
 – mg = ma 

                                 R = m(g + a
                                    = 70 (10 + 5)
                                    = 70 × 15
                                    = 1050 N
Therefore,
Reading on the weighing scale = 1050 over straight g
                                               = 1050 over 10
                                               = 105 kg
(d) When the lift moves freely under gravity, acceleration a =
Using Newton’s second law of motion, we can write the equation of motion as, 
                       R + mg = ma 

                                 R = m(g – a
                                    = m(g – g)
                                    = 0 
∴ Reading on the weighing scale = 0 over straight g= 0 kg 
The man will be in a state of weightlessness.
Question 455
CBSEENPH11020072

Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t> 4 s,0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only). 


Solution
(a)
For t < 0
From the given graph, the position of the particle is coincident with the time axis.
That is, the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.
For t > 4 s 
In the given graph, the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of  3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4 
The position-time graph has a constant slope in the given graph.
Hence, the acceleration produced in the particle is zero.
Therefore, the force acting on the particle is zero.
(b)
At t = 0,
Impulse = Change in momentum 
          = mv – mu 

Mass of the particle, m = 4 kg 
Initial velocity of the particle, u = 0 
Final velocity of the particle, v = 3 over 4 m/s
∴ Impulse = 4 x (  3 over 4- 0)
              = 3 kg m/s
At t = 4 s,
Initial velocity of the particle, u = 3 over 4 m/s 
Final velocity of the particle, v = 0 
∴ Impulse = 4 (0 - 3 over 4 ) = -3 kg m/s 
(a) For t < 0
It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.
For t > 4 s 
It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 
3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4
It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.
(b) At t = 0
Impulse = Change in momentum 
           = mv – mu 

Mass of the particle, m = 4 kg 
Initial velocity of the particle, u = 0 
Final velocity of the particle, v = 3 over 4 m/s 
∴ Impulse = 4 ( 3 over 4 - 0) = 3 kg m/s
At t = 4 s
Initial velocity of the particle, u =  m/s

Final velocity o9f the particle, v = 0
∴ Impulse = 4 (0 - 3 over 4) = -3 kg m/s 
Question 456
CBSEENPH11020073

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to

(i) A, (ii) B along the direction of string.

What is the tension in the string in each case?

Solution
Horizontal force, F = 600 N
Mass of body A, m1 = 10 kg
Mass of body B, m2 = 20 kg
Total mass of the system, m = m1 + m2 = 30 kg
Using Newton’s second law of motion,
                        F = ma 
Acceleration (a) produced in the system can be calculated as, 
         straight a space equals space straight F over straight m = 600 over 30 space equals space 20 space m divided by s squared space
When force is applied on a body A. 
Equation of motion can be written as, 
F - T = m1a
Therefore,
T = F - m1a
   = 600 - 10 x 20
   = 400 N                                ...  (i)
When Force is applied on a body B, we have 
F - T = m2a
i.e., T = F - m2
Therefore, 
T = 600 - 20 x 20 = 200 N         ... (ii)
From (i) and (ii), we can say that the answer is different in both the cases. 
Therefore, the answer depends on which end of mass, the force is applied.

NCERT Book Store

NCERT Sample Papers

Entrance Exams Preparation