Laws of Motion

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

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, 
          = 
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 - m₂a 
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.

Initial momentum of the system (parent nucleus) = 0
Let v₁ and v₂ be the respective velocities of the daughter nuclei having masses m₁ and m₂.
Total linear momentum of the system after disintegration = m₁v₁ + m₂v₂ 

According to the law of conservation of momentum, 
Total initial momentum = Total final momentum 
                                 0 = m₁v₁ + m₂v₂ 

                                v₁ = -m₂v₂ / m₁ 

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Mass of each ball, m = 0.05 kg 
Initial velocity of the ball = 6 m/s 
Magnitude of the initial momentum of the ball, p₁ = 0.05 x 6
    = 0.3 kg m/s
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball, p_f = - 0.3 kg m/s
Impulse imparted to each ball = p_f - p_i = -0.3 -0.3 = -0.6 kg m/s
The negative sign indicates that the impulses imparted to the balls are opposite in direction.