Laws of Motion

Radius of the circle, r = 1.5 m
Number of revolution per second, n =
40 / 60 = 2 / 3 rps
Angular velocity, ω =  = 2πn 
The tension in the string provides the centripetal force. 
i.e.,
T = F_centripetal 

   =  
    = mrω
    = mr(2πn)² 

    = 0.25 × 1.5 × (2 × 3.14 × ()²
    = 6.57 N 
Maximum tension in the string,

         T_max = 200 N 
           _max = mv²_max / r

∴       v_max = (T_max × r  / m)^1/2 

                 = (200 × 1.5 / 0.25)^1/2 

                 = (1200)^1/2 
                 = 34.64 m/s 
Therefore, the maximum speed of the stone is 34.64 m/s. 

(b) The stone flies off tangentially from the instant the string breaks.
When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks. 

Between x = 0 and x = 2 cm, ball will be rebounding between two walls.
After every 2 s,
Magnitude of the impulse received by the ball = 0.08 × 10^–2 kg m/s
The given graph shows that a body changes its direction of motion after every 2 s.
Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm.
The slope of the position-time graph reverses after every 2 s, the ball collides with a wall after every 2 s.
Therefore, ball receives an impulse after every 2 s.
Given, 
Mass of the ball, m = 0.04 kg
The slope of the graph gives the velocity of the ball.
Using the graph,
Initial velocity, (u) = 
Velocity of the ball before collision, u = 10^–2 m/s
Velocity of the ball after collision, v = –10^–2 m/s 
Here, the negative sign arises as the ball reverses its direction of motion.
Magnitude of impulse = Change in momentum 
                                 = | mv - mu |
                                 = | 0.04 (v - u) |
                                 = | 0.04 (-10^-2 - 10^-2) |
                                 = 0.08 × 10^-2 kg m/s 

The free body diagram of the stone at the lowest point is shown in the figure below:

                   

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force.
i.e.,  F_net = T - mg =                     ...(i)
where,
v₁ is the velocity at the lowest point. 
The free body diagram of the stone at the highest point is shown in the following figure. 
                 
Using Newton’s second law of motion, 

        T + mg =                        ...(ii)
where, v₂ is the velocity at the highest point.

From equations (i) and (ii), 
Net force acting at the lowest = (T - mg)
Net force at the highest points = (T + mg)