Laws of Motion

Question 461
CBSEENPH11020078

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Solution
We have, 
Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Number of revolution per second, n =
40 over 60 equals space 2 over 3 space r. p. s 40 / 60 = 2 / 3 rps
Angular velocity, ω = straight v over straight r = 2πn 
The tension in the string provides the centripetal force. 
i.e.,
T = Fcentripetal 

   = mv squared over straight r 
    = mrω
    = mr(2πn)

    = 0.25 × 1.5 × (2 × 3.14 × (open parentheses 2 over 3 close parentheses)2
    = 6.57 N 
Maximum tension in the string,

         Tmax = 200 N 
           max = mv2max / r

∴       vmax = (Tmax × 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. 
Question 462
CBSEENPH11020079

If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: 

(a) the stone moves radially outwards, 

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Solution
(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. 
Question 463
CBSEENPH11020080

Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?


Solution
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) = fraction numerator left parenthesis 2 space minus space 0 right parenthesis space straight x space 10 to the power of negative 2 end exponent over denominator left parenthesis 2 minus 0 right parenthesis end fraction space equals space 10 to the power of negative 2 end exponent space m divided by s
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 
Question 464
CBSEENPH11020081

A stone of mass tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

  Lowest Point Highest Point
a mg – T1 mg + T2
b mg + T1 mg – T2
c mg + T1 – (mv12/ R mg – T2 + (mv12/ R
d mg – T1 – (mv12/ R mg + T2 + (mv12/ R

T1 and V1 denote the tension and speed at the lowest point.

T2 and v2 denote corresponding values at the highest point.

Solution
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.,  Fnet = T - mg = mv subscript 1 squared over straight R                    ...(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 = mv subscript 2 squared over straight R                       ...(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) 

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