Laws of Motion

  • Question 469
    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 470
    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) 
    Question 471
    CBSEENPH11020082

    A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2.

    The crew and the passengers weigh 300 kg.

    Give the magnitude and direction of the, 

    (a) force on the floor by the crew and passengers,

    (b) action of the rotor of the helicopter on the surrounding air,

    (c) force on the helicopter due to the surrounding air.

    Solution

    Given, 
    Mass of the helicopter, mh = 1000 kg 
    Mass of the crew and passengers, mp = 300 kg
    Total mass of the system, m = 1300 kg
    Acceleration of the helicopter, a = 15 m/s
    Using Newton’s second law of motion, the reaction force R,
               R – mpg = ma

                              = mp(g + a)
                              = 300 (10 + 15) 
                              = 300 × 25 
                              = 7500 N 
    The reaction force will be directed upwards, the helicopter is accelerating vertically upwards. 
    According to Newton’s third law of motion, the force on the floor by the crew and passengers  = 7500 N, directed downward.
    (b)
    Using Newton’s second law of motion, the reaction force R’ experienced by the helicopter can be calculated as, 
                     R' - mg = ma 

                                 = m(g + a)
                                 = 1300 (10 + 15)
                                 = 1300 × 25 
                                         
                                 = 32500 N 
    The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
    (c) The force on the helicopter due to the surrounding air is 32500 N, directed upwards. 

    Question 472
    CBSEENPH11020083

    A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectional area 10–2m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

    Solution
    Given,
    Speed of the water stream, v = 15 m/s
    Cross-sectional area of the tube, A = 10–2 m2

    Volume of water coming out from the pipe per second,
                     VAv 
                         = 15 × 10–2 m3/s 
    Density of water, ρ = 103 kg/m

    Mass of water flowing out through the pipe per second = ρ ×V 
               = 150 kg/s
    The water strikes the wall and does not rebound.
    Therefore, according to Newton's second law of motion, 
    Force exerted by the water on the wall,
    F = Rate of change of momentum = fraction numerator increment straight p over denominator increment straight t end fraction
                                                        = mv over straight t
                                                        = 150 × 15
                                                        = 2250 N

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