Laws of Motion

  • Question 465
    CBSEENPH11020076

    A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s–1, what is the recoil speed of the gun?

    Solution
    Given, 
    Mass of the gun, M = 100 kg
    Mass of the shell, m = 0.020 kg
    Muzzle speed of the shell, v = 80 m/s
    Recoil speed of the gun = V

    Both the gun and the shell are at rest initially.
    Initial momentum of the system = 0
    Final momentum of the system = mv – MV

    Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
    Now, as per the law of conservation of momentum. we have
    Final momentum = Initial momentum 
     
              mv – MV = 0 
    ∴                   V = mv / 

                           = 0.020 × 80 / (100 × 1000)
                           = 0.016 m/s
    Question 466
    CBSEENPH11020077

    A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

    Solution

    The given question can be illustrated as in the following fig. below.

    Here, 
    AO = incident path of the ball, 
    OB = Path followed by ball after deflection, 
    < AOB = Angle between the incident and deflected paths of the ball = 45o

    Therefore, 
    ∠AOP = ∠BOP = 22.5° = θ
    Initial velocity of the ball = Final velocity of the ball = v
    On resolving the component of velocity along v, we have
    Horizontal component of the initial velocity = vcos θ, along RO 
    Vertical component of the initial velocity = vsin θ, along PO
    Horizontal component of the final velocity = vcos θ, along OS
    Vertical component of the final velocity = sin θ,  along OP
    The horizontal components of velocities suffer no change.
    The vertical components of velocities are in the opposite directions.
    So, change in linear momentum of the ball gives us the impulse which is imparted to the ball. 
    That is, 
    Impulse = mvCosθ - (-mvCosθ) 
                 =  2mvCosθ 
    Mass of the ball, m = 0.15 kg 
    Velocity of the ball, v = 54 km/h
                                    = 15 m/s
    Therefore, 
    Impulse = 2 x 0.15 x 15 cos 22.5o
                 
    = 4.16 kg m/s 



    Question 467
    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 468
    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. 

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