Laws of Motion

  • Question 457
    CBSEENPH11020074

    A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

    Solution
    Let mm1, and m2 be the respective masses of the parent nucleus and the two daughter nuclei.
    The parent nucleus is at rest.
    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 + m2v

    According to the law of conservation of momentum, 
    Total initial momentum = Total final momentum 
                                     0 = m1v1 + m2v

                                    v1 = -m2v2 / m

    Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.
    Question 458
    CBSEENPH11020075

    Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s–1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
     

    Solution

    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= - 0.3 kg m/s
    Impulse imparted to each ball = pf - pi = -0.3 -0.3 = -0.6 kg m/s
    The negative sign indicates that the impulses imparted to the balls are opposite in direction.

    Question 459
    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 460
    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 



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