the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact

Question

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat

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Thiên Di 5 years 2021-07-24T08:21:24+00:00 1 Answers 11 views 0

Answers ( )

  1. nguyetanh
    0
    2021-07-24T08:23:20+00:00
    Reply

    Solution :

    Given :

    Mass of the baseball, m = 200 g

    Velocity of the baseball, u = -30 m/s

    Mass of the baseball after struck by the bat, M = 900 g

    Velocity of the baseball after struck by the bat, v = 47 m/s

    According to the conservation of momentum,

    Mv+mu=Mv_1+mv_2

    (900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

    36300 =  (900 x v_1) + (200 x v_2)

    9v_1 + 2v_2 = 363 …………..(i)

    9v_1 = 363 - 2v_2

    v_1=\frac{363 - 2v_2}{9}

    The mathematical expression for the conservation of kinetic energy is

    \frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

    \frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    …………….(ii)

    $(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

    21681 = 9v_1^2+2v_2^2

    Substituting (i) in (ii)

    21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

    (363-2v_2)^2+18v_2^2=195129

    (363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

    22v_2^2-145v_2-63360=0

    Solving the equation, we get

    v_2=96 \ m/s, -30 \ m/s

    The negative velocity is neglected.

    Therefore, substituting 96 m/s for v_2 in (i), we get

    v_1=\frac{363-(2 \times 96)}{9}

         = 19

    Thus, only impulse of importance is used to find final velocity.

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