At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a prize. The ball

Question

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a prize. The ball bounces straight back at 30% of its incoming speed, knocking the bottle straight forward.

What is the bottle’s speed, as a percentage of the ball’s incoming speed?

Express your answer using two significant figures.

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Jezebel 5 months 2021-08-13T22:37:31+00:00 2 Answers 193 views 0

Answers ( )

    0
    2021-08-13T22:39:21+00:00

    Answer:

    v_{f2} =6.5%v_{i1}

    Explanation:

    Mass of the ball: m_{1} =0.12kg]

    Initial velocity of the ball:   v_{i1}

    final velocity of the ball: v_{f1} which is -30/100 of v_{i1} =-0.3v_{i1}

    Mass of the bottle: m_{2} =2.4kg

    Initial velocity of the bottle: v_{i2}=0m/s

    final velocity of the bottle: v_{f2} is unknown (to find)

    by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

    m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}

    so since the bottle is at rest firstly, therefore v_{i2} =0

    m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}

    m_{1} v_{i1}  =m_{1} v_{f1} +m_{2} v_{f2}         equation 1

    so now substitute v_{f1} into equation 1

    m_{1} v_{i1}  =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}

    m_{1} v_{i1}  = -0.3m_{1}v_{i1}  +m_{2} v_{f2}

    collect the like terms

    m_{1} v_{i1}   +0.3m_{1}v_{i1}  =m_{2} v_{f2}

    1.3m_{1} v_{i1}   =m_{2} v_{f2}

    divide both  side by m_{2}

    v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }

    Now substitute

    v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2}    =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}

    v_{f2} =6.5%v_{i1}

    0
    2021-08-13T22:39:22+00:00

    Answer:

    v2 = 6.5% of the ball’s initial velocity

    Explanation:

    Let the mass of the ball and bottle be m1 and m2 respectively.

    m1 = 0.12kg , m2 = 2.4kg

    Let the initial and final velocity of ball be u1 and

    v1 = -0.3u1 while that of the bottle be u2 and v2

    The bottle was initially at rest so u2 = 0m/s

    From the principle of conservation of momentum, momentum before collision is equal to momentum after collision.

    m1u1 + m2u2 = m1v1 + m2v2

    0.12u1 + m2×0 = 0.12(–0.3u1) + 2.4v2

    0.12u1 = –0.036u1 + 2.4v2

    Collecting like terms

    2.4v2 = 0.12u1 + 0.036u1

    2.4v2 = 0.156u1

    v2 = 0.156u1/2.4

    v2 = 0.065u1 = 6.5%u1

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