A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target h

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.50 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?

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  1. Answer:

    K = 0.076 J

    Explanation:

    The height of the target, h = 0.860  m

    The mass of the steel ball, m = 0.0120 kg

    Distance moved, d = 1.50 m

    We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

    [tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}[/tex]

    Put all the values,

    [tex]t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s[/tex]

    The velocity of the ball is :

    [tex]v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s[/tex]

    The kinetic energy of the ball is :

    [tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J[/tex]

    So, the required kinetic energy is 0.076 J.

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