A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest. After the co

Question

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest. After the collision, the 4.5 g coin moves to the left at 11.9 cm/s. (a) Find the final velocity of the other coin. 7.886 Incorrect: Your answer is incorrect. cm/s (b) Find the amount of kinetic energy transferred to the 13.5 g coin. 0.42 Incorrect: Your answer is incorrect. J

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Trúc Chi 6 months 2021-08-09T07:36:44+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-08-09T07:37:45+00:00

    Answer:

    (a) 11.9 cm/s or v’ = 0.119 m/s

    (b) 9.56×10⁻⁵ J

    Explanation:

    (a)

    From the law conservation of momentum,

    total momentum before collision = Total momentum after collision

    For elastic collision,

    mu +m’u’ = mv+m’v’………………… Equation 1

    Where m = mass of the first coin, u = initial velocity of the first coin, m’ = mass of the second coin, u’ = initial velocity of the second coin, v = final velocity of the first coin, v’ = final velocity of the second coin

    Note: Since the second coin was initially at rest, u’ = 0 m/s, and m’u’ = 0

    Therefore,

    mu = mv+m’v’

    make v’ the subject of the equation

    v’ = (mu-mv)/m’……………………. Equation 2

    Let: The right direction be positive and the left be negative.

    given: m = 4.5 g, u = 23.8 cm/s, v = -11.9 cm/s (left) m’ = 13.5 g

    Substitute into equation 2

    v’ = [4.5×23.8-4.5×(-11.9)]/13.5

    v’ = (107.1+53.55)/13.5

    v’ = 160.65/13.5

    v’ = 11.9 cm/s or v’ = 0.119 m/s

    Hence the final velocity of the final velocity of the other coin = 0.119 m/s

    (b)

    The amount of kinetic energy transferred = 1/2m'(v’²-u’²)…………… Equation 3

    Given: m’ = 13.5 g = 0.0135 kg, v’ = 0.119 m/s, u’ = 0 m/s (at rest)

    Substitute into equation 3

    The amount of kinetic energy transferred = 1/2(0.0135)(0.119²)

    The amount of kinetic energy transferred = 0.00675(0.014161)

    The amount of kinetic energy transferred = 9.56×10⁻⁵ J

    0
    2021-08-09T07:38:10+00:00

    Answer:

    a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

    Explanation:

    a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

    (4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

    The final velocity is:

    v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

    b) The change in the kinetic energy of the 13.5 g coin is:

    \Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

    \Delta K = 9.559\times 10^{-5}\,J

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