A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find:

Question

A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find:

a. The speed of the skater after throwing the stone.
b. The distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between his skates and the ice is 0.03.

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Huyền Thanh 3 years 2021-08-16T03:54:25+00:00 2 Answers 88 views 1

Answers ( )

  1. Cherry
    -1
    2021-08-16T03:55:34+00:00
    Reply

    Answer:

    Explanation:

    know that there is no external force on skater and the stone so the total momentum of the system will remains constant

    so we will have

    here we have

    so the skater will move back with above speed

    now the deceleration of the skater is due to friction given as

  2. niczorrrr
    0
    2021-08-16T03:56:04+00:00
    Reply

    Answer:

    (a) 3.553 m/s

    (b) 21.46 m

    Explanation:

    (a) Applying the law of of momentum,

    Total momentum before collision = Total momentum after collision

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

    Where m and m’ are the mass of skater and stone respectively,  u and u’ are the initial velocity of skater and stone respectively, v and v’ are the final velocity of the skater and the stone respectively.

    Note, u = 0 m/s, u’ = 0 m/s

    Therefore,

    0 = mv+m’v’

    -mv = m’v’……………. Equation 2

    make v the subject of the equation

    v = -m’v’/m…………. Equation 3

    Given: m = 45 kg, m’ = 7.65 kg, v’ = 20.9 m/s

    Substitute into equation 3

    v = 7.65(20.9)/45

    v = -3.553 m/s

    Hence the speed of the skater = 3.553 m/s

    (b) F = mgμ…………..Equation 4

    But F = ma

    Therefore,

    ma = mgμ

    a = gμ…………… Equation 5

    Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction

    Given: μ = 0.03, g = 9.8 m/s²

    Substitute into equation 5

    a = 0.03(9.8)

    a = 0.294 m/s²

    Using the equation of motion,

    v² = u²+2as…………. Equation 6

    Where s = distance moved by the skater.

    note that u = 0 m/s.

    therefore,

    v² = 2as

    s = v²/2a……………. Equation 7

    Given: v = 3.553 m/s, a = 0.294

    Substitute into equation 7

    s = 3.553²/(2×0.294)

    s = 12.62/0.588

    s = 21.46 m

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