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11 September, 11:30

A 75.2 kg ice skater, moving at 7.7 m/s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 3.85 m/s. Suppose the average force a skater can experience without breaking a bone is 4824 N. If the impact time is 0.103 s, what is the magnitude of the average force each skater experiences?

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Answers (2)
  1. 11 September, 11:35
    0
    2810.8 N, no bone will break

    Explanation:

    We have,

    m1=m2=75.2kg

    vi1=7.7m/s

    vi2=0m/s

    vf1=vf2=3.85m/s

    Fmax = 4824N

    Favg=?

    To calculate the average force we need to calculate the impulse I

    According to the Impulse-Momentum Theorem

    I=Δp=FavgΔt

    We know that momentum is always conserved so Δp1=Δp2

    As average force exerted on both the skaters will be same due to action-reaction phenomena we find impact of force for the second skater.

    I=Δp2=Favg2Δt

    mvf2-mvi2=Favg2Δt

    ∵vi=0 ∴mvi2=0

    Favg2=mvf2/Δt

    Favg2 = (75.2*3.85) / 0.103

    Favg2=2810.8N

    ∵Favg2
    ∴No bone will break
  2. 11 September, 11:36
    0
    F1 = - 2810.87N

    F2 = 2810.87N

    Their bones are safe.

    Explanation:

    Change in momentum = impulse

    ∆p = F∆t

    F = ∆p/∆t

    Given;

    m1 = m2 = 75.2kg

    v1i = 7.7m/s

    v2i = 0m/s

    v1f=v2f = 3.85m/s

    ∆t = 0.103s

    Where;

    m1 and m2 are mass of the skaters

    v1i and v2i are the initial velocity of the skaters

    v1f and v2f are the final velocity of the skaters

    F1 and F2 are the forces experienced by skater 1 and skater 2

    For skater 1

    F1 = ∆p1/∆t

    F1 = m1 (v1f - v1i) / ∆t

    F1 = 75.2kg (3.85-7.7) / 0.103

    F1 = - 2810.87N

    For skater 2

    F2 = ∆p2/∆t

    F2 = m2 (v2f - v2i) / ∆t

    F2 = 75.2kg (3.85-0) / 0.103

    F2 = 2810.87N

    Therefore, their bone is safe since their average force is less than 4824N
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