Physics: Two ice skaters stand at rest in the center of an ice rink.?

When they push off against one another the 70-{rm kg} skater acquires a speed of 0.66 m/s.

If the speed of the other skater is 0.83 m/s, what is this skater's mass?

3 Answers

  • Momentum before = momentum after

    (M1+M2)*V = M1V1 + M2V2

    0 = M1V1 + M2V2

    M1v1 = - M2V2 [Taking 1 to represent 70kg skater, and 2 the other]

    M2 = - (M1V1)/V2

    M2 = - (70x0.66)/0.83

    M2 = ~ - 55.7 [Minus represents motion in opposite direction]

  • conservation of momentum: m1v1 = m2v2. m1 = sixty six kg; v1 = 0.sixty 4 m/s. m2 = x; v2 = 0.80 3 m/s. m1v1 = (sixty six kg)(0.sixty 4 m/s) = 40 two kg·m/s (to 2 significant figures). x·v2 = m1v1, so x = m1v1 / v2 = (40 two kg·m/s) / (0.80 3 m/s) = fifty one kg.

  • they are both 𝒈𝒶𝓎.

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