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Two ice skaters, one with a mass of 45 kg and the other with a mass of 60 kg, push off against one another starting from a stationary position. After they have pushed off, the relative speed at which they separate is 0.600m/s. What is the speed of the 60kg skater?

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boffeemadrid

Answer:

0.25714 m/s

Explanation:

[tex]m_1[/tex] = Mass of 45 kg skater

[tex]m_2[/tex] = Mass of 60 kg skater

[tex]v_1[/tex] = Speed of 45 kg skater

[tex]v_2[/tex] = Speed of 60 kg skater

As the momentum is conserved

[tex]m_1v_1+m_2v_2=0\\\Rightarrow 45v_1+60v_2=0\\\Rightarrow v_1=-\frac{60v_2}{45}\\\Rightarrow v_1=-\frac{4}{3}v_2[/tex]

[tex]v_r[/tex] = Relative speed = 0.6 m/s

[tex]v_r=v_2-v_1\\\Rightarrow v_r=v_2-\left(-\frac{4}{3}v_2\right)\\\Rightarrow 0.6=\frac{7}{3}v_2\\\Rightarrow v_2=\frac{0.6\times 3}{7}\\\Rightarrow v_2=0.25714[/tex]

The speed of the 60kg skater is 0.25714 m/s

Lanuel

The speed of the second ice skaters (60kg skater) is 0.257 m/s.

  • Let the mass of the first ice skaters be [tex]M_1[/tex].
  • Let the mass of the second ice skaters be [tex]M_2[/tex].

Given the following data:

  • Mass of first ice skaters = 45 Kg
  • Mass of second ice skaters = 60 Kg
  • Relative speed = 0.600 m/s.

To find the speed of the second ice skaters, we would apply the law of conservation of momentum:

[tex]M_1V_1 + M_2V_2 = 0\\\\45V_1 + 60V_2 = 0\\\\45V_1 = -60V_2\\\\V_1 = \frac{-60V_2}{45}\\\\V_1 = \frac{-4V_2}{3}[/tex]

Mathematically, the relative speed is given by the formula:

[tex]V_R = V_2 - V_1\\\\V_R = V_2 - (\frac{-4V_2}{3}) \\\\0.6 = \frac{7V_2}{3}\\\\1.8 = 7V_2\\\\V_2 = \frac{1.8}{7.2}[/tex]

Velocity, V2 = 0.257 m/s

Read more: https://brainly.com/question/23092552

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