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Two students on roller skates stand face-toface, then push each other away. One student has a mass of 93 kg and the second student 65 kg. Find the ratio of the magnitude of the first student’s velocity to the magnitude of the second student’s velocity.

Answer :

Answer:

[tex]\frac{v_1}{v_2} = 0.698[/tex]

Explanation:

As we know that the two students are standing on skates

So there is no external force on the system of two students

So we can say that momentum is conserved

So here initially both students are at rest and hence initial momentum is zero

So we have

[tex]P_i = P_f[/tex]

[tex]m_1v_1 + m_2v_2 = 0[/tex]

[tex]\frac{v_1}{v_2} = \frac{m_2}{m_1}[/tex]

[tex]\frac{v_1}{v_2} = \frac{65}{93}[/tex]

[tex]\frac{v_1}{v_2} = 0.698[/tex]

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