Answer :
The correct ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity is 65:93.
To find the ratio of the magnitudes of the velocities of the two students after they push each other away, we can use the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum of a system remains constant.
Let's denote the velocities of the first and second students as[tex]\( v_1 \)[/tex]and [tex]\( v_2 \)[/tex] respectively. Since the students push each other in opposite directions, their momenta will be equal in magnitude but opposite in direction. We can write the conservation of momentum as:
[tex]\[ m_1 \cdot v_1 = m_2 \cdot v_2 \][/tex]
where [tex]\( m_1 = 93 \)[/tex] kg is the mass of the first student and[tex]\( m_2 = 65 \) kg[/tex] is the mass of the second student.
To find the ratio of the velocities, we divide both sides of the equation by[tex]\( m_2 \cdot v_2 \)[/tex]:
[tex]\[ \frac{m_1 \cdot v_1}{m_2 \cdot v_2} = 1 \][/tex]
[tex]\[ \frac{v_1}{v_2} = \frac{m_2}{m_1} \][/tex]
Substituting the given masses:
[tex]\[ \frac{v_1}{v_2} = \frac{65 \text{ kg}}{93 \text{ kg}} \][/tex]
Simplifying the ratio, we get:
[tex]\[ \frac{v_1}{v_2} = \frac{65}{93} \][/tex]
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]
