The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]
the momentum is defined by the product of the mass and the velocity of the body.
p = mv
The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.
If the system is isolated, the forces during the collision are internal and the it is conserved. Let's find the momentum is two instants.
Initial instant. Before crash.
p₀ = m₁ u₀ + 0
Final moment. After crash.
[tex]p_f = m_1 u_f + m_2 v_f[/tex]
The momentum is preserved.
p₀ = [tex]p_f[/tex]
[tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]
Since the collision is elastic, the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex] + ½ m₂ [tex]v_f^2[/tex]
Let's write our system of equations.
[tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]
Let's solve
[tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]
[tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]
[tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
Learn more here: brainly.com/question/8351094
