Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi=0). (Figure 1)After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1>m2, so that after the collision, the two objects move off in the direction of the first object before the collision. What is the final speed vf of block 2?
Answers
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]
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