Answer to Question #274898 in Classical Mechanics for Log

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m₁ = 4.99 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m₂ = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m₁ rises after the elastic collision.

Answer 1

$\frac{m_1v_1^2}2=m_1gh, \implies$

$v_1=\sqrt{2gh}=9.90~\frac ms,$

$m_1v_1-m_1u_1=m_2u_1+m_2v_1,\implies$

$u_1=\frac{m_1-m_2}{m_1+m_2}v_1=-3.52~\frac ms,$

$m_1gh_{max}=\frac{m_1u_1^2}2,\implies$

$h_{max}=\frac{u_1^2}{2g}=0.63~\frac ms.$