Question #274898

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.


1
Expert's answer
2021-12-03T12:45:00-0500

m1v122=m1gh,    \frac{m_1v_1^2}2=m_1gh, \implies

v1=2gh=9.90 ms,v_1=\sqrt{2gh}=9.90~\frac ms,

m1v1m1u1=m2u1+m2v1,    m_1v_1-m_1u_1=m_2u_1+m_2v_1,\implies

u1=m1m2m1+m2v1=3.52 ms,u_1=\frac{m_1-m_2}{m_1+m_2}v_1=-3.52~\frac ms,

m1ghmax=m1u122,    m_1gh_{max}=\frac{m_1u_1^2}2,\implies

hmax=u122g=0.63 ms.h_{max}=\frac{u_1^2}{2g}=0.63~\frac ms.


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