Answer
a) $v_f=\sqrt{2gh(\frac{m_1-m_2}{m_1+m_2})}$
b) $v_f=1.1m/s$
Work Step by Step
(a) We can find the required speed as
$K.E_i+U_i=K.E_f+U_f$
$\implies 0+0=(\frac{1}{2})m_1v_f^2+(\frac{1}{2})m_2v_f^2+m_1gy_1+m_2gy_2$
$\implies 0=(\frac{1}{2})(m_1+m_2)v_f^2+m_1gh+m_2g(-h)$
$\implies (m_2-m_1)gh=(\frac{1}{2})(m_1+m_2)v_f^2$
This simplifies to:
$v_f=\sqrt{2gh(\frac{m_1-m_2}{m_1+m_2})}$
(b) We know that
$v_f=\sqrt{2(9.81)(1.2)(\frac{4.1-3.7}{3.7+4.1})}$
$v_f=1.1m/s$
