Answer to Question #104005 in Mechanics | Relativity for Opatola

Question #104005
Particles of mass m1 and m2 (m2>m1) are connected by a large inextensible string passing over a smooth fixed pulley. Initially both masses hang vertically with mass m2 at a height X above the floor. If the system is released from rest, what speed will mass m2 hit the floor and the mass m1 will rise a further distance (m2-m1)x/m1+m2 after this occurs
1
Expert's answer
2020-02-28T10:30:23-0500
"(m_2+m_1)a=(m_2-m_1)g\\to a=\\frac{(m_2-m_1)}{(m_2+m_1)}g"

"v^2=2ax"

The speed is

"v=\\sqrt{\\frac{(m_2-m_1)}{(m_2+m_1)}2gx}"

For mass m2 we can apply the conservation of energy principle:


"m_2gh=0.5m_2v^2"

"h=\\frac{(m_2-m_1)}{(m_2+m_1)}x"


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