Question

Particles of m1 and m2 (m2>m1) are connected by a line in extensible 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, with what speed will mass m2 hit the floor and the mass will rise a further distance (m2-m1) x) /m1+m2 after this occurs ​

Answer 1

The height is  h = (m2 - m  1  )(m 2  + m  1 ) x

Explanation:

(m2  + m  1  )a = (m  2 −m  1 )g

→ a=  (m  2  - m  1  ) (m  2  + m 1  )  g

V^2 =2ax

The speed is

v = √ (m  2  - m  1  ) (m  2 + m  1  ) 2gx

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

m  2 gh = 0.5 m  2  v^2

h = (m2 - m  1  )(m 2  + m  1 ) x

Thus the height is  h = (m2 - m  1  )(m 2  + m  1 ) x

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