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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 ​
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Answer :

Poltergeist

Answer:

The velocity with which the mass will hit the floor is [tex]v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}[/tex]

Explanation:

If the tension in the string is [tex]T[/tex], for [tex]m_1[/tex] we have

[tex]T- m_1g =m_1a[/tex],

and for the mass [tex]m_2[/tex]

[tex]T -m_2g = -m_2a[/tex]

From these equations we solve for [tex]a[/tex] and get:

[tex]a =(\dfrac{m_2-m_1}{m_2+m_1}) g.[/tex]

The kinematic equation

[tex]v_f^2 = v_0^2+2ax[/tex]

gives the final velocity [tex]v_f[/tex] of a particle, when its initial velocity was [tex]v_0[/tex], and has traveled a distance [tex]x[/tex] while undergoing acceleration [tex]a[/tex].

In our case

[tex]v_0 = 0[/tex] (the initial velocity of the particles is zero)

[tex]a =(\dfrac{m_2-m_1}{m_2+m_1}) g.[/tex]

which gives us

[tex]v_f^2 = 2ax[/tex]

[tex]v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g[/tex]

[tex]\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }[/tex]

which is the velocity with which the mass [tex]m_2[/tex] will hit the floor.

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