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 ​
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Answer 1

Answer:

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

Explanation:

If the tension in the string is $T$, for $m_1$ we have

$T- m_1g =m_1a$,

and for the mass $m_2$

$T -m_2g = -m_2a$

From these equations we solve for $a$ and get:

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

The kinematic equation

$v_f^2 = v_0^2+2ax$

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

In our case

$v_0 = 0$ (the initial velocity of the particles is zero)

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

which gives us

$v_f^2 = 2ax$

$v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g$

$\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }$

which is the velocity with which the mass $m_2$ will hit the floor.