Question

Two masses are connected by a light string passing over a
light, frictionless pulley, as shown in the figure below. The
object of mass m1 is released from rest at height h above the
table. Use the isolated system model to answer the following. (We
assume m1 > m2.)

(a) Determine the speed of m2 just as m1 hits the ground. (Use
any variable or symbol stated above along with the following as
necessary: g.)

v =

(b) Find the maximum height above the ground to which m2
rises. (Use any variable or symbol stated above along with the
following as necessary: g.)

hmax =

Answer #1

Initial energy of the system

= potential energy of the mass 1 = m_{1}gh
---------(1)

Now final energy of the system when mass 1 hits the floor

Final energy

= KInetic energy of mass 1 and 2 + potential energy of mass 2

= (1/2)(m_{1} +m_{2})V^{2} +
m_{2}gh -------------(2)

Applying the energy conservation

Initial = final energy

(1/2)(m_{1} +m_{2})V^{2} + m_{2}gh
= m_{1}gh

(1/2)(m_{1} +m_{2})V^{2} = (m_{1} -
m_{2})gh

(b) Now the kinetic energy of the mass 2 when mass 1 hits the floor
is

K = (1/2)mV^{2}

Now this energy will consumed by increasing the potential energy ,
therefore

At a height say h_{2} it will get consumed, therefore

(1/2)m_{2}V^{2} = m_{2}gh_{2}

h_{2} = (1/2g)V^{2}

Hence the maximum height from the ground

h_{max} = h + h_{2}

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