Question

How long will it take for the heavier mass to reach the floor? What will be the speed of the two masses when the heavier mass hits the floor?
An Atwood’s machine has masses of $100g$ and $110g$ . The lighter mass is on the floor and the heavier mass is $75cm$ above the floor.

Answer 1

An Atwood’s machine works like a pulley having two strings on the two sides. In the given problem, these two strings have two different masses attached to them. The acceleration of the masses has to be calculated first. Thereafter, for the 1st query, the equation of motion for the heavier mass is to be used to find the required time. And, for the 2nd question, the equation of motion of the heavier mass is needed to calculate the final velocity. Note that two equations of motion are needed here: one should contain the time and another should have the final velocity.

Formula used:
If the two masses are ${m_1}$ and ${m_2}$( ${m_2} > {m_1}$),
The acceleration of these masses, $a = \dfrac{{{m_2} - {m_1}}}{{{m_2} + {m_1}}}g$ , $g$ is the acceleration due to gravity.
The equation of motion of the heavier mass having an initial velocity $u$covers $x$ distance in time $t$ with acceleration $a$,
$x = ut + \dfrac{1}{2}a{t^2}$
If the masses have the final velocity $v$,
${v^2} = {u^2} + 2ax$

Complete step-by-step solution:
Atwood's machine works like a pulley that has two strings. Let us consider two masses: ${m_1}$ and ${m_2}$are hung in those strings. ${m_1}$is the lighter mass and ${m_2}$is the heavier mass.
So, the acceleration of these masses, $a = \dfrac{{{m_2} - {m_1}}}{{{m_2} + {m_1}}}g$ , $g$ is the acceleration due to gravity.
Given that, ${m_1} = 100g$, ${m_2} = 110g$
$\therefore a = \dfrac{{110 - 100}}{{110 + 100}} \times 10$[ taking, $g = 10m/{s^2}$]
$\Rightarrow a = \dfrac{{10}}{{210}} \times 10$
$\Rightarrow a = \dfrac{{10}}{{21}}m/{s^2}$
The equation of motion of the heavier mass,
$x = ut + \dfrac{1}{2}a{t^2}$
Given that, initial velocity $u = 0$covers
Distance $x = 75cm = 0.75m$ in time $t$
acceleration $a = \dfrac{{10}}{{21}}$
$\Rightarrow 0.75 = 0 + \dfrac{1}{2} \times \dfrac{{10}}{{21}}{t^2}$
$\Rightarrow {t^2} = \dfrac{{0.75 \times 21 \times 2}}{{10}}$
$\Rightarrow {t^2} = 3.15$
$\Rightarrow t = 1.77\sec$
The heavier mass will reach the floor at the time $t = 1.77\sec$
If the mass has the final velocity $v$,
${v^2} = {u^2} + 2ax$
$\Rightarrow {v^2} = {0^2} + 2 \times \dfrac{{10}}{{21}} \times 0.75$
$\Rightarrow {v^2} = 0.714$
$\Rightarrow v = 0.844m/s$
Therefore, the speed of the two masses when heavier mass hits the floor is $v = 0.844m/s$

Note: An Atwood's Machine is an easy device having a pulley, with two masses attached by a wire that goes through the pulley. In an 'ideal Atwood's Machine,' it is assumed the pulley has negligible mass and is frictionless. The string is non-expandable as well as frictionless, hence a fixed length, and is also massless.