# 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 100 g and 110 g. The lighter mass is on the floor and the heavier mass is 75 cm above the floor.

Mar 7, 2017

$t \approx 1.83 s$

$v \approx 0.82 m {s}^{- 1}$

#### Explanation:

If we say that the heavier mass is ${m}_{2}$ and the lighter is ${m}_{1}$, and we idealise so that, for instance the pulley has no inertia, then the magnitude of the acceleration of the masses is:

$a = g \cdot \frac{{m}_{2} - {m}_{1}}{{m}_{2} + {m}_{1}} = \frac{g}{22}$

In terms of tile, the relevant equation of motion for the heavier mass (at constant acceleration) is:

$x = u t + \frac{1}{2} a {t}^{2}$

$t = \sqrt{\frac{2 x}{a}}$

$t = \sqrt{\frac{2 \cdot 0.75}{\frac{g}{22}}} \approx 1.83 s$

In terms of velocity, the relevant equation of motion for the heavier mass (at constant acceleration) is:

${v}^{2} = {u}^{2} + 2 a x$

$v = \sqrt{2 a x}$

$= \sqrt{2 \left(\frac{g}{22}\right) 0.75} \approx 0.82 m {s}^{- 1}$