Question

Suppose you have a glider and a hanging mass that you wish to know the masses of but you do not have a triple-beam balance available. You set up an experiment like that done in Lab 4: Newton's Second Law, in which a glider moves across an essentially frictionless airtrack, pulled by a light string which passes over a pulley and connects to a hanging mass. You take the following data: H = 75.0 cm t = 1.8 s Where H is the distance the hanging mass falls and t is the time it takes to fall (t has been averaged over five trials). Using methods similar to those in Lab 4, determine which of the following could be the masses of the glider (m1) and the hanging mass (m2).

Answer 1

the choice of a glider to hanging mass is a ratio of 20.2

Step-by-step explanation:

The Newton´s second law for glider is:

Fnet=Ma

T=Ma

The Newton´s second law for hanging mass is:

Fnet=ma

mag-T=ma

Replacing:

mag-Ma=ma (eq. 1)

Clearing a:

a=(mg)/(m+M)

The expression for motion is:

`y=vot+(1)/(2)at^(2)`

`H=0+(1)/(2)at^(2)`

Clearing a:

`a=(2H)/(t^(2) )`

Replacing values:

`a=(2*0.75)/(1.8^(2) ) =0.46`

From eq. 1:

M/m=20.2