Question One

The equation derived in the lab manual is hereby recalled as;

a=(F_M-F_m)/(M+m)

=(Mg-mg)/(M+m)

=[(M-m)/(M+m)]g

Therefore to determine whether this equation makes sense, we will equate each of the masses to zero. Experimentally, when one mass is equated to zero, say, by removing all the pennies in one mass, or by simply cutting the string at one arm of the Atwood machine, then the other mass will fall freely to the graduated at an acceleration equal to the force of gravity, irrespective of the quantity of the mass (In cases where only the gravitational force acts on the mass). Therefore in this case, the acceleration is supposed to be equal to:

Use your promo and get a custom paper on

**"Expansion Questions And Answers".**

a=g

Now, equating the smaller mass to zero theoretically into the equation:

a=[(M-0)/(M+0)]g=M/M g

Giving

a=g

This confirms our expectation that the acceleration during the free fall of the larger mass will be due to the force of gravity only which makes sense.

Question Two

Now, suppose we equate the larger mass to zero. Similarly, this can experimentally be achieved by removing all the pennies in the larger mass, or by simply cutting the string of the arm of the Atwood machine having the smaller mass, the heavier mass will fall freely to the ground at an acceleration equal to the force of gravity. Therefore, similar to the previous case, the acceleration is supposed to be equal to:

a=g

Now, equating the larger mass to zero theoretically into the equation;

a=[(0-m)/(0+m)]g=(-m)/m g

Giving

a=-g

This goes against the expectation where the acceleration in a free fall is expected to be equal the acceleration due to gravity. In this case, the acceleration is seen to be equal to the acceleration due to gravity but towards the negative direction. Since this case does not happen in reality, we can say that equating the larger mass to zero in the equation theoretically does not make sense.