# Finding Acceleration with Unknown Force & Mass

mcdowellmg

## Homework Statement

"A certain force gives an object of mass m1 an acceleration of 12.2 m/s^2 and an object of mass m2 an acceleration of 4 m/s^2. What acceleration would the force give to an object of mass m2-m1, and also of an object of mass m2+m1?"

## Homework Equations

acceleration = force/mass

## The Attempt at a Solution

I deduced that a = F/(m2-m1). From there, I calculated that a*m2 = a*m1, because the F is the same throughout. I worked a little algebra...4*m2 = 12.2*m1, so 4/12.2*m2 = m1. That means that m1 is .32787 of m2, and thus m2 - m1 would leave an "m3" that is .67213 of m2. I tried to apply this to the acceleration by calculating .67213*4...getting an acceleration of 2.7, but that would mean m3 is greater than m2 in mass...which could not be possible. I guess I am getting a little lost in my own (il)logic. Any ideas?

Thanks!

Homework Helper
welcome to pf!

hi mcdowellmg! welcome to pf!

(try using the X2 and X2 icons just above the Reply box )
I deduced that a = F/(m2-m1).

uhh? what is a?

start again, and first convert the following words into two equations:
"A certain force gives an object of mass m1 an acceleration of 12.2 m/s^2 and an object of mass m2 an acceleration of 4 m/s^2.

mcdowellmg
a is acceleration, sorry!

12.2*m1 = F and 4*m2 = F, so I know that 12.2*m1 = 4*m2, because F is the same throughout. After that, I'm stuck...and frustrated!

Thanks!

Homework Helper
a is acceleration, sorry!

12.2*m1 = F and 4*m2 = F, so I know that 12.2*m1 = 4*m2, because F is the same throughout. After that, I'm stuck...and frustrated!

Thanks!
So m1= F/12.2 and m2= F/4. Now m1- m2= F/12.2- F/4= F/?

mcdowellmg
m2-m1 = F/8.2 ?

Am I not able to get an actual number?

Homework Helper
look at the question!!

it doesn't ask for m2 - m1, it only asks for the acceleration of such a mass

which is … ?

mcdowellmg
Oops! So it is as simple as subtracting the accelerations to get 8.2? I find that hard to believe! I guess I made it too complicated. Thanks.