Question

An 8.75 kg point mass and a 13.5 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 18.0 cm from the 8.75 kg mass along the line connecting the two fixed masses.

A.Find the magnitude a of the acceleration of the particle.

B.Find the direction of the acceleration of the particle.

The acceleration of the particle is toward the 8.75 kg mass.

The acceleration of the particle is toward the 13.5 kg mass.

Answer 1

Given

   masses m1 = 8.75 kg, m2 = 13.5 kg

   separated by a distance = 50 cm = 0.5 m

   m1----------m---------------------------m2

   <--18cm-----><-------32----------------->
   <------------------50 cm---------------->

we know that the force of attraction between two masses separated by a distance r is

   F = Gm1*m2/r^2

F1 is ma1 = G*m1*m/r1^2       F2 is ma2 = G*m2*m/(r2^2)

   ma1 = G*8.75*m /0.18^2 ; ma2 = G*13.5*m /0.32^2
   a1 = 6.674*10^-11*8.75/0.18^2 ; a2 = 6.674*10^-11*13.5/0.32^2
   a1 = 1.8023919753086*10^-8 m/s2;    a2 = 8.79873046875*10^-9 m/s2

magnitude of the acceleration is a1+a2

           a = (-1.8023919753086*10^-8 +8.79873046875*10^-9)
           a = (-9.225189284336)*10^-9 m/s2

the dierection is in the -ve x direction towards m1