An 9.00 kg point mass and a 12.0 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 20.0 cm from the 9.00 kg mass along the line connecting the two fixed masses.
a) Find the magnitude of the acceleration of the particle. (in m/s^2)
b) Find the direction of the acceleration of the particle. (either toward the 12 kg mass or the 9kg mass)© BrainMass Inc. brainmass.com December 24, 2021, 9:13 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
What we basically need to do is to first find the gravitational force of attraction between the particle and each of the masses using Newton's law of universal gravitation;
F=(G * m1 * m2) / r^2
where G= 6.67 * 10^-11 is the universal gravitational constant, m1 and m2 are the masses in attraction and r is there separation.
Now, to get the force of attraction between the particle (mass m) and the 9 kg mass, their separation is 20 cm (or 0.2 m), so the gravitational force of attraction between them is
F9 = (6.67 * 10^-11 * m * 9) / 0.2^2 = 1.5 * 10^-8 * m in Newtons
And between the particle (mass m) and the 12 kg mass, the gravitational force of attraction is
F12 = (6.67 * 10^-11 * m * 12) / 0.3^2 = 0.9 * 10^-8 * m in Newtons
NB: the separation was 50cm - 20cm = 30cm (or 0.3m).
Now, the force attracting the particle to the 9kg mass is greater than that attracting it to the 12kg mass, so the resultant force will be in the direction of the 9kg mass, and will have a value of
F= F9 - F20 = (1.5 * 10^-8 * m) - (0.9 * 10^-8 * m) = 0.6 * 10^-8 * m in Newtons
And finally, the magnitude of the acceleration is
a = F/m (This is from Newton's second law of motion)
a = (0.6 * 10^-8 * m) /m = 0.6 * 10^-8 (in m/s^2) OR same as 6 * 10^-9 m/s^2
For the b) part of your question, the direction of the acceleration of the particle is the same as the direction of the resultant force on the particle, that is toward the 9kg mass as mentioned above.© BrainMass Inc. brainmass.com December 24, 2021, 9:13 pm ad1c9bdddf>