Question

Aт 8.0kg point mass and a 15.0kg point mass are held in place 50.0cm apart. A particle of mass m is released from a point between the masses, 20cm from the 8.0kg mass along a line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Answer 1

Let us use notation:

$m_1 = 8 kg, m_2 = 15 kg$ - masses of the particle 1 and 2 respectively, $m$ - the mass of the middle particle.

$r = 50 cm$ - distance between the particles,

$r_1 = 20 cm, r_2 = r - r_1 = 30 cm$ - distances from the middle particle to the first and second respectively.

The first and second particles attract the middle particle with the force $F_{1,2} = G \frac{m_{1,2} m}{r_{1,2}^2}$, according to Newton's law of universal gravity. The absolute value of the net force is $F = |F_1 - F_2| = G m \left| \frac{m_1}{r_1^2} - \frac{m_2}{r_2^2}\right|$ . The value inside the modulus is approximately $33.3 \frac{kg}{m^2}$, i.e. it is positive, so the net force is directed towards the first particle. The absolute value of acceleration is $a = \frac{F}{m} = G \left| \frac{m_1}{r_1^2} - \frac{m_2}{r_2^2}\right| \approx 2.22 \cdot 10^{-9} \frac{m}{s^2}$ .