Question

An 8.00-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 8.00-kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Answer 1

Answer:

Acceleration of the particle is 4.45 m/s² and acceleration direction is towards the 8 kg point mass.  

Explanation:

Let m₁ be 8 kg point mass and m₂ be 12 kg point mass and take direction along m₁ mass as positive x-axis. The separation between twp point masses is 50 cm. According to the problem, a particle of mass m is held between the two point masses m₁ and m₂.

Distance between m₁ and m, r₁ = 20 cm = 0.2 m

Distance between m₂ and m, r₂ = (50 - 20) cm =30 cm = 0.3 m

The particle of mass m experiences an attractive gravitational force due to both the point masses. Thus, the net force on the particle of mass m is :

F = F₁ + F₂

F₁ is acting along the positive x axis and F₂ is acting along the negative x axis.

$F = \frac{Gm_{1}m }{r_{1} ^{2} } - \frac{Gm_{2}m }{r_{2} ^{2} }$

Substitute the values of G, m₁, m₂, r₁ and r₂ in the above equation.

$F = \frac{6.67\times10^{-11} \times8m }{0.2 ^{2} } - \frac{6.67\times10^{-11}\times12m }{0.3^{2} }$

F = 4.45m N/kg

Acceleration of the particle = $\frac{F}{m}$ = 4.45 m/s²

The direction of acceleration of the particle is towards the 8 kg point mass.