Q8E An 8.00鈥塳驳 point mass and a聽1... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q8E (page 425)

An $8.00 鈥� kg$ point mass and a $12.00 鈥� kg$ point mass are held $50.0 鈥� cm$ apart. A particle of mass mis released from a point between the two masses $20.0 鈥� cm$ from the $8.00 鈭� kg$ group along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Short Answer

Expert verified

With the magnitude and direction of the acceleration of the particle $+ 2.22 脳 10^{鈭� 9} 鈥尘 / s^{2}$ areas, the net force is mainly directed towards the first mass.

Step by step solution

Step 1: Identification of the given data

The given data can be listed below as follows,

The point mass is of $8.00 鈭� kg$ and $12.00 鈭� kg$ respectively.
The distance amongst the point mass is $50.0 鈥塩尘$ .
The known value of gravitational constant G is $6.67 脳 10^{鈭� 11} {鈥塏办驳}^{-2} m^{2}$ .
The distance of the particle of mass from the $8.00 鈭� kg$ point mass is $20 cm$ .
The distance of the middle mass from the first and second mass is $50 鈭� 20 = 30 鈥塩尘$ .

Significance of Newton’s law of gravity on the particle of mass

The law elucidates that a particle in the universe attracts the same or different particle with the help of a force that is inversely proportional to the square of the particles' distances and directly proportional to the product of the masses.

The product of the masses, which is divided by the square of the distances of the groups, gives the absolute value of the net force. Moreover, the acceleration of the particle is to be found with the help of the total value of the net force.

Determination of the magnitude and direction of the acceleration of the particle

From Newton鈥檚 law of gravitation, the attraction of the middle particle with the first and second particle is expressed as:

$F = G \frac{m_{1} m_{2}}{r^{2}}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌)

Where Gis the gravitational constant, $m_{1}$ is the first mass and $m_{2}$ is the second mass, and ris the distance amongst the groups.

Furthermore, substituting the equation i) to find out the absolute value of the net force, which is expressed as:

role="math" localid="1655713254679" $\begin{matrix} F_{0} = [F_{1} 鈭� F_{2}] \\ = Gm [\frac{m_{1}}{r_{1}^{2}} 鈭� \frac{m_{2}}{r_{2}^{2}}] \end{matrix}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌颈)

Here $r_{1}$ is the distance of the particle of mass from the first mass and $r_{2}$ the distance of the middle group from the first and second mass.

The absolute value of the acceleration gathered from equation ii) is as follows:

$\begin{matrix} a = \frac{F}{m} \\ = G [\frac{m_{1}}{r_{1}^{2}} 鈭� \frac{m_{2}}{r_{2}^{2}}] \end{matrix}$

Substituting the provided values in the above equation, we get,

$\begin{matrix} F = 6 .67 脳 10^{鈭� 11} {鈥塏办驳}^{-2} m^{2} 脳 [\frac{8 鈥塳驳}{20 {鈥塩尘}^{2}} 鈭� \frac{12 鈥塳驳}{30 {鈥塩尘}^{2}}] \\ 鈮� + 2 .22 脳 10^{鈭� 9} 鈥尘 / s^{2} \end{matrix}$

Thus, the magnitude and direction of the acceleration of the particle $+ 2.22 脳 10^{鈭� 9} 鈥尘 / s^{2}$ areas, the net force is mainly directed towards the first mass.