91Ó°ÊÓ

The given data can be listed below,

• The mass of the particle placed at P is, m = 0.0150 kg
• The angle at which the particle is aligned below x axis is,$∠={45}^{°}$
• The mass of the sphere which is on x and y axis is,role="math" localid="1655806402339" $m_p=1\mathrm{kg}$
• The distance of the particle from point P is, r = 0.50 m.
• The mass of the sphere which is inclined at 45 degree is,$m_{in}=2\mathrm{kg}.$

Mass creates a Gravitational field. If some other particle with mass enters this field, it experiences a gravitational force. The gravitational field's intensity is a continuously varying spatially variable value at any position.

The gravitational force of the particle is given by,

$F_g=\frac{G{m}_{p}m}{r^2}$

Here,$m_p$is the mass of the particle at x or y axis, G is the constant of gravitation whose value is$6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2,m$,m is the mass of the sphere, and r is the distance of the particle from P.

The net gravitational force will be in the direction of an angle of 45° from the x-axis, and the magnitude is given by,

$F_R=Gm\left[\frac{m_{in}}{2r^r}+\frac{m_p}{r^2}\right]$

Substituting all the values in the above equation,

$$\begin{gathered} F_R=6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2(0.0150\mathrm{kg})\left[\frac{2\mathrm{kg}}{2{(0.50\mathrm{m})}^2}+\frac{1\mathrm{kg}}{{(0.50\mathrm{m})}^2}\sin {45}^{°}\right] \\ =9.67×{10}^{-12}\mathrm{N} \\ \mathrm{Thus},\mathrm{the}\mathrm{resultant}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{gravitational}\mathrm{force}\mathrm{is}9.67×{10}^{-12}\mathrm{N}. \end{gathered}$$

The kinetic energy of the particle is given by,

$K.E=\frac{1}{2}m{v}^2$

Here, m is the mass of the particle and v is the velocity of the particle.

Potential energy of the particle is given by,

$U=-\frac{G{m}_{p}m}{r}$

Here, G is the constant of gravitation whose value is $6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2,m_p$, is the mass of the particle, m is the mass of the sphere and r is the distance of particle from P.

$\frac{1}{2}m{v}^2=\frac{G{m}_{p}m}{r}$

The initial displacement is so large that the initial potential energy may be taken to be zero. Substituting the values in the above equation.

$$\begin{gathered} \frac{1}{2}\mathrm{m}{\mathrm{v}}^2=6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\left(\mathrm{m}\right)\left[\frac{2\mathrm{kg}}{\sqrt{2}\left(0.05\mathrm{m}\right)}+\frac{1\mathrm{kg}}{(0.50\mathrm{m})}2\right] \\ \mathrm{v}=\sqrt{2×6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\left[\frac{2\mathrm{kg}}{\sqrt{2}\left(0.05\mathrm{m}\right)}+\frac{1\mathrm{kg}}{(0.50\mathrm{m})}2\right]\left(\frac{1\mathrm{kg}.\mathrm{m}/\mathrm{s}}{1\mathrm{N}}\right)} \\ =3.02×{10}^{-5}\mathrm{m}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{when}\mathrm{it}\mathrm{reaches}\mathrm{its}\mathrm{origin}\mathrm{is}3.02×{10}^{-5}\mathrm{m}/\mathrm{s}. \end{gathered}$$