Question

Two$\mathbf{20}\mathbf{}\mathbf{Kg}$spheres are fixed in place on a yaxis, one at$\mathit{y}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.40}\mathbf{}\mathit{m}$and the other at$\mathit{y}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{.40}\mathbf{}\mathit{m}$. A 10 kg ball is then released from rest at a point on the xaxis that is at a great distance (effectively infinite) from the spheres. If the only forces acting on the ball are the gravitational forces from the spheres, then when the ball reaches the (x, y) point$\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}\mathbf{,}$,what are (a) its kinetic energy and (b) the net force on it from the spheres, in unit-vector notation?

Short answer

1. The kinetic energy of a ball at a point (0.30, 0),$K=5.3\times {10}^{-8}J.$
2. The net gravitational force acting on a ball from the spheres at (0.30,0) in unit vector notation, $F_{net}=-6.4\times {10}^{-8}Ni$.

Step-by-step solution

Listing the given quantities

The mass of each sphere is $M=20kg$

The mass of ball is$m=10kg$

Understanding the concept of conservation of energy

Using the law of conservation of energy, we can find the kinetic energy of a ball at a point (0.30m, 0). From this, we can find the expression for x component of the force. Then inserting the value of theta in it, we can find the net gravitational force acting on the ball from the spheres at (0.30, 0) in unit vector notation.

Formulae:

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ F=\frac{GMm}{r^2} \end{gathered}$$

(a) Calculations of kinetic energy of a ball at a point(0.30, 0)

We have the equation for conservation of energy as

$K_1+U_1=K_2+U_2$

Now, the sphere starts from rest, meaning$v_1=0or{K}_1=0.$

Also, initially the sphere is at infinite distance, so$U_1=0.$

So, the conservation of energy equation becomes

$$\begin{gathered} K_2+U_2=0 \\ {K}_2-2\frac{GMm}{r}=0 \\ {K}_2=2\frac{GMm}{r} \end{gathered}$$

At the point (0.30,0), the distance 'r' of the ball from each sphere is 0.50 m.

So, the kinetic energy becomes$K_2=5.3\times {10}^{-8}J$

The kinetic energy of a ball at a point (0.30, 0),$K=5.3\times {10}^{-8}J.$

(b) Calculations of the net gravitational force acting on ball

Since, the y component of each force will cancel out, the net force is along x direction. It is given by

$$\begin{gathered} 2F_x=2\frac{GMm}{r^2}cos\theta \\ \theta =\left(\frac{0.4}{0.3}\right)=53° \\ {\overrightarrow{F}}_{net}=(-6.4\times {10}^{-8}N)\widehat{i} \end{gathered}$$

The net gravitational force acting on a ball from the spheres at (0.30,0) in unit vector notation, $F_{net}=-6.4\times {10}^{-8}Ni$