91Ó°ÊÓ

The mass of each sphere is$m=60kg$ . The distance between centres of the spheres is$R=3.2m$ . At a particular instant the velocity of one of the spheres is role="math" localid="1656672572359" $\left(0,5,0\right)\mathrm{m}/\mathrm{s}$and the velocity of the other sphere is$\left(0,-5,0\right)\mathrm{m}/\mathrm{s}$ .

A force is a push or pull on an object is because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them .The acted force may be of attraction or of repulsion .The two items no longer feel the force after the interaction ends

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum$\frac{d\stackrel{â‡\P Ä}{p}}{dt}$ and the perpendicular rate of change of momentum$\frac{d\stackrel{â‡\P Ä}{p}}{dt}$are the two elements that we are concerned with.

$$\begin{gathered} F_{net}=\frac{d\stackrel{â‡\P Ä}{p}}{dt} \\ =\frac{d\stackrel{â‡\P Ä}{p}}{d{t}_{ll}}+\frac{d\stackrel{â‡\P Ä}{p}}{dt} \end{gathered}$$

In this case, the force acting on the object is in +y or -y direction, therefore, there is no parallel force and equals zero.

$\frac{d\stackrel{â‡\P Ä}{p}}{dt}=0$

As a result, the rate change is the direction change owing to the perpendicular rate of change.

The net force exerted on the object equals the rate change of the momentum and the magnitude of the perpendicular rate change at speeds much less than the speed of light is given by

$$\begin{gathered} F_{net}=\frac{d\stackrel{â‡\P Ä}{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

Also, the object is under two forces, the tension forceof the string and its weight, so the net force of both forces is

$F_{net}=F_T+mg$

In the outer space, the weight force is neglected, so equation will be in the form

$F_{net}=F_T$

Therefore,

$F_T=2\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

Now put the values for m, v, g and R to get the tension force in the cable

$$\begin{gathered} F_T=2\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ =\frac{\left(60\mathrm{kg}\right){\left(5\mathrm{m}/\mathrm{s}\right)}^2}{3.2\mathrm{m}} \\ =937.5\mathrm{N} \end{gathered}$$

Both spheres have the same magnitude of the velocity where the magnitude of speed for each sphere was calculated by

$$\begin{gathered} {\mathrm{v}}_1=\sqrt{\left(0^2+5^2+0^2\right)} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

And

localid="1656672905409" $$\begin{gathered} {\mathrm{v}}_2=\sqrt{\left(0^2+{\left(-5\right)}^2+0^2\right)} \\ =5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence,${\mathrm{v}}_1={\mathrm{v}}_2=\mathrm{v}=5\mathrm{m}/\mathrm{s}$.

Thus, the tension in the cable is $937.5\mathrm{N}$