Two balls of mass $\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{kg}\mathbf{}\mathbf{and}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{kg}$are connected by a low-mass spring (Figure 3.63). This device is thrown through the air with low speed, so air resistance is negligible. The motion is complicated: the balls whirl around each other, and at the same time the system vibrates, with continually changing stretch of the spring. At a particular instant, the $\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{kg}\mathbf{}$ball has a velocity $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$and the$\mathbf{}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{kg}$ball has a velocity $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{.}\mathbf{}$a) At this instant, what is the total momentum of the device? b) What is the net gravitational (vector) force exerted by the earth on the device? c) At a timelater, what is the total momentum of the device?

The given data can be listed below as,

• The mass of the first ball is$0.3\hspace{0.33em}kg$.
• The mass of the second ball is$0.5\hspace{0.33em}kg$ .
• The first ball has a velocity of$(4,-3,2)\hspace{0.33em}m/s$ .
• The second ball has a velocity of$(2,1,4)\hspace{0.33em}m/s$ .

The momentum law states that the amount of the momentum of a body remains constant that is the product of the mass and the velocity.

The equation of the momentum gives the momentum of the body. Moreover, the equation of force gives the net force on the device.

a) From the principle of conservation of the linear momentum, the momentum of the device is expressed as:

$P=m_1{v}_1+m_2{v}_2$

Here, P is the momentum of the particle, $m_1,v_1,m_2\hspace{0.33em}and{v}_2$are the mass and the velocity of the first and the second particle respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=(0.3\hspace{0.33em}kg)\times \left(\right(4,-3,2)\hspace{0.33em}m/s)+(0.5\hspace{0.33em}kg)\times \left(\right(2,1,4)\hspace{0.33em}m/s) \\ p=(1.2\hat{ı}-0.9\hat{ȷ}+0.6\hat{k})+(\hat{ı}+0.5\hat{ȷ}+2\hat{k}) \\ p=(2.2\hat{ı}-0.4\hat{ȷ}+2.6\hat{k})\mathrm{kgm}/\mathrm{s} \end{gathered}$$

Thus, the total momentum of the device is $(2.2\hat{ı}-0.4\hat{ȷ}+2.6\hat{k})\mathrm{kgm}/\mathrm{s}$.

b) From Newton’s gravitational law, the net gravitational force exerted by the earth on the object is expressed as:

$F=m_{1}g+m_{2}g$

Here, F is the net gravitational force, g is the acceleration due to gravity that is $9.8\mathrm{m}/{\mathrm{s}}^2\text{and}{m}_1\text{and}{m}_2$are the masses of the first and the second particle respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} F=9.8\mathrm{m}/{\mathrm{s}}^2\times (0.3\mathrm{kg}+0.5\mathrm{kg}) \\ F=7.84N \end{gathered}$$

Thus, the net gravitational (vector) force exerted by the earth on the device is $7.84N.$

c) From the law of conservation of momentum, the equation of the total momentum of the device can be expressed as:

$p=Ft$

Here, p is the total momentum, F is the net gravitational (vector) force and t is the time required that is $0.1\hspace{0.33em}s.$

Substituting the values in the above equation, we get-

$$\begin{gathered} p=7.84\hspace{0.33em}N\times 0.1\hspace{0.33em}s \\ p=0.784\hspace{0.33em}N.s \end{gathered}$$

Thus, the total momentum of the device is $0.784\hspace{0.33em}N.s$