A tennis ball of mass $\mathbf{0}\mathbf{.06\hspace{0.33em}}\mathbf{kg}$travelling at a velocity of$\mathbf{(}\mathbf{9}\mathbf{,-2,13)\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$is about to collide with an identical tennis ball whose velocity is$\mathbf{(}\mathbf{4}\mathbf{,5,-10)\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$. a) What is the total momentum of the system of the two-tennis ball? b) What is the velocity of the center of mass of the two tennis balls?

The given data can be listed below as,

• The mass of the first tennis ball is $0.06\mathrm{kg}$.
• The velocity of the first tennis ball is $(9,-2,13)\mathrm{m}/\mathrm{s}$.
• The mass of the second tennis ball is the same as the first tennis ball.
• The velocity of the second tennis ball is $(4,5,-10)\mathrm{m}/\mathrm{s}$.

The law of the conservation of momentum states that the total momentum for a body before and after the collision becomes equal.

The law of the center of mass states that if a rigid object is pushed then it will continue to move as a point mass.

The equation of the momentum gives the total momentum of the system of the two tennis ball and the equation of velocity gives the velocity of the two tennis ball.

a) From the law of conservation of momentum, the equation of the total momentum of the system of the tennis ball can be expressed as:

$\mathrm{p}={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2$

Here, $p$is the total momentum of the system, $m_1\text{and}{m}_2$are the mass of the first and the second tennis ball respectively and $v_1\text{and}{v}_2$are the velocity of the first and the second tennis ball respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=(0.06\mathrm{kg})\times \left(\right(9,-2,13)\mathrm{m}/\mathrm{s})+(0.06\mathrm{kg})\times \left(\right(4,5,-10)\mathrm{m}/\mathrm{s}) \\ p=(0.78,0.18,0.18)\mathrm{kg}·\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the total momentum of the system of the two tennis ball is $(0.78,0.18,0.18)kg.m/s.$

b) From the law of center of mass, the equation of the velocity of the center of mass of the tennis ball is expressed as:

$v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

Substituting the values in the above equation, we get-

$$\begin{gathered} v=\frac{(0.06\mathrm{kg})\times \left(\right(9,-2,13)\mathrm{m}/\mathrm{s})+(0.06\mathrm{kg})\times \left(\right(4,5,-10)\mathrm{m}/\mathrm{s})}{(0.06\mathrm{kg})+(0.06\mathrm{kg})} \\ v=(6.5,1.5,1.5)m/s \end{gathered}$$

Thus, the velocity of the center of mass of the tennis ball is$(6.5,1.5,1.5)\hspace{0.33em}m/s$.