91Ó°ÊÓ

The given data can be listed below as,

• The mass of rock 1 is,$m_1=5\text{ k²µ}$
• The velocity of rock 1 before colliding to rock 2 is,$u_1=〈30,\text{ }45,\text{ }−20〉\text{ m/²õ}$
• The velocity of rock 1 after colliding to rock 2 is,$v_1=〈−10,\text{ }50,\text{ }−5〉\text{ m/²õ}$
• The mass of rock 2 is,$m_2=8\text{ k²µ}$
• The velocity of rock 2 before colliding to rock 2 is,$u_2=〈−9,\text{ }5,\text{ }4〉\text{ m/²õ}$

The total momentum experienced by a system before collision of two objects is same as that of the total momentum of the same system after the collision of two objects. It is expressed as follows,

${\mathbf{m}}_1{\mathbf{u}}_1\mathbf{+}{\mathbf{m}}_2{\mathbf{u}}_2\mathbf{=}{\mathbf{m}}_1{\mathbf{v}}_1\mathbf{+}{\mathbf{m}}_2{\mathbf{v}}_2$

Here, $m_1$is the mass of object 1,$u_1$ is the initial velocity of object 1,$v_1$ is the final velocity of object 1, $m_2$is the mass of object 2,$u_2$ is the initial velocity of object 2,$v_2$ is the final velocity of object 2.

Substitute all the values in the above expression.

$\begin{array}{c}5\left(30,\text{ }45,\text{ }-20\right)+8\left(-9,\text{ }5,\text{ }4\right)=5\left(-10,\text{ }50,\text{ }-5\right)+8v_2\\ 8v_2=5\left(30,\text{ }45,\text{ }-20\right)+8\left(-9,\text{ }5,\text{ }4\right)−5\left(-10,\text{ }50,\text{ }-5\right)\\ 8v_2=\left(128,\text{ }15,\text{ }−43\right)\\ v_2=\frac{\left(128,\text{ }15,\text{ }−43\right)}{8}\\ =\left(16,\text{ }1.875,\text{ }−5.375\right)\text{ m/²õ}\end{array}$

Thus, the final velocity of rock 2 is .$\left(16,\text{ }1.875,\text{ }−5.375\right)\text{ m/²õ}$