91Ó°ÊÓ

The given data can be listed below as,

• The mass of rock 1 is,$m_1=15\text{ k²µ}$
• The velocity of rock 1 before colliding to rock 2 is,$u_1=〈10,\text{ }−30,\text{ }0〉$
• The mass of rock 2 is,$m_2=32\text{ k²µ}$
• The velocity of rock 2 before colliding to rock 2 is,$u_2=〈15,\text{ }12,\text{ }0〉$

The total momentum experienced by a system before the collision of two objects is the 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.

Write the expression of law of conservation of momentum when two objects stuck together.

$m_1{u}_1+m_2{u}_2=\left(m_1+m_2\right)v$

Here, v isthe final velocity of when both the rock stuck together.

Substitute all the values in the above expression.

$\begin{array}{c}15\left(10,\text{ }−30,\text{ }0\right)+32\left(15,\text{ }12,\text{ }0\right)=\left(15+32\right)v\\ 47v=\left(630,\text{ }−66,\text{ }0\right)\\ v=\frac{\left(630,\text{ }−66,\text{ }0\right)}{47}\\ =\left(13.40,\text{ }−1.4042,\text{ }0\right)\text{″¾/²õ}\end{array}$

Thus,the final velocity of when both the rock stuck together is.$\left(13.40,\text{ }−1.4042,\text{ }0\right)\text{″¾/²õ}$