91Ó°ÊÓ

The given data can be listed below as-

• The mass of the first rock before the collision is $9\mathrm{kg}$.

• The velocity of the first tock before the collision is $(4100,−2600,2800)\mathrm{m}/\mathrm{s}$.

• The mass of the other rock before the collision is $6\mathrm{kg}$.

• The velocity of the second rock before the collision is $(-450,1800,23500)\mathrm{m}/\mathrm{s}$.

• The chunk of the first rock that broke and sticked with the second rock is $\text{2kg}$.

• The $7\mathrm{kg}$rock has a velocity of $(1300,200,1800)\mathrm{m}/\mathrm{s}$.

This law elucidates that the total momentum of a body before and also after collision remains constant if no external forces get involved.

The equation of the momentum gives the velocity of the other rock.

From the law of the conservation of momentum, the equation of the velocity of the other rock can be expressed as-

${\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2={\mathrm{M}}_1{\mathrm{V}}_1+{\mathrm{M}}_2{\mathrm{V}}_2$

Here, ${\mathrm{m}}_1$and ${\mathrm{m}}_2$are the mass of the first and the second rock before the collision and ${\mathrm{v}}_1$and ${\mathrm{v}}_2$are the velocities of the first and the second rock before the collision, ${\mathrm{M}}_1$and ${\mathrm{M}}_2$are the masses of the first and the second rock after the collision and ${\text{V}}_1$and ${\mathrm{V}}_2$are the velocities of the first and the second rock after the collision.

Substituting the values in the above equation, we get-

role="math" localid="1658130167025" $\begin{array}{r}\left(9\mathrm{kg}\right)×\left(\right(4100,−2600,2800)\mathrm{m}/\mathrm{s})+\left(6\mathrm{kg}\right)×\left(\right(−450,1800,3500)\mathrm{m}/\mathrm{s})=\\ \left(7\mathrm{kg}\right)×\left(\right(1300,200,1800)\mathrm{m}/\mathrm{s})\\ +\left(8\mathrm{kg}\right)×\left({\mathrm{V}}_2\mathrm{m}/\mathrm{s}\right)\end{array}$

$\begin{array}{r}\left(\right(36900,−23400,25200)\mathrm{kg}×m/\mathrm{s})+\left(\right(−2700,10800,21000)\mathrm{kg}×\mathrm{m}/\mathrm{s})=\\ \left(\right(9100,1400,12600)\mathrm{kg}×\mathrm{m}/\mathrm{s})\\ +8V_2\mathrm{kg}×\mathrm{m}/\mathrm{s}\end{array}$

$\begin{array}{c}\left(\right(34200,−12600,46200)\mathrm{kg}×m/\mathrm{s})−\left(\right(9100,1400,12600)\mathrm{kg}×m/\mathrm{s})=\\ 8{\mathrm{V}}_2\mathrm{kg}×m/\mathrm{s}\\ {\mathrm{V}}_2=(3137.5,−1750,4200)\mathrm{m}/\mathrm{s}\end{array}$

Thus, the velocity of the other rock, whose mass is $8\mathrm{kg}$is $(3137.5,−1750,4200)\mathrm{m}/\mathrm{s}$.