Question

In outer space a small rock with mass $\mathbf{5}\mathbf{}\mathbf{kg}$ traveling with velocity$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1800}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$ strikes a stationary large rock head-on and bounces straight back with velocity role="math" localid="1658131357647" $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{1500}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. After the collision, what is the vector momentum of the large rock?

Short answer

The vector momentum of the large rock is $(0,1600,0)\raisebox{1ex}{$\mathrm{kg}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the small rock is $5\mathrm{kg}$.

• The velocity of the small rock is $(0,1800,0)\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

• The velocity of the small rock after colliding with the large rock is $(0,-1500,0)\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

Significance of the law of conservation of momentum for the large rock

This law illustrates that the total momentum of a body before and also after collision remains equal if no external forces are involved.

The equation of the law of conservation of momentum gives the vector momentum of the large rock.

Determination of the vector momentum of the large rock

From the law of conservation of the momentum, the equation of the momentum of a large rock can be expressed as-

${\mathrm{p}}_1={\mathrm{p}}_2+{\mathrm{p}}_3$

Here, ${\mathrm{p}}_1$is the initial momentum of the small rock, ${\mathrm{p}}_2$and ${\mathrm{p}}_3$are the final momentum of the small and the large rock respectively .

Substituting the values in the above equation, we get-

${\mathrm{p}}_3={\mathrm{p}}_1-{\mathrm{p}}_2$

${\mathrm{p}}_3=\left(\right(0,1800,0)\mathrm{m}/\mathrm{s}\times (5\mathrm{kg}\left)\right)-\left(\right(0,-1500,0)\mathrm{m}/\mathrm{s}\times (5\mathrm{kg}\left)\right)$

${\mathrm{p}}_3=(0.16500,0)\mathrm{kg}\times \mathrm{m}/\mathrm{s}$

Thus, the vector momentum of the large rock is $(0.16500,0)\mathrm{kg}\times \mathrm{m}/\mathrm{s}$.