In outer space, far from other objects, two rocks collide and stick together. Before the collision their momenta were $\mathbf{(}\mathbf{-}\mathbf{10}\mathbf{,}\mathbf{20}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}$and$\mathbf{(}\mathbf{8}\mathbf{,}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{12}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}$. What was their total momentum before the collision? What must be the momentum of the combined object after the collision?

The given data can be listed below as-

• The momenta of the first object before collision is $(-10,20,-5)\hspace{0.33em}kg.m/s.$.
• The momenta of the second object after collision is$(8,-6,12)\hspace{0.33em}kg.m/s.$ .

This law illustrates that the momentum before collision is equal to the total momentum after a collision for a particular object without the involvement of an external force.

The equation of the momentum gives the total momentum before and after the collision for the objects.

From the law of conservation of momentum, the equation of the total momentum before the collision is expressed as-

$p=p_1+p_2$

Here, p is the total momentum before the collision, $p_{1}and{p}_2$are the momenta of the first and the second object respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=(-10,20,-5)\hspace{0.33em}kg.m/s+(8,-6,12)\hspace{0.33em}kg.m/s \\ p=(-2,14,7)kg.m/s \end{gathered}$$

According to the law of conservation of momentum, the total momentum before and after collision is same. Hence, the total momentum of the combined objects after collision is same as the total momentum before the collision.

collision is same as the total momentum before the collision.

Thus, the total momentum before the collision is localid="1658062049571" $(-2,14,7)kg.m/s$. The total momentum after the collision of the combined objects is also $(-2,14,7)kg.m/s$.