A spaceship moving with constant speed ${v_0}$ in gravity free space along the $ + Y$ axis suddenly shoots out one third of its part with speed $2{v_0}$ along the $ + X$ axis. The speed of the remaining part is \[\dfrac{{\sqrt {10 + x} }}{2}{v_0}\] . Find $x$ .

Hint: Consider the whole spaceship to be a system. No external force is applied on this system as the spaceship shoots its part suddenly. Linear momentum of a system is conserved when no external force is applied on the system. Linear momentum of a body of mass $m$ moving with velocity $\vec v$ is the product of mass and velocity and is given by $\vec p = m\vec v$ .

Complete step by step answer:
Let us consider the whole spaceship to be a system and no external force is applied on this system as the spaceship shoots its part suddenly.
So, we can apply conservation of linear momentum on this system which states that linear momentum of a system is conserved when no external force is applied on the system and we know that the linear momentum of a body of mass $m$ moving with velocity $\vec v$ is the product of mass and velocity and is given by $\vec p = m\vec v$ .
Let the mass of the spaceship be $M$ .
Initially, the spaceship is moving with constant speed ${v_0}$ in gravity free space along the $ + Y$ axis .
So, the initial momentum of the system is
${\vec p_i} = M{v_0}\hat j$
Now, the spaceship suddenly shoots out one third of its part $\left( {\dfrac{M}{3}} \right)$ with speed $2{v_0}$ along the $ + X$ axis.
Let the final velocity of the remaining part be ${v_x}\hat i + {v_y}\hat j$ .
Then the final momentum of the system will be
${\vec p_f} = \dfrac{M}{3} \times 2{v_0}\hat i + \dfrac{{2M}}{3}\left( {{v_x}\hat i + {v_y}\hat j} \right)$
Now, by applying conservation of linear momentum on the system i.e. ${\vec p_i} = {\vec p_f}$ we have
$M{v_0}\hat j = \dfrac{{2M}}{3}{v_0}\hat i + \dfrac{{2M}}{3}{v_x}\hat i + \dfrac{{2M}}{3}{v_y}\hat j$
Now by equating the x-components from both sides we have
$\dfrac{{2M}}{3}{v_0} + \dfrac{{2M}}{3}{v_x} = 0$
On simplifying we have
${v_x} = - {v_0}$
Now by equating the y-components from both sides we have
$M{v_0} = \dfrac{{2M}}{3}{v_y}$
On simplifying we have
${v_y} = \dfrac{{3{v_0}}}{2}$
So, the final velocity of the remaining part be ${v_x}\hat i + {v_y}\hat j = - {v_0}\hat i + \dfrac{{3{v_0}}}{2}\hat j$
Therefore, the speed of the remaining part is given by $\sqrt {{v_x}^2 + {v_y}^2} = \sqrt {{v_0}^2 + {{\left( {\dfrac{{3{v_0}}}{2}} \right)}^2}} = \dfrac{{\sqrt {13} }}{2}{v_0}$
Now, comparing this speed with the given form \[\dfrac{{\sqrt {10 + x} }}{2}{v_0}\] , we get the value of $x$ as
$x = 3$ which is the final answer.

Note: When objects collide with each other or eject some part of it, they exert force on each other. But these are internal forces which get balanced by other internal forces by adjustment in their magnitudes and direction. We know that the force is directly proportional to the rate of change of momentum. Thus the change in momentum caused by these internal forces get cancelled out by that of other internal forces. Hence, the linear momentum is conserved for the system. But if there is any external force acting on the system, there will be no force to balance the external force. Thus the linear momentum will not be conserved for that system.