Centre of Mass and Collision Part 1

Given, The mass of the first ball, $m_{1}=10\,\text{kg}$ The mass of the second ball, $m_{2}=20\,\text{kg}$ The initial velocity of the first ball, $u_{1}=10\sqrt{3}\,\text{m/s}$ The initial velocity of the second ball, $u_{2}=0\,\text{m/s}$ The final velocity of the first ball, $v_{1}=0\,\text{m/s}$ The final velocity of the first piece of the second ball, $v_{2}=10\,\text{m/s}$ Let's consider the final velocity of the second piece of the second ball $=v_{2}$ As shown in the figure, The net external force on the system is to be zero. Hence, we can use the law of conservation of linear momentum in both directions. $\ln x$-direction the linear momentum conserved, In $x$-direction the linear momentum conserved, $\Rightarrow\;\;v_{3}=20\,\text{m/s}$ Hence, the velocity of the ball moving at $30^{\circ}$ with respect to the $X$-axis is $20\,\text{m/s}$. So, the value of $x$ to the nearest integer is 20 .