A bullet of mass $'x'$ moves with a velocity $'y'$, hits a wooden block of mass $'x'$at rest and gets embedded in it. After collision, the wooden block and bullet it moves, with the velocity:
$
  A.\dfrac{{xy}}{{x + z}} \\
  B.\dfrac{{x + y}}{x} \\
  C.\dfrac{{zy}}{{x + y}} \\
  D.\dfrac{{x + y}}{z}y \\
 $

Hint: Here we will proceed by using the approach of law of conservation of momentum for inelastic collision. Then by using the approach given in the question, we will get our answer.
Law of conservation of momentum – It is defined as the total linear momentum of a system of particles in the absence of the external forces is constant in magnitude and direction independent of any reactions among the other components of the system.

Complete step-by-step answer:

It is given that, Mass of the bullet before hitting a wooden block ${m_1} = x$. And, the velocity of the bullet before hitting the wooden block $ = y$.
Also, Mass of the wooden block on which the bullet gets strike $'{m_2}' = z$
After collision, the velocity of the bullet becomes zero, as it comes to the rest.

That is ${\mu _2} = 0$

It is also given that after collision, then the bullet gets embedded in it and the wooden block and bullet starts moving.

And, we have to find that velocity $'v'$in which the wooden block and the embedded bullet in it moves.

Since, the momentum in this case is inelastic.

So, we will apply the law of conservation of momentum for inelastic collision,
That is ${m_1}{\mu _1} + {m_2}{\mu _2} = v\left( {{m_1} + {m_2}} \right)$ ……. (1)
Here we have
$
  {m_1} = x,{\mu _1} = y \\
  {m_2} = z,{\mu _2} = 0 \\
 $
V=?

By putting all the values in equation (1) we get

$
  x \times y + z \times d = v\left( {x + z} \right) \\
   \Rightarrow xy = v\left( {x + z} \right) \\
   \Rightarrow \dfrac{{xy}}{{x + z}} = v \\
   \Rightarrow v = \dfrac{{xy}}{{x + z}} \\
 $

Therefore, A is the correct option.

Note: Whenever we come up with this type of question, it should be noted that in case of inelastic collision there is some loss of energy and the final velocity of the combined objects depends on the masses and the velocities of the two objects that collided.