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**Hint:**The kinetic energy (KE) of an object is the energy it possesses because of its motion. It is defined as the work necessary to accelerate from rest to the specified velocity of a body of a given mass. Having obtained this energy through its acceleration, until its speed varies, the body retains this kinetic energy.

**Complete step by step solution:**

Kinetic energy-energy type which is caused by an entity or a particle's movement. The mass accelerates and thereby gains kinetic energy if work is performed on a mass with the help of a net force. Kinetic energy is one of the properties of a moving object or particle and depends on its movement as well as mass. The movement type can be translation (or movement along a route between places), rotation around an axis, vibration or some mixture of motions.

A body's translational kinetic energy is equal to half its mass $m$, and its velocity, $v$ or $\dfrac {1} {2} m {v^2}.$

This formula is only valid for low to relatively high speeds; it gives too small values for very high speed particles. As an object's speed exceeds light, it changes its mass and it must be subjected to the rules of relative efficiency. Relativistic kinetic energy is analogous to an increase in particle mass in the remainder compounded by the square of light velocity.

The bullet emerges with only $10\% $ of its kinetic energy.

Therefore,

$\dfrac {1} {2} m {v^2} = 0.1(K.E) $

Where, $m$ is mass, $v$ is velocity and $K.E$ is kinetic energy.

$ = 0.1 \times \dfrac{1}{2}m{u^2}$

It is given that $m = 50g$or $m = 50 \times {10^ {- 3}} kg$

$v = 200m{s^ {- 1}} $

Therefore, from the above equations,

$\dfrac {1} {2} \times 50 \times {10^ {- 3}} \times {u^2} = 0.1 \times [\dfrac{1}{2} \times 50 \times {10^ {- 3}} \times {20^2}] $

${u^2} = 0.1 \times {200^2} $

$u = 20\sqrt {10} m{s^ {- 1}} $

**The emergence speed of bullet is $20\sqrt {10} m{s^ {- 1}}.$**

**Note:**The cumulative kinetic energy of a body or mechanism equals the sum of the kinetic energies of each movement form.

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