Answer

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**Hint:-**Here, we have to apply the law of conservation of momentum. In the subject of Physics, there are several quantities that are conserved. So, what is meant by a conserved quantity?

Conserved quantity is a function of the dependent variables in a dynamically changing system whose value remains equal along the entire phase and the path that the system undergoes.

Examples of the conserved quantities are mass, charge, energy, and momentum.

**Complete step-by-step solution:-**

The momentum is defined as the product of mass and the velocity of the body.

Momentum, $p = mv$

where m = mass and v = velocity.

However, this is only the mathematical definition. The momentum of a body represents the effect or impact that it possesses by the virtue of its mass and the velocity at which it is moving. We can understand that momentum represents the consequence of the body moving.

Since, we have seen that the quantity momentum is conserved here, let us understand the law of conservation of momentum.

It states that,

In the collision of two elastic bodies, the total momentum of the system before the collision is equal to the total momentum after the collision after the collision has happened.

Consider 2 bodies – object 1 and object 2 of masses ${m_1}$and ${m_2}$ respectively, moving towards each other at initial velocity of ${v_1}\& {v_2}$ respectively and collide. After collision, their new masses are ${M_1}\& {M_2}$ and their new velocities are ${V_1}\& {V_2}$.

Initial momentum, $p = {m_1}{v_1} + {m_2}{v_2}$

Final momentum after collision, $P = {M_1}{V_1} + {M_2}{V_2}$

As per the conservation of momentum,

$p = P$

$ \to {m_1}{v_1} + {m_2}{v_2} = {M_1}{V_1} + {M_2}{V_2}$

Here, let us consider object 1 as the bullet and object 2 as the block.

Since the block is stationery, we can say, ${v_2} = 0$

Thus, ${m_1}{v_1} = {M_1}{V_1} + {M_2}{V_2}$

Since the mass of the bullet does not change after the collision as there is no destruction or change of shape in the bullet,

Mass of bullet, ${m_1} = {M_1} = m = 20g = 0 \cdot 02kg$

Initial velocity of the bullet, ${v_1} = 600m{s^{ - 1}}$

Mass of the block, ${M_2} = M = 4kg$

Thus, the equation will be –

$m{v_1} = m{V_1} + M{V_2}$

$ \Rightarrow 0 \cdot 02 \times 600 = 0 \cdot 02{V_1} + 4{V_2}$

$ \Rightarrow 12 = 0 \cdot 02{V_1} + 4{V_2}$

To calculate the final speed of bullet, ${V_1}$, we have to calculate the final speed of the block, ${V_2}$

To find ${V_2}$:

The bullet is travelling at a high speed and hits the block. This causes the bullet to transfer its kinetic energy to the block which aids it in lifting the block up to a height h, thereby, converting it to potential energy in the block due to the height h. This is because of the law of conservation of energy, where –

Kinetic energy = Potential energy

$\dfrac{1}{2}MV_2^2 = Mgh$

Taking acceleration due to gravity, $g = 10m{s^{ - 2}}$ and given, height, $h = 0 \cdot 2m$, substituting we get –

$\dfrac{1}{2} \times 4 \times V_2^2 = 4 \times 10 \times 0 \cdot 2$

$ \Rightarrow V_2^2 = \dfrac{{8 \times 2}}{4} = 4$

$ \Rightarrow {V_2} = 2m{s^{ - 1}}$

Substituting in the equation of momentum, we get the final velocity of the bullet, ${V_1}$:

$12 = 0 \cdot 02{V_1} + 4{V_2}$

$ \Rightarrow 12 = 0 \cdot 02 \times {V_1} + 4 \times 2$

$ \Rightarrow 0 \cdot 02 \times {V_1} = 12 - 8 = 4$

$ \Rightarrow {V_1} = \dfrac{4}{{0 \cdot 02}} = \dfrac{{400}}{2} = 200m{s^{ - 1}}$

Hence, the final velocity of the bullet is 200 m/s.

Hence, the correct option is Option A.

**Note:-**In this problem, we have assumed the case of elastic collision. Elastic collision is the type of collision where the total energy of the system before the collision remains the same after the collision. However, it is not practical in real life to have elastic collisions. There is always loss of energy in the system due to collisions which come out in the form of sound, heat, light and so on.

This is why, one of the most major consequences of a collision is the loud sound that it produces.

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