Answer

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**Hint:-**In this problem, the mass of the bullet is given and the velocity of the bullet is given and the mass of the sandbag is also given, then the velocity gained by the bag can be determined by the conserving the momentum of the system.

**Useful formula:**

Momentum of an object is equal to the product of mass and velocity,

$p = m \times v$

Where, $p$ is the momentum of the object, $m$ is the mass of the object and $v$ is the velocity of the object

**Complete step by step solution:**

Given that,

Mass of bullet, ${m_b} = 25\,g$,

Velocity of the bullet, ${v_b} = 400\,m{s^{ - 1}}$,

The mass of the sandbag, ${m_s} = 4.975\,kg$.

The momentum of the bullet is equal to the momentum of the sandbag, then

${m_b} \times {v_b} = \left( {{m_b} + {m_s}} \right) \times {v_s}\,...............\left( 1 \right)$

In RHS the mass of the bullet is added with the mass of the sand because after firing the bullet is put into the sand bag, so the mass of the bullet is added with the mass of the sandbag.

Now substituting the mass of the bullet, velocity of the bullet and the mass of the sandbag in the equation (1), then

$0.025 \times 400 = \left( {0.025 + 4.975} \right) \times {v_s}$

On multiplying the terms in LHS and adding the terms in RHS, then

$10 = 5 \times {v_s}$

By keeping the term ${v_s}$ in one side and the other terms in others side, then

${v_s} = \dfrac{{10}}{5}$

On dividing the above equation, then

${v_s} = 2\,m{s^{ - 1}}$

Thus, the above equation shows the velocity of the sandbag.

**Hence, the option (D) is correct.**

**Note:**The mass of the bullet is added with the mass of the sand bag because after firing the bullet, the bullet from the gun hits the sandbag and gets inside the sandbag, so the mass of the bullet is also considered for the momentum of the sandbag.

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