{A bullet of mass }10{ g is horizontally fired with velocity }300{ m s}^{-1}{ from a pistol of mass }1{ kg. What is the recoil velocity of the pistol?}

Correct Answer is : 3 m s−13{ m s}^{-1}3 m s−1

Correct Answer is : 3 m s−13{ m s}^{-1}3 m s−1

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UPSC NDA 2 2022 General Ability Solved Paper

Question : 149 of 150

Marks: +1, -0

\text{A bullet of mass }10\text{ g is horizontally fired with velocity }300\text{ m s}^{-1}\text{ from a pistol of mass }1\text{ kg. What is the recoil velocity of the pistol?}

Solution:

Concept:

The principle of conservation of linear momentum states that the total momentum of an isolated system remains constant.

Explanation:

Given: mass of bullet

m_1 = 10\text{ g} = 0.01\text{ kg}

Velocity of bullet

v_1 = 300\text{ m s}^{-1}

Mass of pistol

m_2 = 1\text{ kg}

Before firing, both pistol and bullet are at rest, so initial total momentum is zero.

According to conservation of momentum, after firing:

m_1 v_1 + m_2 v_2 = 0

Substitute values:

0.01 \times 300 + 1 \times v_2 = 0

3 + v_2 = 0

→

v_2 = -3\text{ m s}^{-1}

The negative sign indicates that the recoil velocity of the pistol is opposite to the direction of the bullet's velocity.

Answer:

Option C:

-3\text{ m s}^{-1}

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