- This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 \text{ g} = 0.01 \text{ kg}$, $v_b = 300 \text{ m/s}$, $m_p = 1 \text{ kg}$. - Substituting the values: $(0.01 \text{ kg})(300 \text{ m/s}) + (1 \text{ kg}) v_p = 0$ - $3 \text{ kg} \cdot \text{m/s} + v_p \text{ kg} = 0$ - $v_p = -3 \text{ m/s}$