Question

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. The recoil velocity of the gun is (Acceleration due to gravity $g = 10$ ms$^{-2}$):

• A. $0.6$ ms$^{-1}$
• B. $0.8$ ms$^{-1}$
• C. $0.2$ ms$^{-1}$
• D. $0.4$ ms$^{-1}$

Hint:

For projectile motion problems with recoil, apply the conservation of linear momentum, ensuring that both momentum components (horizontal and vertical) are considered separately.

Answer 1

The Correct Option is D

Step 1: Compute the Time of Flight Using free-fall motion, $$t = \sqrt{\frac{2h}{g}}.$$ Substituting values: $$t = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \text{ s}.$$
Step 2: Compute Horizontal Velocity of the Ball The ball moves horizontally 400 m in this time: $$v_b = \frac{\text{horizontal distance}}{\text{time}} = \frac{400}{10} = 40 \text{ ms}^{-1}.$$
Step 3: Apply Conservation of Momentum Using the law of conservation of momentum: $$m_b v_b = m_g v_g.$$ where: - $m_b = 1$ kg (mass of the ball), - $v_b = 40$ ms$^{-1}$ (velocity of the ball), - $m_g = 100$ kg (mass of the gun), - $v_g$ is the recoil velocity of the gun. Solving for $v_g$: $$100 v_g = 1 \times 40.$$ $$v_g = \frac{40}{100} = 0.4 \text{ ms}^{-1}.$$ % Final Answer Thus, the correct answer is option (4): $0.4$ ms$^{-1}$.