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\(^{-1}\).

Answer 2

Step 1: Calculate the Duration of the Fall
The ball is dropped from a height \( h = 500 \, \text{m} \), and we want to find the time it takes to reach the ground.
Using the kinematic equation for free fall without initial velocity: \[ t = \sqrt{\frac{2h}{g}}. \] Substitute the values \( h = 500 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \): \[ t = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \text{ seconds}. \]
Step 2: Determine Horizontal Velocity of the Ball
While the ball is falling, it also moves horizontally a distance of 400 m.
Since horizontal motion and vertical motion are independent, the horizontal velocity \( v_b \) remains constant during the fall and is given by: \[ v_b = \frac{\text{horizontal distance}}{\text{time}} = \frac{400}{10} = 40 \text{ m/s}. \]
Step 3: Apply the Principle of Conservation of Momentum
When the ball is fired from the gun, the gun recoils backward due to conservation of linear momentum.
Before firing, total momentum is zero, so after firing: \[ \text{momentum of ball} + \text{momentum of gun} = 0. \] Mathematically, \[ m_b v_b = m_g v_g, \] where \( m_b = 1 \, \text{kg} \) (ball mass), \( v_b = 40 \, \text{m/s} \) (ball velocity), \( m_g = 100 \, \text{kg} \) (gun mass), and \( v_g \) is the recoil velocity of the gun.
Rearranging for \( v_g \): \[ v_g = \frac{m_b v_b}{m_g} = \frac{1 \times 40}{100} = 0.4 \text{ m/s}. \]
Final Conclusion:
Therefore, the recoil speed of the gun is \[ \boxed{0.4 \text{ ms}^{-1}}. \] This shows how momentum conservation helps determine recoil velocity in firing scenarios.