Question

A bullet when fired into a target loses half of its velocity after penetrating 20 cm. Further distance of penetration before it comes to rest is

• A. 6.66 cm
• B. 3.33 cm
• C. 12.5 cm
• D. 10 cm
• E. 5 cm

Hint:

A useful shortcut for this specific problem: If a bullet loses \(1/n\) of its velocity after distance \(s\), it travels a further distance of \(s / (n^2 - 1)\) before stopping. Here, \(n=2\), so distance = \(20 / (2^2 - 1) = 20/3\).

Answer 1

The Correct Option is A

Concept: This problem is solved using the third equation of motion, assuming the target exerts a constant retarding force (constant deceleration).
• Equation of Motion: \(v^2 = u^2 + 2as\).

Step 1: Find the deceleration in the first stage.
Initial velocity = \(u\). After distance \(s_1 = 20 \text{ cm}\), velocity \(v_1 = u/2\). \[ (u/2)^2 = u^2 + 2a(20) \] \[ u^2/4 - u^2 = 40a \] \[ -3u^2/4 = 40a \implies a = -3u^2/160 \]

Step 2: Calculate the further distance of penetration.
For the second stage, initial velocity is \(u/2\) and final velocity is \(0\). Let further distance be \(s_2\). \[ 0 = (u/2)^2 + 2 \left( \frac{-3u^2}{160} \right) s_2 \] \[ 0 = u^2/4 - \frac{3u^2}{80} s_2 \] \[ \frac{u^2}{4} = \frac{3u^2}{80} s_2 \] \[ s_2 = \frac{80}{4 \times 3} = \frac{20}{3} \approx 6.66 \text{ cm} \]