Question

A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further \( D \times 10^{-3} \) m before coming to rest. The value of \( D \) is :

• A. 2
• B. 5
• C. \(32\times10^{-3}\)
• D. 4

Answer 1

The Correct Option is C

Step 1 — Symbols and given data
Let the initial speed be \(v_i\). After traveling \(s = 4\ \text{cm} = 0.04\ \text{m}\), the speed reduces to \(v_1 = \frac{2}{3}v_i\). This reduction implies a constant acceleration (deceleration), denoted by \(a\), where \(a<0\). The objective is to find the additional distance, \(D\), required for the speed to decrease from \(v_1\) to zero.

Step 2 — Use the kinematic equation to find acceleration
Applying the kinematic equation \(v_f^2 = v_i^2 + 2as\) with \(v_f = v_1\) and \(s = 4\ \text{cm}\): \[ v_1^2 = v_i^2 + 2 a s. \] Substituting \(v_1 = \frac{2}{3}v_i\): \[ \left(\frac{2}{3}v_i\right)^2 = v_i^2 + 2 a s \quad\Longrightarrow\quad \frac{4}{9}v_i^2 - v_i^2 = 2 a s. \] Simplifying the equation yields: \[ -\frac{5}{9}v_i^2 = 2 a s. \] Solving for acceleration: \[ a = -\frac{5\,v_i^2}{18\,s}. \] This negative value for \(a\) confirms deceleration.

Step 3 — Distance to stop from \(v_1\)
Using the same kinematic equation, with initial speed \(v_1\), final speed \(v_f = 0\), and distance \(D\): \[ 0 = v_1^2 + 2 a D. \] Solving for \(D\): \[ D = -\frac{v_1^2}{2a}. \] Substituting \(v_1^2 = \frac{4}{9}v_i^2\) and the expression for \(a\): \[ D = -\frac{\tfrac{4}{9}v_i^2}{2\left(-\dfrac{5\,v_i^2}{18\,s}\right)} = \frac{\tfrac{4}{9}v_i^2}{\dfrac{10\,v_i^2}{18\,s}} = \frac{4}{9}\cdot\frac{18\,s}{10} = \frac{4s}{5}. \] The initial speed \(v_i\) is not required for this calculation; \(D\) is dependent only on the initial segment distance \(s\).

Step 4 — Numeric evaluation
Given \(s = 0.04\ \text{m}\): \[ D = \frac{4 \times 0.04}{5} = 0.032\ \text{m}. \] Converting to other units:

• \(0.032\ \text{m} = 32\ \text{mm}\).
• \(0.032\ \text{m} = 3.2\ \text{cm}\).

The bullet penetrates an additional 0.032 m (3.2 cm) before stopping.

Step 5 — Total penetration (optional)
The total distance penetrated from the start until rest is the sum of the initial 4 cm and the additional distance \(D\): \[ \text{Total} = 4\ \text{cm} + 3.2\ \text{cm} = 7.2\ \text{cm} = 0.072\ \text{m}. \]

Quick summary / final answer

Additional distance before coming to rest: \( \boxed{0.032\ \text{m} = 3.2\ \text{cm} = 32\ \text{mm}}\) = \(32\times10^{-3}\).

Notes: (1) The solution assumes constant deceleration and is independent of the initial speed \(v_i\). (2) The kinematic formula \(v^2 = u^2 + 2as\) was used throughout.