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 the first 4 cm the speed becomes \[ v_1 = \frac{2}{3}\,v_i \] (since it loses one-third of its velocity). The distance covered during this reduction is \[ s = 4\ \text{cm} = 0.04\ \text{m}. \] Assume constant acceleration \(a\) (deceleration, so \(a<0\)). We need the additional distance \(D\) from the point where the speed is \(v_1\) until the speed becomes zero.

Step 2 — Use the kinematic equation to find acceleration

Use the basic relation for uniform acceleration: \[ v_1^2 = v_i^2 + 2 a s. \] Substitute \(v_1 = \tfrac{2}{3}v_i\): \[ \left(\frac{2}{3}v_i\right)^2 = v_i^2 + 2 a s \quad\Longrightarrow\quad \frac{4}{9}v_i^2 - v_i^2 = 2 a s. \] Simplify the left side: \[ -\frac{5}{9}v_i^2 = 2 a s \quad\Longrightarrow\quad a = -\frac{5}{9}\frac{v_i^2}{2s} = -\frac{5\,v_i^2}{18\,s}. \] (Note: \(a\) is negative, as expected for deceleration.)

Step 3 — Distance to stop from \(v_1\)

From speed \(v_1\) to rest (final speed 0), use \[ 0 = v_1^2 + 2 a D \quad\Longrightarrow\quad D = -\frac{v_1^2}{2a}. \] Substitute \(v_1^2 = \tfrac{4}{9}v_i^2\) and the expression for \(a\): \[ D = -\frac{\tfrac{4}{9}v_i^2}{2\left(-\dfrac{5\,v_i^2}{18\,s}\right)} = \frac{\tfrac{4}{9}v_i^2}{\dfrac{10\,v_i^2}{18\,s}} = \frac{4}{9}\cdot\frac{18\,s}{10} = \frac{4s}{5}. \] Remark: Notice the initial speed \(v_i\) cancels out — the further distance \(D\) depends only on the known segment \(s\).

Step 4 — Numeric evaluation

With \(s = 0.04\ \text{m}\), \[ D = \frac{4 \times 0.04}{5} = 0.032\ \text{m}. \] Convert to common units:

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

So the bullet will penetrate an additional 0.032 m (3.2 cm) before coming to rest.

 

Step 5 — Total penetration (optional)

If you want the total penetration from the start until rest, add the initial 4 cm: \[ \text{Total} = 4\ \text{cm} + 3.2\ \text{cm} = 7.2\ \text{cm} = 0.072\ \text{m}. \]

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 result is independent of the initial speed \(v_i\), provided the deceleration is constant. (2) All steps used the standard kinematic formula \(v^2 = u^2 + 2as\).

Answer 2

Given:

The bullet loses one-third of its velocity after traveling \(s = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m}\), with final velocity \(v = \frac{2u}{3}\).

Using the kinematic equation:

\(v^2 - u^2 = 2a \cdot s\)

Substituting the values:

\(\left(\frac{2u}{3}\right)^2 - u^2 = 2a \cdot (4 \times 10^{-2})\)

Simplifying:

\(\frac{4u^2}{9} - u^2 = 2a \cdot (4 \times 10^{-2})\)

\(\frac{4u^2}{9} - \frac{9u^2}{9} = 2a \cdot (4 \times 10^{-2})\)

\(-\frac{5u^2}{9} = 2a \cdot (4 \times 10^{-2})\)

\(a = \frac{-5u^2}{72 \times 10^{-2}}\)

Now, for the bullet to come to rest:

\(v^2 - u^2 = 2a \cdot D\)

Substitute \(v = 0\), \(a = \frac{-5u^2}{72 \times 10^{-2}}\), and solve for \(D\):

\(0 - u^2 = 2 \cdot \left(\frac{-5u^2}{72 \times 10^{-2}}\right) \cdot D\)

\(u^2 = \frac{10u^2}{72 \times 10^{-2}} \cdot D\)

\(D = \frac{72 \times 10^{-2}}{10}\)

\(D = 32 \times 10^{-3} \, \text{m} = 32 \, \text{mm}.\)

The Correct answer is: 32mm