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

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