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