Question:

A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further D×103 D \times 10^{-3} m before coming to rest. The value of D D is :

Updated On: Jan 27, 2025
  • 2
  • 5
  • 32×10332\times10^{-3}
  • 4
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The Correct Option is C

Solution and Explanation

Given:

The bullet loses one-third of its velocity after traveling s=4cm=4×102ms = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m}, with final velocity v=2u3v = \frac{2u}{3}.

Using the kinematic equation:

v2u2=2asv^2 - u^2 = 2a \cdot s

Substituting the values:

(2u3)2u2=2a(4×102)\left(\frac{2u}{3}\right)^2 - u^2 = 2a \cdot (4 \times 10^{-2})

Simplifying:

4u29u2=2a(4×102)\frac{4u^2}{9} - u^2 = 2a \cdot (4 \times 10^{-2})

4u299u29=2a(4×102)\frac{4u^2}{9} - \frac{9u^2}{9} = 2a \cdot (4 \times 10^{-2})

5u29=2a(4×102)-\frac{5u^2}{9} = 2a \cdot (4 \times 10^{-2})

a=5u272×102a = \frac{-5u^2}{72 \times 10^{-2}}

Now, for the bullet to come to rest:

v2u2=2aDv^2 - u^2 = 2a \cdot D

Substitute v=0v = 0, a=5u272×102a = \frac{-5u^2}{72 \times 10^{-2}}, and solve for DD:

0u2=2(5u272×102)D0 - u^2 = 2 \cdot \left(\frac{-5u^2}{72 \times 10^{-2}}\right) \cdot D

u2=10u272×102Du^2 = \frac{10u^2}{72 \times 10^{-2}} \cdot D

D=72×10210D = \frac{72 \times 10^{-2}}{10}

D=32×103m=32mm.D = 32 \times 10^{-3} \, \text{m} = 32 \, \text{mm}.

The Correct answer is: 32mm

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