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Physics Question on Newtons Laws of Motion

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

A

2

B

5

C

32×10−332\times10^{-3}

D

4

Answer

32×10−332\times10^{-3}

Explanation

Solution

Given:

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

Using the kinematic equation:

v2−u2=2a⋅sv^2 - u^2 = 2a \cdot s

Substituting the values:

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

Simplifying:

4u29−u2=2a⋅(4×10−2)\frac{4u^2}{9} - u^2 = 2a \cdot (4 \times 10^{-2})

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

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

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

Now, for the bullet to come to rest:

v2−u2=2a⋅Dv^2 - u^2 = 2a \cdot D

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

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

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

D=72×10−210D = \frac{72 \times 10^{-2}}{10}

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

The Correct answer is: 32mm