JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Work, Power & Energy

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The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at

(4 + x)

cm inside the block. The value of

x

is:

(A)2.0
(B)1.0
(C)0.5✓ Correct
(D)1.5

Answer: (C)

Solution

Using

v^{2} = u^{2} - 2 a s

, after 4 cm:

(\frac{V}{3})^{2} = V^{2} - 2 a (4)

, giving

a = \frac{V ^{2}}{9}

. When the bullet stops:

0 = V^{2} - 2 a (4 + x)

. Substituting

a

V^{2} = 2 \times \frac{V ^{2}}{9} \times (4 + x)

, so

9 = 2 (4 + x)

, giving

4 + x = 4.5

, hence

x = 0.5

cm.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option

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