JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Work, Power & Energy

A bullet of mass 20 g has an initial speed of

1 ms^{- 1}

just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistances of

2.5 \times 1 0^{- 2} N

, the speed of the bullet after emerging from the other side of the wall is close to

(A) $0.7 ms^{- 1}$ ✓ Correct
(B) $0.3 ms^{- 1}$
(C) $0.1 ms^{- 1}$
(D) $0.4 ms^{- 1}$

Answer: (A)

Solution

As the bullet ploughs through the wall the resisting force does negative work, and the work-energy theorem says

\frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - F d

. The speed-drop term is

\frac{2 F d}{m} = \frac{2 \times ( 2.5 \times 1 0 ^{- 2} ) \times 0.20}{0.020} = 0.5 m^{2} s^{- 2}

, so

v = v_{0}^{2} - \frac{2 F d}{m} = 1^{2} - 0.5 = 0.5 \approx 0.7 m s^{- 1}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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