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A bullet of $20 g$ strikes sand bag at a speed of $200 m / s$ and gets embedded after travelling $2 cm$ , then the resistive force exerted by the sand on the bullet is

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2 \times 10^{3} N

2 \times 10^{4} N

2 \times 10^{6} N

2 \times 10^{5} N

answer is B.

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Detailed Solution

A bullet of

20 g

strikes sand bag at a speed of

200 m / s

and gets embedded after travelling

2 cm

, and then the resistive force exerted by the sand on the bullet is

2 \times 10^{4} N .

Given, mass of bullet,

m = 20 g = 20 \times 10^{- 3} kg

,
Distance travelled,

s = 2 cm = 0.02 m

Initial velocity,

u = 200 m / s

, Final velocity,

v = 0 m / s

From the third equation of motion,

v^{2} = u^{2} + 2 as

⇒

a = {(v}^{2} - u^{2}) / 2 s = [{(0)}^{2} - {(200)}^{2}] / (2 \times 0.02)

⇒

a = - 10^{6} m / s^{2}

Resistive force =

F = ma

\Rightarrow F = 20 \times 10^{- 3} \times {(- 10)}^{6}

\Rightarrow F = - 2 \times 10^{4}

N
(Negative sign shows resistance)

∴

The resistive force exerted by sand on the bullet is

2 \times 10^{4}

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