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A bullet of $10 g$ strikes a sand bag at a speed of $10^{3} m / s$ and gets embedded after travelling $5 cm$ . The resistive force exerted by the sand on the bullet is

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- 10^{8} N

- 10^{7} N

- 10^{4} N

- 10^{3} N

answer is C.

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Detailed Solution

The resistive force exerted by the sand on the bullet is

- 1 0^{4} N

.
Given,
Mass of the bullet,

m = 10 g

In SI unit, the mass of the bullet can be written as

m = 10 g = 0.001 kg

Initial velocity,

u

{= 10}^{3} m / s

Final velocity,

v = 0 m / s

Distance travelled by the bullet,

s = 5 cm

In SI unit the distance travelled can be witten as

s = 0.05 m

As we know the relation of

u, v

and

s

are

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

Now the putting the values we have

\Rightarrow 0 - {(1 0^{3})}^{2} = 2 a \times (0.05)

\Rightarrow a = - 1 0^{6} / 0.1 = - 1 0^{7}

\Rightarrow a = - 1 0^{7} m / s^{2}

Now, the required resistive force is given by the relation

F = ma

On substituting the values, we get the force as

\Rightarrow F = 0.001 \times - 1 0^{7}

\Rightarrow F = - 1 0^{4} N

Hence, the resistive force exerted by the sand on the bullet is

- 1 0^{4} N

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