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A bullet of $10 g$ strikes a sandbag 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^{6} N

- 10^{5} N

- 10^{2} N

answer is C.

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Detailed Solution

A bullet of

10 g

strikes a sandbag 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

- 10^{4} N

.
According to the given information, we have,

mass (m) = 10 g = 0.01 kg

initial velocity (u) = 10^{3} m / s

final velocity (v) = 0

[As it stops finally]
Distance travelled by the bullet,

s = 5 cm

= 0.05 m

As we know the relation of

u

v

and

s

by the third equation of motion

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

Now the putting the values we have

\Rightarrow 2 as = 0 - {(10}^{3})^{2}

\Rightarrow

2 \times a \times 0.05 = - 1 0^{6}

\Rightarrow a = - 10^{6} / 0.1

\Rightarrow

a = - 10^{7}

\Rightarrow

a = - 10^{7} m / s^{2}

Hence the required resistive force

F = ma

\Rightarrow F = 0.01 \times (- 10^{7})

\Rightarrow F = - 1 0^{5} N

Here the negative sign of force represents a resistive force.

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