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A bullet of mass $20 g$ moving with a speed of $120 m / s$ hits a thick muddy wall and penetrates it. It takes $0.03 s$ to stop in the wall. Then the force exerted by the bullet on the wall is ____ $N .$

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Detailed Solution

The force exerted by the bullet on the wall is

80 N

.
The velocity of the bullet as it hits the wall is (initial velocity)

u = 120 m / s

And final velocity

v = 0

Mass of the bullet

m = 20 g

Time

t = 0.03 s

By using

v = u + at

;
We get

0 = 120 + a \times 0.03

Acceleration,

a = \frac{- 120}{0.03}

\Rightarrow

a = - 4000 m / s^{2}

We know the relation

F = ma

Here

F, m,

and

a

denote the force, mass and acceleration, respectively.
So, put the values of

m

and

a

in the above equation, we get

F = 0.02 kg \times (- 4000) m / s^{2}

(since

20 g = \frac{20}{1000} kg = 0.02 kg

)

F = - 80 N

Here,

F = - 80 N,

the force exerted by the wall on the bullet is in a direction opposite to that of the velocity. But from Newton’s third law, the force exerted by the bullet on the wall is also 80 N, in the direction of the velocity.

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