Question

ABullet of mass 10 gram travelling horizontally with a velocity of 150 metre per second strikes a stationary wooden block and comes to rest in 0.03 second the distance of peniteration of the bullet into the block

Answer 1

Explanation:

Answer:Given, Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg

Initial velocity of bullet, u = 150 m/s

Since bullet comes to rest, thus final velocity, v =0

Time, t = 0.03 s

Distance of penetration, i.e. Distance, covered (s)=?

Magnitude of force exerted by wooden block =?

we know that,

$v = u + at \\ = 0 = 150ms { -}^{1} + a \times 0.03s \\ = - 150mps = a \times 0.03s \\ = a = - \frac{150mps}{0.03s} = - 5000ms - {}^{2}$

we know that,

$s = ut + \frac{1}{2} at {}^{2} \\ = s = 150mps \times 0.03s + \frac{1}{2} ( - 5000ms - {}^{2} \\ = s = 4.5m - 2500ms - {}^{2} \times 0.0009s {}^{2} \\ = s = 4.5m - 2.25m \\ = s = 2.25m$

magnitude of force exerted by wooden block

we know that, Force = mass x

acceleration

or,

$f = 0.01kg \times - 5000ms {}^{ - 2} = - 50n$

Therefore, Penetration of bullet in wooden block = 2.25 m

Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

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