A bullet of mass 10, g travelling horizontall | vedclass

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Question

(C) Given:Initial velocity of the bullet,$u = 150\, m/s$Final velocity of the bullet,$v = 0\, m/s$ (as it comes to rest)Time taken,$t = 0.03\, s$Mass

Vedclass · Accepted Answer

$2.25\, m, 50\, N$

Vedclass · Answer

(C) Given:Initial velocity of the bullet,$u = 150\, m/s$Final velocity of the bullet,$v = 0\, m/s$ (as it comes to rest)Time taken,$t = 0.03\, s$Mass of the bullet,$m = 10\, g = 0.01\, kg$Using the first equation of motion,$v = u + at$:$0 = 150 + a(0.03)$$a = -150 / 0.03 = -5000\, m/s^2$Using the second equation of motion to find the distance of penetration $(s)$:$s = ut + (1/2)at^2$$s = (150 	imes 0.03) + (1/2 	imes -5000 	imes (0.03)^2)$$s = 4.5 - (2500 	imes 0.0009)$$s = 4.5 - 2.25 = 2.25\, m$Using Newton's second law of motion to find the force $(F)$:$F = ma$$F = 0.01\, kg 	imes 5000\, m/s^2 = 50\, N$Thus,the distance of penetration is $2.25\, m$ and the force exerted is $50\, N$.

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