A bullet of mass 0.04; kg moving with a speed | vedclass

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(A) Given: Mass of the bullet $m = 0.04\; kg$,initial velocity $u = 90\; m/s$,final velocity $v = 0\; m/s$,and distance $s = 60\; cm = 0.6\; m$.Using

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(A) Given: Mass of the bullet $m = 0.04\; kg$,initial velocity $u = 90\; m/s$,final velocity $v = 0\; m/s$,and distance $s = 60\; cm = 0.6\; m$.Using the third equation of motion,$v^2 = u^2 + 2as$,we find the acceleration $a$:$0^2 = (90)^2 + 2 	imes a 	imes 0.6$$0 = 8100 + 1.2a$$a = -\frac{8100}{1.2} = -6750\; m/s^2$.The negative sign indicates retardation.According to Newton's second law of motion,the average resistive force $F$ is given by $F = m 	imes |a|$.$F = 0.04\; kg 	imes 6750\; m/s^2 = 270\; N$.Thus,the average resistive force exerted by the block on the bullet is $270\; N$.

$A$ bullet of mass $0.04\; kg$ moving with a speed of $90\; m/s$ enters a heavy wooden block and is stopped after a distance of $60\; cm$. What is the average resistive force exerted by the block on the bullet (in $; N$)?

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