A bullet of mass 20 , g travelling horizontal | vedclass

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(A) Mass of bullet $m = 20 \, g = 0.02 \, kg$.Initial velocity of bullet $u_1 = 500 \, m/s$.Mass of block $M = 10.0 \, kg$.Initial velocity of block $

Vedclass · Accepted Answer

$0.16$

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(A) Mass of bullet $m = 20 \, g = 0.02 \, kg$.Initial velocity of bullet $u_1 = 500 \, m/s$.Mass of block $M = 10.0 \, kg$.Initial velocity of block $u_2 = 0 \, m/s$.Final velocity of bullet $v_1 = 100 \, m/s$.Let the final velocity of the block be $v_2$.Applying the law of conservation of linear momentum:$m u_1 + M u_2 = m v_1 + M v_2$$0.02 	imes 500 + 10 	imes 0 = 0.02 	imes 100 + 10 	imes v_2$$10 = 2 + 10 v_2$$10 v_2 = 8 \implies v_2 = 0.8 \, m/s$.Now,the block slides $d = 20 \, cm = 0.2 \, m$ before coming to rest. Using the work-energy theorem:Change in kinetic energy = Work done by friction$0 - \frac{1}{2} M v_2^2 = -f_k 	imes d$$-\frac{1}{2} 	imes 10 	imes (0.8)^2 = -(\mu M g) 	imes 0.2$$5 	imes 0.64 = \mu 	imes 10 	imes 10 	imes 0.2$$3.2 = \mu 	imes 20$$\mu = \frac{3.2}{20} = 0.16$.Thus,the coefficient of friction is $0.16$.

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