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GOLD Gold benefits c Secure https://session.masteringphysics.com/myct/itemView?assignmentProblemlD=791 27480 mework 7 Problem 9.51 Resources « previous | 8 o roblem 9.51 Part A What is the initial speed of the bullet? A bullet of mass 1.1x10-3 kg embeds itself in a wooden block with mass 0.985 kg, which then compresses a spring (k = 170 N/m ) by a distance 3.0x10-2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.56 Express your answer using two significant figures. Vin = 568.35 m/s Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part B What fraction of the bullet's initial kinetic e temperature, etc.) in the collision between the b et and the block? Express your answer using two significant figures. nergy is dissipated (in damage to the wooden block, rising AK/K-.070 888 Fa 3 5 option command command option

Explanation / Answer

from momentum conservation

momentum before collision = momentum after collision

m*vi = (M+m)*vf

after collsiion

initial kinetic energy Ki = (1/2)*(M+m)*vf^2

work done by friction Wf = -uk*(M+m)*g*x

energy stored Uf = (1/2)*k*x^2

W = dKE + dU

-uk*(m+M)*g*x = (1/2)*(m+M)*(0^2 - vf^2) + (1/2)*k*x^2

-0.56*(0.0011+0.965)*9.8*0.03 = -(1/2)*(0.0011+0.965)*vf^2 + (1/2)(170*0.003^2

vf = 0.575 m/s

m*vi = (M+m)*vf

vi = (M+m)*vf/m

Vi = (0.965+0.0011)*0.575/0.0011

vi = 505 m/s <<<--------ANSWER
================

dK = (1/2)*m*(vf^2-vi^2)

K (1/2)*m*vi^2

dK/K = (vf^2-vi^2)/vi^2

dK/K = (0.575^2-505^2)/505^2 = 0.99 = 99%

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