Physics
posted by Brittanie .
A 26 bullet strikes and becomes embedded in a 1.35 block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.23, and the impact drives the block a distance of 9.9 before it comes to rest, what was the muzzle speed of the bullet?

There will be two parts to this question.
1) the bullet strikes and stays inside the wood
2) both bullet and wood block travel as as a system for 9.9m, under the friction of 0.23.
Mass of bullet= Mb, Vb=speed of bullet
Mass of wood bloc=Mw, Vw= speed of wood
Mass of bullet and wood block=Ms, Vs= speed of system
Step 1:
1) MaVa+ MwVw= (Ma+Mw) Vs; Vw at rest initially.
MaVa=(Ma+Mw)Vs
Step 2
1) Friction force= u.Ms.g= 0.23x1.376x9.8 =3.10N
Friction work= F.distance= 3.10N x 9.9m=30.7J
Step 3
(we need to calculate the Kinetic energy before and after of the system, as it travels actoss 9.9m with friction.
1) KEinitial= KE final
Work friction+ KE initial = KE final
Wf + 0.5MsVsinital(squared) 0.5MsVsfinal
((Vs final is zero after 9.9m. So we're left))
Wf + O.5MsVsinitial(squared )=0
Wf= 0.5MsVsinitial(squared)
Vsinitial(squared)= 2Wf/Ms
= 2(30.7J)/1.376kg
Vsinitial= 6.68m/s
Step 4
(plug the Vs initial answer from step 3 into the Vs in step 1)
MaVa=(Ma+Mw)6.68m/s
Va=(Ma+Mw)6.68/Ma
=(1.376x 6.68/ 0.026kg
= 353.5m/s < (ANSWER)
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