Question

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?

Answer 1

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)