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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?

  • Physics -

    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.

    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)

    =(1.376x 6.68/ 0.026kg
    = 353.5m/s <- (ANSWER)

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