physics q:?


A bullet of mass m is fired horizontally into a wooden block of mass M lying on a table. The bullet remains in the block after the collision. The coefficient of friction between the block and table is μ, and the block slides a distance d before stopping. Find the initial speed v0 of the bullet in terms of M, m, μ, g, and d.

2 Answers

  • Anonymous
    2 months ago
    Favorite Answer


    conserve momentum for the collision

    find the post-collision KE, and convert that to friction work.

    In reverse!

    friction work

    W = µ(M+m)gd

    and this is equal to the post-collision KE:

    ½(M+m)v² = µ(M+m)gd

    v² = 2µgd

    v = √(2µgd)

    is the post collision velocity

    now conserve momentum:

    m*v0 = (M+m)v

    v0 = (M+m)*√(2µgd) / m

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  • Anonymous
    2 months ago

    conservation of energy shall apply :


    *(M+m) = (M+m)*Vw^2

    m+M cross

    Vw = √2*d*g*μ

    conservation of momentum shall apply :

    Vw*(m+M) = Vo*m

    Vo = Vw*(m+M)/m = (√2*d*g*μ)*(M+m)/m

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