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Physics

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a 9.50-g bullet. moving horizontally with an initial speed vi , embeds itself in a 1.45-kg pendulum bob that is initially at rest. The length of the pendulum is L=0.745m. After the collision, the pendulum swings to one side and comes to rest when it has gained a vertical height of 12.4 cm. Find the initial speed of the bullet.

  • Physics - ,

    from the final height, the PE gained was

    (m+M)gh (m is the bullet mass).

    So the initial velocity of the bullet/block must be

    1/2 (m+M)v^2=(m+M)gh or
    V=sqrt(2gh)

    Now the law of momentum conservation holds when the bullet hits the block

    (m+M)V=mVi

    (m+M)sqrt(2gh)=mVi

    and solve for Vi

  • Physics - ,

    Thanx ,, but I got something equals to 239.51 m/s .. and I don't think it is right ..
    I did it like this
    1/2 ( 0.0095 + 1.45 ) sqrt ( 2*9.8*0.124) = 0.0095 vi

  • Physics - ,

    I got my mistake ,, thank you very much for the help :)

  • Physics - ,

    No I didn't get it right :S ,, I did it like this exactly
    ( 0.0095 + 1.45 ) sqrt ( 2*9.8*0.124) = 0.0095 vi

  • Physics - ,

    Thanks the answer was right, I related it to the small mass and it made sense ,, I submitted it , and it was right

  • Physics - ,

    239.51

  • Physics - ,

    239.51m/s

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