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A ballistic pendulum consists of a rectangular wooden block of mass 1.5 kg suspended from a ceiling. A bullet mass 0.025 kg undergoes a completely inelastic collision with the ballistic pendulum. Due to the collision the pendulum rises to a height of 45 cm. Calculate the speed with which bullet hits the pendulum.

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    While swinging up H = 0.45 m elevation, the block and bullet convert kinetic energy to potential energy, without heat loss. This tells you that
    V = sqrt(2gH) = 2.97 m/s
    is the velocity of the block and bullet just after impact.

    Next, apply conservation of momentum to the inelastic bullet-impact process, assuming that the pendulum does not swing during that brief interval. Let v be the bullet speed, m the bullet mass and M be the block's mass. Momentum conservation tells you that

    m v = (M + m) V

    Solve for v

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  • physics -

    i like 44

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