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18 December, 11:55

7. A bullet of mass 10g strikes a ballistic pendulum of mass 2kg. The center of mass of the pendulum rises a vertical distance of 12cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

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  1. 18 December, 12:40
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    m = mass of the bullet = 10 g = 10 x 10⁻³ kg = 0.01 kg (since 1 g = 10⁻³ kg)

    M = mass of pendulum = 2 kg

    h = height to which pendulum rises = 12 cm = 0.12 m

    V = velocity of the pendulum-bullet combination after collision = ?

    Using conservation of energy

    kinetic energy of the combination just after collision = Potential energy gained due to raise in height of the center of mass

    (0.5) (m + M) V² = (m + M) gh

    V = sqrt (2gh)

    inserting the values

    V = sqrt (2 x 9.8 x 0.12)

    V = 1.5 m/s

    v = velocity of the bullet before the collision

    using conservation of momentum

    momentum of bullet before collision = momentum of bullet-pendulum combination after collision

    m v = (m + M) V

    (0.01) v = (0.01 + 2) (1.5)

    v = 301.5 m/s

    hence initial speed of the bullet is 301.5 m/s
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