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1 July, 09:25

500 kg cannon and is at rest on a horizontal, frictionless surface on a clear and sunny day. When a crazy cool 100 kg clown decided to show how the cannon works they decided to get fired horizontally from the barrel of the cannon and the cannon recoils with a speed of 5 m/s. A very inquisitive Physics student figured out how fast the clown was traveling when leaving the cannon. What value did they come up with?

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  1. 1 July, 11:53
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    Given that,

    Mass of cannon

    M1 = 500kg and initially at rest

    U1 = 0m/s

    Mass of clown

    M2 = 100kg

    so it was initial at rest before this time, therefore, U2 = 0

    Recoils speed of cannon V1 = 5m/s, the recoils speed is after the cannon has left the barrel.

    Using construction of linear momentum

    Momentum before collision is equal to momentum after collision

    The initial momentum is zero since the two bodies are until at rest

    And the final momentum is

    M1•V1 + M2•V2

    Then,

    P (initial) = P (final)

    0 = M1•V1 + M2•V2

    0 = 500 * 5 + 100 * V2

    0 = 2500 + 100•V2

    100•V2 = - 2500

    V2 = - 2500/100

    V2 = - 25m/s

    So, the final velocity of the clown is - 25m/s, opposite direction of the cannon
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