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9 May, 20:19

A 0.439 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.307 kg puck moving initially along the x axis with a speed of 2.19 m/s. After the collision, the 0.307 kg puck has a speed of 1.19 m/s at an angle of 37◦ to the positive x axis. Determine the magnitude of the velocity of the 0.439 kg puck after the collision

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  1. 9 May, 20:24
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    u2 = 0.266 m/s

    Explanation:

    Let the left Puck mass at rest = m1 = 0.307 Kg

    mass of the right puck m2 = 0.439 kg

    velocity of m1 before collision v1 = 2.19 m/s

    velocity of m2 before collision v2 = 0m/s

    velocity of m1 after collision u1 = 1.19 m/s

    velocity of m2 after collision u2 = ? m/s

    θ = 37°

    Solution:

    Before collision:

    Momentum (y-axis) before collision = 0 Kgm/s

    Momentum (x-axis) before collision = m1v1 + m2v2 = 0.307 Kg x 2.19 m/s + 0

    = 0.672 Kgm/s

    After collision:

    Momentum (y-axis) after collision = m1u1 sinθ + m2u2 sinθ

    = 0.307 x 1.19 m/s sin 37 ° + 0.439 x u2 sin 37°

    = 0.22 + 0.26 u2

    Momentum (x-axis) after collision = m1u1 cosθ + m2u2 cos θ

    = 0.307 x 1.19 m/s cos 37 ° + 0.439 x u2 cos 37°

    = 0.29 + 0.35 u2

    According to law of conservation momentum

    momentum before collision = momentum after collision.

    0 + 0.672 Kgm/s = 0.22 Kgm/s + 0.26 kg u2 + 0.29 Kgm/s + 0.35 kg u2

    0.672 Kgm/s = 0.51 Kgm/s + 0.61 u2

    u2 = 0.266 m/s
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