A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis wit

Question

A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of ? = 52.0

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Orla Orla 4 years 2021-08-31T18:13:08+00:00 1 Answers 49 views 0

Answers ( )

  1. Ladonna
    0
    2021-08-31T18:14:46+00:00
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    Answer:

    u2 = 0.897 m/s

    Explanation:

    Note: Question is incomplete ( upon the missing of velocity of the 0.300 kg puck after collision — i have tried to find as below)

    Let the left Puck mass at rest = m1 =0.300 Kg

    mass of the right puck m2 = 0.200 kg

    velocity of m1 before collision v1= 2.00 m/s

    velocity of m2 before collision v2 = 0m/s

    velocity of m1 after collision u1 =1.00 m/s

    velocity of m2 after collision u2 = ? m/s

    θ = 52°

    Solution:

    Before collision:

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

    Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.300 Kg x 2.00 m/s + 0

    = 0.600 Kgm/s

    After collision:

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

    = 0.300 x 1.00 m/s sin 52 °  + 0.2 x u2 sin 52°

    = 0.236 + 0.158 u2

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

    = 0.300 x 1.0 m/s cos 52 ° + 0.200 x u2 cos 52°

    = 0.184 + 0.123 u2

    According to law of conservation momentum

    momentum before collision = momentum after collision.

    0 + 0.60 Kgm/s =  0.236 Kgm/s  + 0.158 kg u2 + 0.184 Kgm/s + 0.123 kg u2

    0.672 Kgm/s = 0.420 Kgm/s + 0.281 u2

    u2 = 0.897 m/s

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )