Yoyo asked in Science & MathematicsPhysics · 1 month ago

Conservation of Momentum?

A hockey puck with mass 0.170 kg traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at 1.50 m/s strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is 30.0° away from the blue line at a speed of 0.750 m/s (see the figure). What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision?

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  • Ash
    Lv 7
    1 month ago
    Favourite answer

    Conservation of momentum along blue line

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

    (0.170)(1.5) + (0.170)(0) = (0.170)(0.750 cos30) + (0.170)(v₂cosθ)

    0.255 = 0.110 + (0.170 v₂cosθ)

    0.170 v₂cosθ = 0.145

    v₂cosθ = 29/34 ....(1)

    Conservation of momentum along perpendicular to blue line

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

    (0.170)(0) + (0.170)(0) = (0.170)(0.750 sin30) + (0.170)(v₂sinθ)

    0 = 0.0638 + (0.170 v₂sinθ)

    0.170 v₂sinθ = - 0.0638

    v₂sinθ = -319/850 .....(2)

    (2) divided by (1) gives

    tanθ = (-319/850)/(29/34)

    θ = -24°

    plug in (1)

    v₂cos(-24) = 29/34

    v₂ = 0.934 m/s

    The 2nd puck has magnitude of 0.934 m/s and direction of 24° on the other side of the blue line

    KE conservation check !

    KE before collision= ½m₁u₁² +½ m₂u₂² = ½m(u₁² + u₂²) = ½(0.170)(1.5² + 0²) = 0.191 J

    KE after collision = ½(0.170)(0.750² + 0.934²) = 0.122 J

    Since the KE was lost after collision, this is inelastic collision

  • Whome
    Lv 7
    1 month ago

    Let m be the common mass. m₁ = m₂ = 0.170 kg

    Conservation of momentum in y direction

    m₁(0) + m₂(0) = m₁(0.750sin30) + m₂(v₂sinθ)

    v₂sinθ = -0.375 m/s

    Conservation of momentum in x direction

    m₁(1.50) + m₂(0) = m₁(0.750cos30) + m₂(v₂cosθ)

    1.50 - 0.6495 = v₂cosθ = 0.85048

    v₂sinθ / v₂cosθ  = -0.375 / 0.85048

    tanθ = -0.44092...

    θ = -23.7939... ≈ -23.8°  ANSWER 

    v₂sin-23.8 = -0.375

    v₂ = 0.92948... ≈ 0.929 m/s  ANSWER 

    KEi = ½(0.170)(1.50²) + ½(0.170)(0²) = 0.19125 J

    KEf = ½(0.170)(0.750²) + ½(0.170)(0.929²) = 0.121170...J

    As KEf ≠ KEi, it was not an elastic collision.  ANSWER 

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