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A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30 degrees with respect to the original line of motion. a. Find the velocity (magnitude and direction) of the second ball after the collision. b. Was the collision inelastic or elastic? (make a claim and support with evidence.)

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Answer

given,

initial speed of the billiard ball,u = 5 m/s

initial speed of second ball, u' = 0 m/s

after collision

final speed of first billiard, v = 4.33

at 30° from the original line

second velocity of the second billiard = ?

b)Collision between the two block is elastic.

mass of both the balls are not same

so,

θ₁ + θ₂ = 90°

 θ₁ = 30°

θ₂ = 90° - 30°

θ₂ = 60°

a) using conservation of momentum

  m u + m u' = m v cos θ₁ + m v' cos θ₂

  m x 5  + 0 = m x 4.33 x cos 30° + m x v' x cos 60°

  0.5 v' = 1.25

    v' = 2.5 m/s

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onyebuchinnaji

The final velocity of the second ball after collision is 1.25 m/s.

The collision is elastic since the final velocities of the two balls are different.

The given parameters:

  • Initial velocity of the first ball, u₁ = 5.0 m/s
  • Final velocity of the first ball, v₁ = 4.33 m/s at angle 30 degrees

Let the mass of both balls = m

The final velocity of the second ball after collision is determined by applying the principle of conservation of linear momentum;

[tex]m_1 u_1 + m_2 u_2 = m_1v_1 + m_2 v_2\\\\5m + m(0) = m(4.33 \times cos(30)) + mv_2\\\\5m = 3.75 m + mv_2\\\\1.25 m = mv_2\\\\1.25 \ m/s = v_2[/tex]

Thus, we can conclude that the collision is elastic since the final velocities of the two balls are different.

Learn  more about conservation of linear momentum here: https://brainly.com/question/7538238

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