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Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. The mass of the eight ball, m2, is 6 g with a velocity of 1 m/s. After the balls collide, they bounce off each other. The number three ball moves off with a velocity of 5 m/s. What is the final velocity and direction of the eight ball? +8.6 m/s +6.0 m/s –6.0 m/s –8.6 m/s

Respuesta :

iModern

The other answer is correct, however the answer on edg is +5.7

onyebuchinnaji

The final velocity of the second ball (eight ball) is + 6.0 m/s

The given parameters;

  • mass of the first ball, m₁ = 5 g = 0.005 kg
  • initial velocity of the first ball, u₁ = 3 m/s
  • mass of the second ball, m₂ = 6 g = 0.006 kg
  • initial velocity of the second ball, u₂= 1 m/s
  • final velocity of the first ball, v₁ = 5 m/s
  • final velocity of the second ball, v₂ = ?

The final velocity of the second ball is determined by applying the principle of conservation of linear momentum.

Assume the following direction;

--------->(-)      (+)<--------------  =  (+)<------------   ---------->(-)

m₁(-u₁) + m₂u₂ = m₁v₁  +  m₂(-v₂)

-(0.005 x 3)  +  (0.006 x 1) = (0.005 x 5) - 0.006v₂

-0.009= 0.025 - 0.006v₂

0.006v₂ = 0.025 + 0.009

0.006v₂ = 0.034

[tex]v_2 = \frac{0.034}{0.006} \\\\v_2 = 5.67 \ m/s \ \ \approx \ +\ 6.0\ m/s[/tex]

Thus, the final velocity of the second ball (eight ball) is + 6.0 m/s

Learn more here:https://brainly.com/question/24900666

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