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chubhunter [2.5K]

2 years ago

9

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. Th

e 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 5. 7 m/s â€"5. 7 m/s â€"8. 6 m/s.

1 answer:

Ymorist [56]2 years ago

4 0

This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

$v_2=\frac{34\ g.m/s}{6\ g}\\\\$

<u>v₂ = 5.7 m/s</u>

<u></u>

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

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