A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The first ball stops moving, and the second ball then moves forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?

36 m/s

4 m/s

30 m/s

1.44 m/s

36 m/s

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The original person had to wait 2 years to get the answer.. wow I feel bad for them honestly

Answer is 36 m/s

Anyone?

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Why did the ball go to school? Because it wanted to get an education on velocity! Okay, joke time over. Let's calculate the velocity of the third ball.

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision, the momentum of the two balls with a mass of 0.6 kg and a velocity of 12 m/s is given by:
Total momentum = (0.6 kg + 0.6 kg) × 12 m/s = 14.4 kg·m/s

After the collision, the first ball comes to a stop. Therefore, its momentum becomes zero. The momentum of the second ball with a mass of 0.6 kg can be calculated as:
Momentum of second ball = 0.6 kg × velocity of second ball

Finally, we can use the conservation of momentum to calculate the velocity of the second ball:
Total momentum = Momentum of second ball + Momentum of third ball
14.4 kg·m/s = 0.6 kg × velocity of second ball + 0.2 kg × velocity of third ball

Now we can solve this equation to find the velocity of the third ball:
14.4 kg·m/s = 0.6 kg × velocity of second ball + 0.2 kg × velocity of third ball

Let's do some math and find out the answer!
velocity of third ball = 1.44 m/s

So, the velocity of the third ball is 1.44 m/s. Keep rolling with those physics questions!

To find the velocity of the third ball, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

First, let's calculate the momentum of the two balls before the collision. The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v).

The momentum of the first ball:
p₁ = m₁ * v₁
p₁ = 0.6 kg * 12 m/s
p₁ = 7.2 kg*m/s

The momentum of the second ball will be equal to the negative of the momentum of the first ball since the first ball is brought to a stop:
p₂ = -7.2 kg*m/s

Now, let's consider the collision of the second ball with the third ball. Since the system is closed, the momentum of the second ball right before the collision will be the same as the momentum of both the second and third balls after the collision.

Let's denote the velocity of the third ball as v₃.

The momentum of the second ball before the collision:
p₂ = -7.2 kg*m/s

The momentum of the second ball after the collision:
p₂' = m₂ * v₂
p₂' = 0.6 kg * v₃

The momentum of the third ball after the collision:
p₃' = m₃ * v₃
p₃' = 0.2 kg * v₃

Now, applying the conservation of momentum principle:

Total momentum before the collision = Total momentum after the collision
p₂ = p₂' + p₃'
-7.2 kg*m/s = 0.6 kg * v₃ + 0.2 kg * v₃
-7.2 kg*m/s = 0.8 kg * v₃

Now, we can solve for v₃:

v₃ = (-7.2 kg*m/s) / (0.8 kg)
v₃ = -9 m/s

Since the velocity cannot be negative in this case (since it represents the direction), we take the magnitude of the velocity:

|v₃| = 9 m/s

Therefore, the velocity of the third ball is 9 m/s. None of the given options match this result, so it seems there might be a mistake either in the given options or in the calculations.