# A ball with a mass of #6 kg# moving at #3 m/s# hits a still ball with a mass of #8 kg#. If the first ball stops moving, how fast is the second ball moving?

Momentum is conserved, so if the first ball stops from the collision, then the second ball will have the same momentum as the first ball did.

The momentum of the firs ball is:

If that ball stops, then all the momentum is transferred to the second ball,

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(6 X 3)/8 = 2.25 m per sec in the same direction

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To find the velocity of the second ball after the collision, you can use the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision.

(m_1 v_1 + m_2 v_2 = m_1 v_{1f} + m_2 v_{2f})

Where:

- (m_1 = 6 , \text{kg}) (mass of the first ball)
- (v_1 = 3 , \text{m/s}) (initial velocity of the first ball)
- (m_2 = 8 , \text{kg}) (mass of the second ball)
- (v_2 = 0 , \text{m/s}) (initial velocity of the second ball, since it's still)
- (v_{1f} = 0 , \text{m/s}) (final velocity of the first ball, since it stops moving)
- (v_{2f}) (final velocity of the second ball, which we need to find)

By plugging in the values and solving for (v_{2f}), we get:

(6 \times 3 + 8 \times 0 = 6 \times 0 + 8 \times v_{2f})

(18 = 8 \times v_{2f})

(v_{2f} = \frac{18}{8} = 2.25 , \text{m/s})

So, the second ball is moving at (2.25 , \text{m/s}) after the collision.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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