# A ball with a mass of #12 kg# moving at #3 m/s# hits a still ball with a mass of #21 kg#. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

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To solve for the final velocity of the second ball, we can use the principle of conservation of momentum:

[m_1v_1 + m_2v_2 = (m_1 + m_2)v_f]

Given:

- (m_1 = 12 , \text{kg}) (mass of the first ball)
- (v_1 = 3 , \text{m/s}) (initial velocity of the first ball)
- (m_2 = 21 , \text{kg}) (mass of the second ball)
- (v_2 = ?) (initial velocity of the second ball, assumed to be zero)
- (v_f = 0) (final velocity of both balls)

Using the conservation of momentum equation:

[12 , \text{kg} \times 3 , \text{m/s} + 21 , \text{kg} \times 0 = (12 , \text{kg} + 21 , \text{kg}) \times 0]

[36 , \text{kg}\cdot\text{m/s} = 33 , \text{kg}\cdot\text{m/s}]

[v_2 = \frac{36 , \text{kg}\cdot\text{m/s}}{21 , \text{kg}}]

[v_2 ≈ 1.71 , \text{m/s}]

To find the kinetic energy lost as heat in the collision, we can use the principle of conservation of mechanical energy:

[KE_{\text{initial}} - KE_{\text{final}} = \text{heat energy lost}]

Given:

- (KE_{\text{initial}} = \frac{1}{2}m_1v_1^2) (initial kinetic energy of the first ball)
- (KE_{\text{final}} = \frac{1}{2}(m_1 + m_2)v_f^2) (final kinetic energy of both balls, assumed to be zero)
- (m_1 = 12 , \text{kg}) (mass of the first ball)
- (v_1 = 3 , \text{m/s}) (initial velocity of the first ball)
- (m_2 = 21 , \text{kg}) (mass of the second ball)
- (v_f = 0) (final velocity of both balls)

Using the kinetic energy equations:

[KE_{\text{initial}} = \frac{1}{2}(12 , \text{kg})(3 , \text{m/s})^2] [KE_{\text{initial}} = 54 , \text{J}]

[KE_{\text{final}} = \frac{1}{2}(12 , \text{kg} + 21 , \text{kg}) \times 0^2] [KE_{\text{final}} = 0]

[\text{heat energy lost} = 54 , \text{J} - 0] [\text{heat energy lost} = 54 , \text{J}]

Therefore, the second ball is moving at approximately (1.71 , \text{m/s}), and (54 , \text{J}) of kinetic energy was lost as heat in 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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- What is the momentum of a #400"kg"# object moving with velocity #8"m"//"s"#?
- Which has more momentum, an object with a mass of #3kg# moving at #14m/s# or an object with a mass of #12kg# moving at #6m/s#?
- What is the law of conservation of momentum?
- How does impulse relate to work?

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