# A ball with a mass of #15 kg# moving at #6 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?

Conservation of momentum states that before and after a physical reaction there must always be the same amount of momentum.

Momentum is the product of mass and velocity, or

Before the reaction, there is a total momentum of

Therefore, after the reaction,

The answer is rounded to the first decimal place.

For the second part of the question, kinetic energy is given by the formula

Before the reaction, total kinetic energy would therefore be

After the reaction, we would have

The change in kinetic energy is

Feel free to check my working, there's some room for error-carried-forward due to rounding. Hope it's somewhat clear.

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To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum:

[m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot u_1 + m_2 \cdot u_2]

where:

- (m_1 = 15 , \text{kg}) (mass of the first ball)
- (m_2 = 21 , \text{kg}) (mass of the second ball)
- (v_1 = 0 , \text{m/s}) (final velocity of the first ball)
- (u_1 = 6 , \text{m/s}) (initial velocity of the first ball)
- (u_2 = 0 , \text{m/s}) (initial velocity of the second ball)
- (v_2) (final velocity of the second ball)

Solving for (v_2): [m_1 \cdot u_1 + m_2 \cdot u_2 = m_1 \cdot v_1 + m_2 \cdot v_2] [15 , \text{kg} \times 6 , \text{m/s} + 21 , \text{kg} \times 0 , \text{m/s} = 15 , \text{kg} \times 0 , \text{m/s} + 21 , \text{kg} \times v_2] [90 , \text{kg} \cdot \text{m/s} = 21 , \text{kg} \times v_2] [v_2 = \frac{90 , \text{kg} \cdot \text{m/s}}{21 , \text{kg}}] [v_2 \approx 4.29 , \text{m/s}]

To find the kinetic energy lost as heat in the collision, we can use the equation:

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

where:

- (KE_{\text{initial}}) is the initial kinetic energy of the system (before the collision)
- (KE_{\text{final}}) is the final kinetic energy of the system (after the collision)

[KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2] [KE_{\text{initial}} = \frac{1}{2} \times 15 , \text{kg} \times (6 , \text{m/s})^2 + \frac{1}{2} \times 21 , \text{kg} \times (0 , \text{m/s})^2] [KE_{\text{initial}} = \frac{1}{2} \times 15 , \text{kg} \times 36 , \text{m}^2/\text{s}^2 + 0] [KE_{\text{initial}} = \frac{1}{2} \times 15 , \text{kg} \times 36 , \text{m}^2/\text{s}^2] [KE_{\text{initial}} = 270 , \text{J}]

[KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2] [KE_{\text{final}} = \frac{1}{2} \times 15 , \text{kg} \times (0 , \text{m/s})^2 + \frac{1}{2} \times 21 , \text{kg} \times (4.29 , \text{m/s})^2] [KE_{\text{final}} = 0 + \frac{1}{2} \times 21 , \text{kg} \times (4.29 , \text{m/s})^2] [KE_{\text{final}} = 0 + \frac{1}{2} \times 21 , \text{kg} \times 18.3841 , \text{m}^2/\text{s}^2] [KE_{\text{final}} = 194.016 , \text{J}]

[KE_{\text{lost}} = KE_{\text{initial}} - KE_{\text{final}}] [KE_{\text{lost}} = 270 , \text{J} - 194.016 , \text{J}] [KE_{\text{lost}} \approx 75.984 , \text{J}]

Therefore, the velocity of the second ball after the collision is approximately (4.29 , \text{m/s}), and the kinetic energy lost as heat in the collision is approximately (75.984 , \text{J}).

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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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