A ball with a mass of #3 kg# moving at #8 m/s# hits a still ball with a mass of #18 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?

To find the final velocity of the second ball, use the principle of conservation of momentum. The equation is:

(m_1 \times v_1 + m_2 \times v_2 = m_1 \times u_1 + m_2 \times u_2)

where:

• (m_1 = 3 , \text{kg}) (mass of the first ball)
• (v_1 = 0 , \text{m/s}) (final velocity of the first ball)
• (m_2 = 18 , \text{kg}) (mass of the second ball)
• (v_2) (final velocity of the second ball)
• (u_1 = 8 , \text{m/s}) (initial velocity of the first ball)
• (u_2 = 0 , \text{m/s}) (initial velocity of the second ball)

Solving for (v_2):

(3 \times 0 + 18 \times v_2 = 3 \times 8 + 18 \times 0)

(18v_2 = 24)

(v_2 = \frac{24}{18} = \frac{4}{3} , \text{m/s})

To find the kinetic energy lost as heat, use the equation:

(KE_{\text{lost}} = KE_{\text{initial}} - KE_{\text{final}})

(KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2)

(KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2)

(KE_{\text{initial}} = \frac{1}{2} \times 3 \times 8^2 = 96 , \text{J})

(KE_{\text{final}} = \frac{1}{2} \times 3 \times 0^2 + \frac{1}{2} \times 18 \times (\frac{4}{3})^2 = 0 + 12 , \text{J})

(KE_{\text{lost}} = 96 - 12 = 84 , \text{J})

The final velocity of the second ball is ( \frac{4}{3} , \text{m/s}), and the kinetic energy lost as heat in the collision is (84 , \text{J}).

I can only answer the first part.

We use the law of conservation of momentum, which states that,

Plugging in the values, we get,