A ball with a mass of #2 kg# moving at #14 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?

Since the momentum is preserved,

Additionally, energy is conserved.

lost in the collision, after that

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 is equal to the total momentum after the collision.

(m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot v_1' + m_2 \cdot v_2')

Where: (m_1 = 2 , \text{kg}) (mass of the first ball) (v_1 = 14 , \text{m/s}) (initial velocity of the first ball) (m_2 = 21 , \text{kg}) (mass of the second ball) (v_2 = 0 , \text{m/s}) (initial velocity of the second ball)

Solving for (v_2'): (2 \cdot 14 + 21 \cdot 0 = 2 \cdot 0 + 21 \cdot v_2') (28 = 21 \cdot v_2') (v_2' = \frac{28}{21} , \text{m/s} = \frac{4}{3} , \text{m/s})

To find the kinetic energy lost as heat in the collision, you can use the equation: (KE_{\text{lost}} = KE_{\text{initial}} - KE_{\text{final}})

Where: (KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2) (initial kinetic energy of the first ball) (KE_{\text{final}} = \frac{1}{2} m_1 v_1'^2) (final kinetic energy of the first ball)

(KE_{\text{initial}} = \frac{1}{2} \cdot 2 \cdot (14)^2) (KE_{\text{final}} = \frac{1}{2} \cdot 2 \cdot (0)^2)

(KE_{\text{lost}} = \frac{1}{2} \cdot 2 \cdot (14)^2 - \frac{1}{2} \cdot 2 \cdot (0)^2) (KE_{\text{lost}} = 196 , \text{J})