# A ball with a mass of 9 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?

Jan 22, 2016

Use conservation of momentum and conservation of energy
${v}_{2} ' = \frac{9}{21} 3 = \frac{9}{7} \frac{m}{s}$
energy lost to collision = 40.5-17.36 = 23.14 J

#### Explanation:

Momentum of the system before collision:
${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = {m}_{1} {v}_{1} ' + {m}_{2} {v}_{2} '$
m_1 = 9 kg; m_2 = 21 kg; v_1 = 3 m/s; v_2 = 0; v_1' = 0
${v}_{2} ' = {m}_{1} / {m}_{2} {v}_{1}$ substitute
${v}_{2} ' = \frac{9}{21} 3 = \frac{9}{7} \frac{m}{s}$
now kinetic energy before
KE = 1/2mv^2; KE_i = 1/2 9 (3)^2 ; KE_f = 1/2 21 (9/7)^2
KE_i = 1/2 9*9 = 40.5 J; KE_f = 1/2 3 (9)^2/7 = 17.36 J
energy lost to collision = 40.5-17.36 = 23.14 J