Step 1

**$\mathbf{KE}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{KE}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}(0.17\mathrm{kg}){(4.7\frac{m}{s})}^{\mathbf{2}}$**

**KE = 1.8777 J**

The kinetic energy of a rolling billiard ball is given by KE = 1/2mv^{2}. Suppose a 0.17-kg billiard ball is rolling down a pool table with an initial speed of 4.7 m/s. As it travels, it loses some of its energy as heat. The ball slows down to 3.8 m/s and then collides head-on with a second billiard ball of equal mass. The first billiard ball completely stops and the second one rolls away with a velocity of 3.8 m/s. Assume the first billiard ball is the system.

Calculate q for the process.

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