A $250 \mathrm{~g}$ sticky hockey puck slides along a frictionless, horizontal ice surface at $75 \mathrm{~m} / \mathrm{s}$ in the $+\mathrm{x}$ direction. It collides with a $500 \mathrm{~g}$ sticky puck originally sliding at $25 \mathrm{~m} / \mathrm{s}$ in the -y direction. They collide perfectly inelastically. How much kinetic energy is lost in this collision?

Question

Quizplus · Accepted Answer

The total kinetic energy before the collision is:$$K_i = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}(0.25 \mathrm{~kg})(75 \mathrm{~m/s})^2 + \frac{1}{2}(0.5 \mathrm{~kg})(25 \mathrm{~m/s})^2 = 1406.25 \mathrm{~J}$$The total kinetic energy after the collision is:$$K_f = \frac{1}{2}(0.75 \mathrm{~kg})v^2$$where v is the velocity of the combined pucks after the collision.Using the conservation of momentum, we have:$$0.25 \mathrm{~kg}(75 \mathrm{~m/s}) + 0.5 \mathrm{~kg}(-25 \mathrm{~m/s}) = (0.75 \mathrm{~kg})v$$$$v = 25 \mathrm{~m/s}$$Therefore, the kinetic energy lost in the collision is:$$K_i - K_f = 1406.25 \mathrm{~J} - \frac{1}{2}(0.75 \mathrm{~kg})(25 \mathrm{~m/s})^2 = 521 \mathrm{~J}$$

A 250 g250 \mathrm{~g}250 g Sticky Hockey Puck Slides Along a Frictionless, Horizontal Ice Surface

A $250 \mathrm{~g}$ Sticky Hockey Puck Slides Along a Frictionless, Horizontal Ice Surface