Classification π1 = 0.2 ππ π2 = 0.25 ππ 5π π£1 = π π£2 = 0 Solution: π1 π£1 + π2 π£2 = π1 π£1π + π2 π£2π (0.2ππ) (5π/π ) + (0.25ππ) (0π/π ) = (0.2ππ) (0) + (0.25ππ) (π£2π ) (1.0ππ − π/π ) + 0 = 0 + (0.25ππ) (π£2π ) (0.25ππ)(π£2π ) 1.0 ππ − π/π = (0.25ππ) (0.25ππ) (1. 0 kg – m/s) / 0.25ππ = π£2π ππ/π = π£2π π£1π = 0 π£2π = ? π1 π£1 = π2 π£2 π£= π1 + π£2 GIVEN: π1 = 20 ππ π2 = 30 ππ π π£1 = 5 π π£2 = 2 π/π π π 20 ππ 5 + (30 ππ)(2 ) π π π£= 20 ππ + 30 ππ 100ππ β π/π + 60ππ β π/π π£= 50ππ 160ππ β π/π π£= 50ππ π π = π. ππ π Thus, after Anne collides, the combined velocity of Anne and her brother is 3.20m/s. The coefficient of restitution is actually a measure of the “restitution” (i.e., what you give back) of a collision between two objects, or in other words, how much of the kinetic energy remains after two objects collide. Let’s suppose a rubber ball bounces on a flat, hard surface. Obviously, the rubber ball will rebound off the surface, but with only a fraction of its original energy, because all real collisions are inelastic. (Note: If this collision were elastic, then the ball would have bounced back with the same amount of energy it had before striking the surface.) You see, when you deform something by colliding it with something else (say, when you bounce a basketball on the ground), a fraction of its original energy is lost. That’s why the basketball bounces lower with every collision – as its energy gets converted to heat/vibrations. As the ball bounces, it keeps losing energy and becomes less and less ‘bouncy’. In this case, you can think of the coefficient of restitution as an entity that tells you how efficient the“bouncing” process is. The more efficient it is, the more ‘bouncy’ the basketball shall be. 2 πΆππππππππππ‘ ππ π ππ π‘ππ‘π’π‘πππ = √(2πβ) √2πβ β πΆππππππππππ‘ ππ π ππ π‘ππ‘π’π‘πππ = √ 2 π»
