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Momentum: The quantity of motion of a moving body, measured as a product of its mass and velocity Momentum • • • • • • • • • • Represented by the variable P. P = mv, where m = mass, kg & v = velocity, m/s Momentum is conserved; it cannot be created or destroyed. Any changes in momentum of a body will generate changes in momentum elsewhere in its system. Momentum is a different quantity from kinetic energy in that its velocity is not squared. If two objects collide, their respective momentums will both change. A collision is a classic type of demonstration of momentum. Two types of collisions: Elastic and non-elastic Elastic: Objects bounce off each other with no energy loss. Non-elastic: Objects stick together when they collide. • Changes in momentum are associated with applied forces. Remember Newton’s 1st Law: “An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.” • When two objects exchange momentum (collide), we represent it mathematically as: Ptotal = (P1i + P2i) = (P1f + P2f); (m1v1i)+ (m2v2i)= (m1v1i)+ (m2v2i ) • • • • • • Where: P1 = momentum of 1st object; P2 = momentum of 2nd object, i = initial, before collision; f = final, after collision Note: velocities change but masses do not. When an outside force is applied to an object in motion, its momentum is changed. (Force)(time) = (mass)(change in velocity): F t = m Δv The quantity Force • time is known as impulse. Impulse = Change in momentum And since the quantity m • v is the momentum, the quantity m•Δv will be the change in momentum. Recall that velocity is a VECTOR, so it has both magnitude and direction. Example of Impulse: When a batter hits a pitch, an impulse is applied to the bat and the ball, which changes their respective momentums. Example Problems 1) What is the magnitude of momentum of a 1.5 kg mass traveling at a rate of 5.0 m/s? P = mv = (1.5 kg)(5.0 m/s) = 7.5 kg m/s 2) A ball with mass 50 g is thrown with a velocity of 7.0 m/s at a stationary, 10 g aluminum can. The ball strikes the can and continues forward at 4.7 m/s (elastic collision). What is the velocity of the can after the ball strikes it? Pib = (0.050 kg)(7.0 m/s) = 0.35 kg m/s; Pic = 0 (the initial velocity = 0) Pfb = (0.050 kg)(4.7 m/s) = 0.235 kg m/s; Pfc = (0.010 kg) vfc 0.35 + 0 = 0.235 + 0.010 vfc (0.35 ‒ 0.235) / 0.01 = vfc vfc = 11.5 m/s 3) Next, consider problem #2, except this time the ball and the can stick together upon colliding (non-elastic collision). What is the velocity for the combined objects after the collision? Pib = (0.050 kg)(7.0 m/s) = 0.35 kg m/s; Pic = 0 (the initial velocity = 0) Ptot = (0.05 + 0.01) kg vf-tot (Note: The two masses are now combined.) 0.35 kg m/s = 0.06 kg vf-tot vf-tot = 5.8 m/s 4) A football with mass = 425 g (yep, I looked this up) is kicked from a tee such that its horizontal velocity = 10 m/s. a. What horizontal impulse was applied to the ball? Impulse = m Δv = (0.425 kg)(10 – 0) m/s = 4.25 kg m/s b. If this impulse’s time duration = 0.20 s, what force was applied? m Δv = F t; t = (4.25 kg m/s) / 0.20 s = 21.25 kg m2/s2 = 21.25 N