Equilribrium, action & Reaction Momentum quantity of motion p = mv p:momentum (kg ·m/s) m:mass (kg) v:velocity (m/s) m p v.

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Presentation transcript:

Equilribrium, action & Reaction Momentum quantity of motion p = mv p:momentum (kg ·m/s) m:mass (kg) v:velocity (m/s) m p v

A. Momentum Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s. GIVEN: p = ? m = 280 kg v = 3.2 m/s WORK : p = mv p = (280 kg)(3.2 m/s) p = 896 kg·m/s m p v

A. Momentum The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg? GIVEN: p = 675 kg·m/s m = 300 kg v = ? WORK : v = p ÷ m v = (675 kg·m/s)÷(300 kg) v = 2.25 m/s m p v

B. Conservation of Momentum Law of Conservation of Momentum The total momentum in a group of objects doesnt change unless outside forces act on the objects. p before = p after

B. Conservation of Momentum Elastic Collision KE is conserved Inelastic Collision KE is not conserved

Conservation of Momentum Conservation of Momentum A 5-kg cart traveling at 1.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision. BEFORE Cart 1: m = 5 kg v = 4.2 m/s Cart 2 : m = 2 kg v = 0 m/s AFTER Cart 1 + 2: m = 7 kg v = ? p = 21 kg·m/s p = 0 p before = 21 kg·m/sp after = 21 kg·m/s m p v v = p ÷ m v = (21 kg·m/s) ÷ (7 kg) v = 3 m/s

Conservation of Momentum Conservation of Momentum A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon? BEFORE Clown: m = 50 kg v = 0 m/s Cannon: m = 250 kg v = 0 m/s AFTER Clown: m = 50 kg v = 20 m/s Cannon: m = 250 kg v = ? m/s p = 0 p before = 0 p = 1000 kg·m/s p after = 0 p = kg·m/s

Conservation of Momentum Conservation of Momentum So…now we can solve for velocity. GIVEN: p = kg·m/s m = 250 kg v = ? WORK : v = p ÷ m v = (-1000 kg·m/s)÷(250 kg) v = - 4 m/s (4 m/s backwards) m p v