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Physics

Physics. Conservation of Momentum. Momentum and Impulse. When a mass is moving at a specific velocity it has momentum. Momentum = mass x velocity ρ = Greek letter “ro” is the symbol for momentum ρ = m x v = kg x m/s = N ● s. Basic Calculations.

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Physics

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  1. Physics Conservation of Momentum

  2. Momentum and Impulse • When a mass is moving at a specific velocity it has momentum. • Momentum = mass x velocity • ρ = Greek letter “ro” is the symbol for momentum • ρ = m x v • = kg x m/s = N ● s

  3. Basic Calculations • A 3400 kg car is traveling at a velocity of 26 m/s. What is the momentum? • ρ = m x v • = 3400 kg x 26 m/s • = 8.84 x 104 N●s Formula Substitution Answer

  4. Change in Momentum • F = ma • ρ = mv • F = ρa/v • F = ρ v/t • F = Δρ m = F/a Solve for mass ρ = (F/a)v Substitute F/a for mass Solve for force a = v/t Acc = velocity/time Substitute v/t for a Cancel like variables. v Δρ = FΔt Δt FΔt = Δρ = mvf - mvi

  5. Momentum and Force, Impulse • A 1500 kg car moving eastward with a velocity of 16 m/s collides with a utility pole and is brought to rest in 0.25 s. Find the magnitude of the force exerted on the car during the collision. • Given: m = 1500 kg vi = -16 m/s Δt = 0.25 s vf = 0 m/s FΔt = Δρ = mvf - mvi = (1500 kg)(0m/s) -(1500 kg)(-16m/s) F = mvf - mvi 0.25 s Δt = 9.6 x 104 N

  6. Let’s Dive into it • A 67.5 kg woman dives off a 12.5 m cliff into a lake. Her body comes to rest in 0.67 s after reaching the water. What force does the water exert on her? • Diagram: • PE =mgh =(67.5 kg)(9.81 m/s2)(12.5 m) =8.28 x 103 j PE= mgh 12.5 m PE = KE = ½ mv2

  7. Continue • PE =mgh =(67.5 kg)(9.81 m/s2)(12.5 m) =8.28 x 103 j • KE = 1/2 mv2, Solve for v, v =√2KE/m • = √ 2(8.28 x 103 j)/(67.5 kg) = 15.7 m/s • F = mv/t = (67.5 kg) ( 15.7 m/s)/ (0.67 s) = 710 N

  8. Stopping Distance • A 1780 kg motorbike traveling at 17.8 m/s applies the brakes which exert a force of 6780 N. How far does the bike travel before it comes to a complete stop. FΔt = Δρ = mvf – mvi Δt = mvf - mvi m = 1780 kg vi = -17.8 m/s Δt = ? vf = 0 m/s F = 6780 N Δx = ? F Δt = (1780 kg)(0m/s) - (1780 m/s)(-17.8 m/s)/6780 N = 4.20 s Δx = 1/2 (vi + vf ) Δt = 1/2 (17.8 m/s + 0m/s)4.20s = 37.4 m

  9. Conservation of Momentum • Conservation of momentum occurs when the initial total momentum is equal to the final total momentum. The object with the largest mass moves slower than the object with the smaller mass if they have the same momentum.

  10. Conservation Equation • ρAi + ρBi = ρAf + ρBf • m1v1,i + m2v2,i = m1v1,f + m2v2,f • Final Velocity: solve for v2f • v2f = m1v1,i + m2v2,i - m1v1,f • m2

  11. Conservative Problem • A 82 kg man steps off a 1.2 kg skateboard with a velocity of 3.4 m/s to the right, what is the final velocity of the skateboard in the opposite direction? • Given: m1= 82 kg , m2= 1.2 kg v1i = 0 m/s, v2i= 0 m/s v1f= 3.4 m/s,v2f= ? m1v1,i + m2v2,i = m1v1,f + m2v2,f, Since the initial velocity’s are = 0 m2v2,f = - m1v1f v2f = -m1v1f m2 = -(82 kg) (3.4 m/s) = 232 m/s 1.2 kg

  12. Inelastic Collisions • An inelastic collision occurs when two masses collide and stick together and move in some final direction and a particular velocity Large mass Small mass m1v1i + m2v2i = ( m1 + m2 )vf Both masses move at the same velocity in the same direction.

  13. Bang • A 1860 kg car is stopped at a light and another car with a mass of 1450 kg strikes it from the rear traveling at 24.0 m/s. How far will they travel into the intersection if they come to a stop in 1.4 s? m1v1i + m2v2i = ( m1 + m2 )vf Since the velocity of the stopped car is = 0 m2v2i = ( m1 + m2 )vf vf = m2v2= (1450 kg)(24.0 m/s) m1 + m2 1450 kg + 1860 kg = 10.5 m/s Δx = vΔt = (10.5 m/s)(1.4s) = 14.7 m

  14. Elastic Collisions • Elastic collisions occur when two object collide and both move in opposite directions after the collision. • m1v1i + m2v2i = m1v1f + m2v2f • What is the recoil velocity of a 1.8 kg gun that shoots a 25g bullet at 167 m/s? • vgun = mb vb = (0.025 kg) ( 167 m/s ) • mgun 1.8 kg • = 2.3 m/s

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