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Momentum and Collision Momentum Momentum: A vector quantity defined as the product of an object’s mass and velocity. p = momentum momentum = mass x velocity p = mv Units of momentum: kg m/s Sample Problem A 2250 kg truck has a velocity of 25 m/s to the east. What is the momentum of the truck? p = mv p = (2250 kg)(25 m/s) p = 5.6 x 104 kg m/s to the east Impulse From Newton’s Second Law: F = ma F = mΔv Δt F = Δp Δt FΔt = Δp or FΔt = mvf – mvi Impulse Impulse: A force applied over a period of time. Impulse is measured in Newton seconds, Ns. force Impulse = FΔt = Δp Area under curve is the impulse time Impulse-Momentum Theorem The impulse on an object is equal to the object’s final momentum minus the object’s initial momentum. FΔt = pf – pi FΔt = mvf – mvi Impulse Problem (FΔt = Δp) A driver accelerates a 240 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.0 m/s to 28.0 m/s over a time interval of 60 s. What is the change in momentum What is the impulse on the snowmobile What is the average force that is exerted on the snowmobile Impulse Problem (FΔt = Δp) A driver accelerates a 240 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.0 m/s to 28.0 m/s over a time interval of 60 s. What is the change in momentum Δp = mvf – mvi = 240 kg(28 m/s) – 240 kg(6 m/s) = 5280 kg m/s Impulse Problem (FΔt = Δp) A driver accelerates a 240 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.0 m/s to 28.0 m/s over a time interval of 60 s. What is the impulse on the snowmobile FΔt = Δp Δp =5280 kg m/s FΔt = impulse Impulse = 5280 kg m/s Impulse Problem (FΔt = Δp) A driver accelerates a 240 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.0 m/s to 28.0 m/s over a time interval of 60 s. What is the average force that is exerted on the snowmobile FΔt = Δp F(60s) = 5280 kg m/s F = 88 N Impulse Problem (FΔt = Δp) Suppose a 60 kg person is traveling in a vehicle moving at 32 m/s. The car hits a concrete wall and the car comes to a stop in 0.2 s. What is the force exerted on the person during the crash? FΔt = Δp F(.20s) = mvf – mvi = 60kg(0m/s) – 60kg(32m/s) F(.20s) = -1920 kg m/s F = -9600 N Impulse Problem (FΔt = Δp) Suppose that this person is now in a specially engineered car that crumples upon impact and takes 0.8 s to come to a complete stop. What is the force on the passenger now? FΔt = Δp = mvf – mvi = 60kg(0m/s) – 60kg(32m/s) = -1920 kg m/s FΔt = Δp F(.80s) = -1920 kg m/s F = -2400 N Collisions Collisions: When two or more objects come into contact with each other and exchange energy and momentum. Elastic Collision: total kinetic energy of the system before the collision equals the total kinetic energy after the collision. Momentum is also conserved. Elastic Collisions • No energy is dissipated as heat energy • No energy is spent on deforming the bodies • Occurs when two objects “bounce” apart after they collide • Close examples: – Bumper cars – Pool table balls – Rubber balls colliding Elastic Collision m1v1,f + m2v2,f = m1v1,i + m2v2,i m1 m2 m1 m2 ½ m1v1,f2 + ½ m2v2,f 2= ½ m1v1,i2 + ½ m2v2,i2 Elastic Collision Problem Ball A (0.355 kg) moves to the right on a frictionless surface at 0.095 m/s and hits ball B (0.710 kg) moving in the same direction at a speed of 0.045 m/s. After the collision the speed of ball A is 0.035 m/s. What is the speed of ball B? mava,f + mbvb,f = mava,i + mbvb,i .355 kg(.035 m/s) + .71 kg(vb,f) = 0.355 kg(.095m/s) + .71kg(.045m/s) Vb,f = +0.075 m/s Elastic Collision Problem 2 Two balls are moving towards each other. The blue ball is moving to the right with a mass of 0.2 kg and a velocity of 1 m/s. The green ball, 0.5 kg, is moving to the left with a velocity of 3 m/s. If the green ball after the collision is moving to the left with a velocity of 2 m/s, what is the velocity of the blue ball? 0.2 kg 1 m/s 3 m/s 0.5 kg Problem 2 0.2 kg 1 m/s 3 m/s 0.5 kg mbluevf + mgreenvf = mbluevi + mgreenvi 0.2kg(vf) +0.5kg(-2 m/s) = 0.2 kg(1 m/s) + .5 kg( -3 m/s) 0.2vf – 1 kg m/s = -1.3 kg m/s 0.2vf = -0.3 kg m/s Vf(blue) = -1.5 m/s Inelastic Collisions Inelastic Collision: A collision between bodies where momentum is conserved but the total kinetic energy is not. Part of the kinetic energy is transformed into another energy form. Usually colliding objects are distorted in some way. Examples: - car collisions - two balls of clay colliding and sticking together - asteroid colliding with a planet - bullet embedding itself in a block of wood Inelastic Collision m1v1,f + m2v2,f = m1v1,i + m2v2,i m1 m2 m1 m2 If objects stick together after collision v1,f = v2,f vf(m1 + m2) = m1v1,i + m2v2,i Inelastic Collision Problem 1 A 35 g bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off together at 8.6 m/s. What was the original velocity of the bullet? 0.035 kg 8.6 m/s 5.0 kg ? m/s Inelastic Collision Problem 1 0.035 kg 8.6 m/s 5.0 kg ? m/s vf(m1 + m2) = m1v1,i + m2v2,i 8.6 m/s(5.0kg + 0.035kg) = 5.0kg(0) + 0.035kg(v) 43.301 kg m/s = 0.035kg(v) v = 1237 m/s Inelastic Collision Problem 2 A 0.105 kg hockey puck moving at 48 m/s is caught by a 75 kg goalie at rest. With what speed does the goalie slide on the ice? vf(m1 + m2) = m1v1,i + m2v2,i vf(0.105kg+ 75 kg) = 75kg(0) + 0.105kg(48m/s) vf(75.105kg) = 5.04 kg m/s vf = 0.067 m/s