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## Linear Momentum

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**Momentum is a measure of how hard it is to stop or turn a**moving object. • What characteristics of an object would make it hard to stop or turn? • For one particle • p = mv • Note that momentum is a vector with the same direction as the velocity! • NO VELOCITY = NO MOMENTUM • For a system of multiple particles • p = Σpi--- add up the vectors • The unit of momentum is… • kg m/s or Ns**p = mv**p = 65 Kg ·10 m/s = 65 kg·m/s east • Calculate the momentum of a 65-kg sprinter running east at 10 m/s. • Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s. Px A2 + b2 = c2 System: 2 runners Py P12 + P22 = P2 P=√(P12 + P22) P=√(ΣPx)2 + (ΣPy)2) p = mv P=√((mxvx)2 + Σ(myvy)2) P1 + P2 ≠ P P=√((10m/s·65 kg)2 + (9.5 m/s·75 kg)2) = 964.45 N·s**p = 65 Kg ·10 m/s = 65 kg·m/s east**p = mv • Calculate the momentum of a 65-kg sprinter running east at 10 m/s. • Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s. Px Now need direction because momentum is a vector System: 2 runners Py θ θ= Tan-1 (O/A) θ= Tan-1 (ΣPx / ΣPx) p = mv θ= Tan-1 (mxvx / myvy) P1 + P2 ≠ P θ= Tan-1 (65kg · 10m/s/ 75kg · 9.5m/s ) = 42.37º E of N**Like any change, change in momentum is calculated by looking**at final and initial momentums. • Δp = pf – pi • Δp: change in momentum • pf: final momentum • pi: initial momentum • Using only a meter stick, find the momentum change of each ball when it strikes the desk from a height of exactly one meter. • Which ball, tennis or golf, has the greatest change in momentum?**In which case is the magnitude of the momentum change**greatest?**Impulse is the product of an external force and time, which**results in a change in momentum of a particle or system. • J = F t and J = ΔP • Therefore Ft = ΔP • Units: N s or kg m/s (same as momentum)**To increase momentum: exert more force or exert force for a**longer period of time • Ex: follow through in sports – tennis, hockey, etc. • To decrease momentum: force must be in opposite direction of motion (negative) • Ex: friction, air resistance, tree slowing down a car • Changes in momentum caused by cushy objects have less force and more time than those caused by hard items. • Double time, cut force in half**Ex: Jump off a 3 meter platform onto concrete – time to**slow down your motion is very small once you contact the cement so the force is great. • Jump off a 3 meter platform onto a thick mat – time to slow down your motion is extended because the matt gives so the force is lessened. • Ex: Car crumple zones increase the time of impact so that less force is exerted causing fewer injuries to passengers. • Ex: bending legs when landing, moving away from a punch, stretchy cord when bungee jumping**Usually high magnitude, short duration.**Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the magnitude of the momentum change? What is the change in the magnitude of the momentum?**F(N)**Impulse on a graph 4000 3000 Area under curve is the impulse 2000 1000 0 t(ms) 1 2 3 4 5 0**Vf=Vi + at**• Suppose a 1.5-kg brick is dropped on a glass table top from a height of 20 cm. • What is the magnitude and direction of the impulse necessary to stop the brick? • If the table top doesn’t shatter, and stops the brick in 0.01 s, what is the average force it exerts on the brick? • What is the average force that the brick exerts on the table top during this period? J = ΔP d = 0.5at2 t= √(2·0.2m/9.8m/s2) = 0.202s Vf=0m/s + (-9.8 m/s2· 0.202s) = -1.98 m/s t= √(2d/a) J = 1.5kg·0m/s – (-1.5kg· 1.98m/s)= +2.97 kg· m/s J = Pf - Pi J = mvf – mvf Ft = ΔP F= ΔP/t F= -2.97 kg· m/s / 0.01s = +296.94 N F= 2.97 kg· m/s / 0.01s = -296.94 N**F(N)**3000 2000 1000 0 0 0.2 0.4 0.6 0.8 t(s) This force acts on a 1.2 kg object moving at 120.0 m/s. The direction of the force is aligned with the velocity. What is the new velocity of the object?**F = 2500 N m = 1.2 kg t = 0.1s vi = 120m/s vf = ?**Ft = mv v = Ft/m v = 2500N· 0.1s/1.2Kg • A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man begins to move forward in the canoe at 0.50 m/s relative to the shore, what happens to the canoe? v = 208.3 m/s + 120m/s =328.3 m/s v = 208.3 m/s System After Before Vperson Vboat**mman = 75kg mcanoe = 120kg vman = 0.5m/s vcanoe = ?**ΣPi = ΣPf 0=vcanoemcanoe + vmanmman vcanoe = - vmanmman / mcanoe • External forces:forces coming from outside the system of particles whose momentum is being considered. External forces change the momentum of the system. • Internal forces:forces arising from interaction of articles within a system. Internal forces cannot change momentum of the system. vcanoe = - 0.5 m/s· 75 kg/ 120kg = - 0.3215 m/s**EXTERNAL FORCE EXAMPLE**System Pi ≠Pf • The club head exerts an external impulsive force on the ball and changes its momentum. • The acceleration of the ball is greater because its mass is smaller.**INTERNAL FORCE EXAMPLE**• The forces the balls exert on each other are internal and do not change the momentum of the system. • Since the balls have equal masses, the magnitude of their accelerations is equal. Pi = Pf System**Explosions**• When an object separates suddenly, as in an explosion, all forces are internal. • Momentum is therefore conserved in an explosion. • There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion. • Recoil • Guns and cannons “recoil” when fired. • This means the gun or cannon must move backward as it propels the projectile forward. • The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.**Collision**• When two moving objects make contact with each other, they undergo a collision. • Conservation of momentum is used to analyze all collisions. • Newton’s Third Law is also useful. It tells us that the force exerted by body A on body B in a collision is equal and opposite to the force exerted on body B by body A.**Collisions Continued**• During a collision, external forces are ignored. • The time frame of the collision is very short. • The forces are impulsive forces (high force, short duration). System**Collision Types**• Elastic collisions • Also called “hard” collisions • No deformation occurs, no kinetic energy lost • Inelastic collisions • Deformation occurs, kinetic energy is lost • Perfectly Inelastic (stick together) • Objects stick together and become one object • Deformation occurs, kinetic energy is lost**(Perfectly) Inelastic Collision**• Simplest type of collisions. • After the collision, there is only one velocity, since there is only one object. • Kinetic energy is lost. • Explosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained!**An 80-kg roller skating grandma collides inelastically with**a 40-kg kid. • What is their velocity after the collision? • How much kinetic energy is lost? System (Grand mother and child • Example Problem Inelastic collision (stick together) a. No external force = 0 ΣPi = ΣPf vf = (80 kg · 6 m/s+ 40 kg ·0m/s)/ (80 kg + 40 kg) = 4 m/s m1iv1i + m2iv2i = m1fv1f + m2fv2f m1iv1i + m2iv2i = (m1f + m2f)vf vf = m1iv1i + m2iv2i / (m1f + m2f)**An 80-kg roller skating grandma collides inelastically with**a 40-kg kid. • What is their velocity after the collision? • How much kinetic energy is lost? System (Grand mother and child • Example Problem Inelastic collision (stick together) b. ΔK = Kf - Ki ΔK = 1/2mfv2f – 1/2miv2i ΔK = 1/2(m1+m2)fv2f – 1/2mi1v2i ΔK = 1/2(80 kg+40 kg)· (4 m/s)2 – 1/2· 80 kg· (6 m/s)2 =-480J**A car with a mass of 950 kg and a speed of 16 m/s to the**east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? C Collide V**A car with a mass of 950 kg and a speed of 16 m/s to the**east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? P = √((mxvx)2 + (myvy)2) ΣPi = Σ Pf P = √((1300kg · 21 m/s)2 + (950kg · 16 m/s )2) = 31246.28 Ns Sum your momentum in the x and y direction mcvc + mvvv = v(mc+mv) P Px v = (mcvc + mvvv) / (mc+mv) (mcvc + mvvv) = P = 31248.28 Ns Py v = (mcvc + mvvv) / (mc+mv) v = P / (mc+mv) P2 = Px2 + Py2 P = √(Px2 + Py2) v = 31248.28 Ns / (950 kg+1300 kg) = 13.88 m/s**A car with a mass of 950 kg and a speed of 16 m/s to the**east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? Solution 2 v vy = 12.13333 m/s ΣPi = Σ Pf ΣPix = Σ Pfx ΣPiy = Σ Pfy vx = 6.7556 m/s mcvc = (mc+mv)v mvvv = (mc+mv)v v2 = vx2 + vy2 vx = mcvc / (mc+mv) vy = mvvv / (mc+mv) v = √(vx2 + vy2) v = √((6.7556 m/s)2 + (12.13333 m/s)2) = 13.89 m/s vx = 950 Kg · 16 m/s / (950 kg + 1300 kg) = 6.7556m/s vy = 1300 Kg · 21 m/s / (950 kg + 1300 kg) = 12.1333 m/s**A car with a mass of 950 kg and a speed of 16 m/s to the**east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? Solution 2 v vy = 12.13333 m/s P Px = myvy vx = 6.7556 m/s Py = mxvx Tan θ= O/A Tan θ= O/A θ= Tan -1 (O/A) θ= Tan -1 (O/A) θ= Tan -1 (myvy / mxvx) θ= Tan -1 (12.13333 m/s / 6.7556 m/s) = 60.89º θ= Tan -1 (1300 kg· 21 m/s/ 950 kg · 16 m/s) = 60. 89º**Conservation of Momentum**• For a closed system (no external forces), total momentum remains the same. • Momentum cannot be created or destroyed, but can be transferred from one object to another. • Ex: • Cannon with cannon ball ready to fire has 0 kgm/s of momentum. • Cannon firing –-> cannon exerts force on cannon ball, cannon ball exerts same force back on the cannon.**These forces occur over the same time period.**• Cannon ball and cannon have equal and opposite impulses () – they add up to zero. • Because the cannon has more mass, it accelerates less than the cannon ball. • The acceleration of the cannon is called recoil – it moves backwards because the cannon ball is shot out forward.**ΣPi = Σ Pf**ΣPi = 0 Σ Pf = mlavla + mpvp 0= mlavla + mpvp mlavla = - mpvp • Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher? • An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity? vla = - mpvp / mla vla = - 0.209 kg · 350 m/s / 5.0kg = 14.63 m/s ΣPx = mxvx Explosion ΣPy = myvy**An exploding object breaks into three fragments. A 2.0 kg**fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity? P Tan θ= O/A Px = myvy θ= Tan -1 (O/A) θ= Tan -1 (myvy / mxvx) Py = mxvx θ= Tan -1 (2.0 kg· 200 m/s/ 4.0 kg · 100 m/s) = P2 = Px2 + Py2 P3 frag = 565.69 Ns P = √(Px2 + Py2) 45º P = √((mxvx)2 + (myvy)2) P = √((4 kg · 100 m/s)2 + (2.0 kg · 200 m/s)2) = 565.69 Ns P3 frag = m3v3 v3 = P3 frag / m3 v3 = 565.69 Ns / 3.0 kg = 188.56 m/s**In elastic collisions, there is no deformation of colliding**objects, and no change in kinetic energy of the system. Therefore, two basic equations must hold for all elastic collisions • ΣPb = Σ Pa (momentum conservation) • Σ Kb = Σ Ka (kinetic energy conservation • A 500-g cart moving at 2.0 m/s on an air track elastically strikes a 1,000-g cart at rest. If the 500 g cart has a velocity of 1 m/s after the collision, what are the resulting velocities of the two carts? m2av2a = m1bv1b - m1av1a ΣPb = Σ Pa v2a = (m1bv1b - m1av1a )/ m2a m1bv1b + m2bv2b = m1av1a + m2av2a v2a = (0.5 kg· 2 m/s – 0.5 kg· 1 m/s)/ 1 kg = 0.5 m/s V2b = 0 m/s m1bv1b = m1av1a + m2av2a**2 – D collisions**• Momentum in the x-direction is conserved. • ΣPx (before) = ΣPx (after) • Momentum in the y-direction is conserved. • ΣPy (before) = ΣPy (after) • Treat x and y coordinates independently. • Ignore x when calculating y • Ignore y when calculating x • Let’s look at a simulation: • http://surendranath.tripod.com/Applets.html**2 m/s**y y 3 m/s 2 kg 8 kg • Calculate velocity of 8-kg ball after the collision. θ1f= 50º x x θ2f= ? 2 kg 0 m/s v2f = ? 8 kg Break down velocity of ball 1 in x and y components v = 2 m/s vy1 = vsin θ vy1 =2m/s sin 50º = 1.53 m/s θ= 50º vx1 = vcos θ vx1 =2m/s cos 50º = 1.29 m/s**2 m/s**y y 3 m/s 2 kg 8 kg • Calculate velocity of 8-kg ball after the collision. θ1f= 50º x x θ2f= ? 2 kg 0 m/s v2f = ? 8 kg vy1 = 1.53 m/s vx1 = 1.29 m/s Determine v2yf Determine v2xf m1bv1b + m2bv2b = m1av1a + m2av2a m1bv1b + m2bv2b = m1av1a + m2av2a V2b = 0m/s V1b = 0m/s V2b = 0m/s v2a = (- m1av1a ) / m2a v2a = (m1bv1b - m1av1a ) / m2a v2a = (– 2 kg ·1.53 m/s ) / 8 kg = - 0.3825 m/s v2a = (2 kg·3 m/s – 2 kg ·1.29 m/s ) / 8 kg = 0.4275 m/s**2 m/s**y y 3 m/s 2 kg 8 kg • Calculate velocity of 8-kg ball after the collision. θ1f= 50º x x θ2f= ? 2 kg 0 m/s v2f = ? 8 kg v2x = 0.4275 m/s θ= ? v = ? v2y = 0.3825 m/s Tan θ = O/A θ = Tan -1 (v2y/v2x) v2 = v2x2 + v2y2 θ = Tan -1 (0.3825 m/s / 0.4275 m/s) = 41.82º v =√( v2x2 + v2y2 ) v =√( (0.4275 m/s)2 + (0.3825 m/s)2 ) = 0.574 m/s