5Sample ProblemCalculate the momentum of a 65-kg sprinter running east at 10 m/s.
6Sample ProblemCalculate 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.
7Change in momentumLike any change, change in momentum is calculated by looking at final and initial momenta.Dp = pf – piDp: change in momentumpf: final momentumpi: initial momentum
8In which case is the change in momentum greatest?
9Momentum change demonstration 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, Bouncy or Lazy, has the greatest change in momentum?Please document the activity in a lab report.(mass = kg; mass = kg)
10Impulse (J)Impulse (J) is the product of an external force and time, which results in a change in momentum of a particle or system.J = F tJ = PUnits: N s or kg m/s (same as momentum)
11Impulsive Forces Usually high magnitude, short duration. Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the momentum change?What is the impulse on the ball from the bat?
12area under curve Impulse (J) on a graph F(N) t (ms) 3000 2000 1000 1 2 1234t (ms)
13Sample ProblemSuppose 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?
14Solution a)Find the velocity of the brick when it strikes the table using conservation of energy.mgh = ½ mv2v = √(2gh) = √(2*9.8 m/s2*0.20 m) = 2.0 m/sCalculate the brick’s momentum when it strikes the table.p = mv = (1.5 kg)(2.0 m/s) = 3.0 kg m/s (down)The impulse necessary to stop the brick is the impulse necessary to change to momentum to zero.J = Dp = pf – pi = 0 – 3.0 kg m/s = -3.0 kg m/sor 3.0 kg m/s (up)
15Solution b) and c)b) Find the force using the other equation for impulse.J = Ft3.0 N s = F (0.01 s)F = 300 N (upward in the same direction as impulse)c) According the Newton’s 3rd law, the brick exerts an average force of 300 N downward on the table.
16Sample Problem F(N) 2,000 1,000 0.20 0.40 0.60 0.80 t(s) This force acts on a 1.2 kg object moving at m/s. The direction of the force is aligned with the velocity. What is the new velocity of the object?
17Solution Find the impulse from the area under the curve. A = ½ base * height = ½ (.1 s)(2500 N) = 125 NsJ = 125 N sSince impulse is equal to change in momentum and it is in the same direction as the existing momentum, the momentum increases by 125 kg m/s.Dp = 125 kg m/sDp = pf - pi = mvf - mvimvf = mvi + Dp= (1.2 kg)(120 m/s) kg m/s = 269 kg m/svf = (269 kg m /s) / (1.2 kg) = 224 m/s
19Law of Conservation of Momentum If the resultant external force on a system is zero, then the vector sum of the momenta of the objects will remain constant.SPbefore = SPafter
20Sample problemA 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?
21SolutionThe momentum before the man moves is equal to the momentum after the man moves.Spb = Spa0 = mmvm + mcvc0 = (75 kg)(0.50 m/s) + (120 kg)vv = - (75 kg)(0.50 m/s)/(120 kg)v = m/sThe canoe slips backward in the water at m/s
22External versus internal forces 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 particles within a system.Internal forces cannot change momentum of the system.
23An external force in golf Consider the collision between the club head and the golf ball in the sport of golf.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.
24An internal force in pool Consider the collision between two balls in pool.The forces they 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 acceleration is equal.
25ExplosionsWhen 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.
26Recoil 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.
27Sample problemSuppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?
28Solution Momentum conservation is used to calculate recoil speed. Spb = Spa0 = mpvp + mlvl0 = (0.209 kg)(350 m/s) + (5.0 kg)vv = - (0.209 kg)(350 m/s)/(5.0 kg)v = m/s
29Sample ProblemAn 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?
30Solution v = p/m = 566/3 400 kg m/s = 189 m/s due SW (4002 + 4002)1/2 The momentum before is zero, so the momentum after is zero.This is a vector addition problem. Each fragment has a momentum magnitude of 400 kg m/s according to the formula p = mv.v = p/m= 566/3= 189 m/sdue SW400 kg m/s400 kg m/s( )1/2566 kg m/sdue southwest400 kg m/s
31Sample problemAn 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?
33CollisionsWhen 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.
34Collisions 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).
35Collision Types Elastic collisions Inelastic collisions “Elastic” means ‘returns to pre-collision state.’No deformation occurs, no kinetic energy lostInelastic collisionsDeformation occurs, kinetic energy is lostPerfectly Inelastic (stick together)Objects stick together and become one objectkinetic energy is lost
36Collisions Inelastic Elastic Objects collide and move separately after the collision.Energy is conserved.Momentum is conserved.InelasticThe collision results in deformation of the objects.Perfectly InelasticTwo objects collide and stick together such that their final velocities are the same after the collision.No deformation occurs.
37ExplosionsExplosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained!
38Sample ProblemAn 80-kg roller skating grandma collides inelastically with a 40-kg kid. What is their velocity after the collision?
40Sample ProblemA train of mass 4m moving 5 km/hr couples with a flatcar of mass m at rest. What is the velocity of the cars after they couple?
41Sample ProblemA fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner?
42Sample ProblemA 500-g cart on an air track strikes a 1,000-g cart at rest. What are the resulting velocities of the two carts? (Assume the collision is elastic, and the first cart is moving at 2.0 m/s when the collision occurs.)
43Solution before after m1v1 = m1v1 + m2v2 1.0 = 0.50v1 + v2 Solve simultaneouslyv1 = m/sv2 = 1.33 m/s
44Sample ProblemSuppose three equally strong, equally massive astronauts decide to play a game as follows: The first astronaut throws the second astronaut towards the third astronaut and the game begins. Describe the motion of the astronauts as the game proceeds. Assume each toss results from the same-sized "push." How long will the game last?
462D-Collisions Momentum in the x-direction is conserved. SPx (before) = SPx (after)Momentum in the y-direction is conserved.SPy (before) = SPy (after)Treat x and y coordinates independently.Ignore x when calculating yIgnore y when calculating x
47Sample problem Calculate velocity of 8-kg ball after the collision. 2 m/sy2 kgy3 m/s50oxx2 kg8 kg0 m/s8 kgvAfterBefore