Presentation on theme: "Chapter 9. Impulse and Momentum"— Presentation transcript:
1 Chapter 9. Impulse and Momentum Chapter Goal: To introduce the ideas of impulse and momentum and to learn a new problem-solving strategy based on conservation laws.
2 Ch. 9 Student Learning Objectives • To understand interactions from the new perspective of impulse and momentum.• To learn what is meant by an isolated system.• To apply conservation of momentum in simple situations.• To understand the basic ideas of inelastic collisions, explosions, and recoil.
3 Impulsive Forces and Newton’s 2nd Law An impulsive force can be defined as a large force exerted during a small interval of time.Collisions and explosions are examples of impulsive forces.Until now, we’ve been using Newton’s Law (and kinematic equations) for constant forces and accelerations.Impulsive forces are not constant.
4 Impulsive Forces and Newton’s 2nd Law Fnet-x = maxThis law applies to non-constant forces as wellFx (t) = max(t) so Fx (t) = m dvx /dtmultiplying through by dt:Fx (t) dt = m dvx. Now take the integral of both sides:=
5 Impulsive Forces and Newton’s 2nd Law This can be written as:= mvf – mviwhich is how Newton actually wrote it in the first place.You can solve for the velocity of an object, even if the acceleration is not constant.
6 Impulse The impulse upon a particle is defined as: Impulse has units of N-sNewton’s Law tells us that the impulse exerted on an object is equal to ∆ mvImpulsive forces are often short! Time is sometimes given in units of ms or µs.
7 Impulse Problem (EOC #5) What value of Fmax gives an impulse of 6 N-s?.75 N1.5 N750 N1500 N
8 Impulse Momentum Theorem (an alternate version of Newton’s 2nd Law) Jx = ∆pxwherechange in momentum = ∆px = mfvf – miviMomentum and change in momentum are vector quantities.
9 The cart’s change of momentum is 30 kg m/s.10 kg m/s.–10 kg m/s.–20 kg m/s.–30 kg m/s.STT9.1Answer: A
10 They exert equal impulses because they have equal momenta. A 10 g rubber ball and a 10 g clay ball are thrown at a wall and have equal impact speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall?They exert equal impulses because they have equal momenta.The clay ball exerts a larger impulse because it gives all its momentum to the wall.Neither exerts an impulse on the wall because the wall doesn’t change velocity.The rubber ball exerts a larger impulse because it bounces.STT9.2Answer: D
12 Graphical explanation of using average force to replace a non-constant impulsive force
13 A bouncing ball (EOC #29)A g ball is dropped from a height of 2.00 m, bounces and rebounds to a height of 1.50 m. The figure shows the impulse received from the floor. What is Fmax to 3 sig figs?Hint: don’t forget to include FG in your calculations, even though it is not shown in the graph!Normal force of floor on ball
14 Sketch a before, after (and if necessary during) representation to show objects before/after interaction.Coordinate system, symbols for knowns and unknowns.Time usually isn’t important. Position is necessary only if needed to find velocity (as in this case).
15 Answers:viy = m/sv2y = 5.42 m/sFmax = 9.38 x 102 N
16 Impulse approximation The figure shows the force on the ball due to the floor. Is it reasonable to neglect the force on the ball due to the earth when calculating Fmax?Calculate Fmax and compare answersForce of floor on ball
17 Impulse approximation The figure shows the force on the ball due to the floor. Is it reasonable to neglect the force on the ball due to the earth when calculating Fmax?Without including the impulse due to gravity, Fmax = 934 N.If the ball were smaller, it would be easier to ignore.Force of floor on ball
18 Conservation of Momentum The change in momentum of ball 1 is due to the impulse of ball 2 on ball 1.The change in momentum of ball 2 is due to the impulse of ball 1 on ball 2.If we added the impulse of ball 1 on ball 2 to the impulse of ball 2 on ball 1, what would we get? Why?
19 Conservation of Momentum If we added the change in momentum of ball 1 to the change in momentum of ball 2, what would we get?Why?
20 Conservation of Momentum Stated mathematically, the law of conservation of momentum for an isolated system isThe total momentum after an interaction is equal to the total momentum before the interaction.
21 vf is greater than v2, but less than v1. vf = v1. The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick together and continue on with velocity vf. Which of these statements is true?vf = v2.vf is less than v2.vf is greater than v2, but less than v1.vf = v1.vf is greater than v1.STT9.5Answer: C
22 Collisions ProblemA 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity?
23 Collisions Problem - Answer A 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity (include direction)?1.5 m/s west
24 An explosion in a rigid pipe shoots out three pieces An explosion in a rigid pipe shoots out three pieces. A 6 g piece comes out the right end. A 4 g piece comes out the left end with twice the speed of the 6 g piece. From which end does the third piece emerge?Right endLeft endSTT9.4Answer: A
25 Explosion Problem - Radioactivity The nucleus of an atom travels east at 1.0 x 106 m/s. The nucleus undergoes radioactive decay and breaks into two masses the smaller being 1/10th the mass of the original nucleus. The smaller fragment travels east at 2.0 x 107 m/s. Determine the velocity of the larger fragment.
26 Explosion Problem - Radioactivity The nucleus of an atom travels east a 1.0 x 106 m/s. The nucleus undergoes radioactive decay and breaks into two masses the smaller being 1/10th the mass of the original nucleus. The smaller fragment travels east at 2.0 x 107 m/s. Determine the velocity of the larger fragment.Answer: 1.1 x 106 m/s going west.
27 Momentum RaceIdentical constant net forces continuously push blocks A and B from the starting point to the finish. Both blocks start at rest. Block A has 4 times the mass of Block B. Which block has the larger change in momentum?A, since it has a greater mB, since it will have a greater vEqual, the mass and velocity change will make the quantities equal.Too little information given to determine
28 vA > vC vA < vC vA = vC Objects A and C are made of different materials, with different “springiness,” but they have the same mass and are initially at rest. When ball B collides with object A, the ball ends up at rest. When ball B is thrown with the same speed and collides with object C, the ball rebounds to the left. Compare the velocities of A and C after the collisions. Is vA greater than, equal to, or less than vC?vA > vCvA < vCvA = vCSTT9.3Answer: B
29 EOC# 47In a ballistics test, a 25-g bullet traveling horizontally at 1200 m/s goes through a 30.0-cm thick, 350 kg, initially stationary target and emerges with a speed of m/s. The target is free to slide on a frictionless surface. Assuming that the target does not move until after the bullet has passed through, calculate:How long the bullet was in the targetThe average force the bullet exerted on the targetThe target’s speed after the bullet emerges.
30 EOC #472.86 x sFavg = 2.6 x 104 Nv target = .021 m/s (assume impulse approximation and ignore gravitational force on the bullet).
31 EOC #68A 20.0-kg ball hangs on a 2.0-m wire. The maximum tension the wire can withstand without breaking is N. A 1.0-kg projectile, traveling horizontally, hits and embeds itself in the ball. The ball swings up in a circle. What is the largest speed this projectile can have and not break the wire?
32 EOC #68 - AnswerA 20.0-kg ball hangs on a 2.0-m wire. The maximum tension the wire can withstand without breaking is N. A 1.0-kg projectile, traveling horizontally, hits and embeds itself in the ball. What is the largest speed this projectile can have and not break the wire? Vprojectile = 90.3 m/s