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# TEST 1

TEST 1. Tuesday May 19 9:30 am CNH-104 Kinematics, Dynamics &amp; Momentum. Momentum. Newton’s original “quantity of motion” a conserved quantity a vector. Newton’s Second Law in another form Momentum and Momentum Conservation. Télécharger la présentation ## TEST 1

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1. TEST 1 Tuesday May 19 9:30 amCNH-104Kinematics, Dynamics & Momentum Physics 1D03 - Lecture 25

2. Momentum • Newton’s original “quantity of motion” • a conserved quantity • a vector • Newton’s Second Law in another form • Momentum and Momentum Conservation Physics 1D03 - Lecture 25

3. Definition: The linear momentump of a particle is its mass times its velocity: pmv Momentum is a vector, since velocity is a vector. Units: kg m/s (no special name). (We say “linear” momentum to distinguish it from angular momentum, a different physical quantity.) Physics 1D03 - Lecture 25

4. Example 1 & 2 • A car of mass 1500kg is moving with a velocity of 72km/h. What is its momentum ? • Two cars, one of mass 1000kg is moving at 36km/h and the other of mass 1200kg is moving at 72km/h in the opposite direction. What is the momentum of the system ? Physics 1D03 - Lecture 25

5. Concept Quiz Which object will have the smallest momentum ? • A 1.6x10-19kg particle moving at 1x103m/s • A 1000kg car moving at 2m/s • A 100kg person moving at 10m/s • A 5000kg truck at rest Physics 1D03 - Lecture 25

6. Newton’s Second Law If mass is constant, then the rate of change of (mv) is equal to m times the rate of change of v. We can rewrite Newton’s Second Law: or Net external force = rate of change of momentum This is how Newton wrote the Second Law. It remains true in cases where the mass is not constant. Physics 1D03 - Lecture 25

7. v F Example 3 Rain is falling vertically into an open railroad car which moves along a horizontal track at a constant speed. The engine must exert an extra force on the car as the water collects in it (the water is initially stationary, and must be brought up to the speed of the train). Calculate this extra force if: v = 20 m/s The water collects in the car at the rate of 6 kg per minute Physics 1D03 - Lecture 25

8. v F Solution Plan: The momentum of the car increases as it gains mass (water). Use Newton’s second law to find F. v is constant, and dm/dt is 6 kg/min or 0.1 kg/s (change to SI units!), so F = (0.1 kg/s) (20 m/s) = 2.0 N F and p are vectors; we get the horizontal force from the rate of increase of the horizontal component of momentum. Physics 1D03 - Lecture 25

9. The total momentum of a system of particles is the vector sum of the momenta of the individual particles: ptotal = p1 + p2 + ... = m1v1 + m2v2 + ... Since we are adding vectors, we can break this up into components so that: px,Tot = p1x + p2x + ….Etc. Physics 1D03 - Lecture 25

10. Example 4 • A particle of mass 2kg is moving with a velocity of 5m/s and an angle of 45o to the horizontal. Determine the components of its momentum. Physics 1D03 - Lecture 25

11. F12 m1 F21 = -F12 m2 Newton’s 3rd Law and Momentum Conservation Two particles interact: Dp1= F12Dt Dp2= F21Dt Newton’s 3rd Law: F21= -F12 The momentum changes are equal and opposite; the total momentum: p = p1 + p2=0 doesn’t change. The fine print:Only internal forces act. External forces would transfer momentum into or out of the system. eg: particles moving through a cloud of gas Physics 1D03 - Lecture 25

12. Conservation of momentum simply says that the initial and final momenta are equal:pi=pf Since momentum is a vector, we can also express it in terms of the components. These are independently conserved: pix=pfx piy=pfy piz=pfz Physics 1D03 - Lecture 25

13. Before After Concept Quiz A subatomic particle may decay into two (or more) different particles. If the total momentum before is zero before the decay before, what is the total after? • 0 kg m/s • depends on the masses • depends on the final velocities • depends on b) and c) together Physics 1D03 - Lecture 25

14. 10 min rest Physics 1D03 - Lecture 25

15. Collisions • Conservation of Momentum • Elastic and inelastic collisions Physics 1D03 - Lecture 25

16. v1,i v2,i m1 m2 F1 F2 = -F1 v1,f v2,f Collisions A collision is a brief interaction between two (or more) objects. We use the word “collision” when the interaction time Δt is short relative to the rest of the motion. During a collision, the objects exert equal and opposite forces on each other. We assume these “internal” forces are much larger than any external forces on the system. We can ignore external forces if we compare velocities just before and just after the collision, and if the interaction force is much larger than any external force. Physics 1D03 - Lecture 25

17. Elastic and Inelastic Collisions Momentum is conserved in collisions. Kinetic energy is sometimes conserved; it depends on the nature of the interaction force. A collision is called elastic if the total kinetic energy is the same before and after the collision. If the interaction force is conservative, a collision between particles will be elastic (eg: billiard balls). If kinetic energy is lost (converted to other forms of energy), the collision is called inelastic (eg: tennis ball and a wall). A completely inelastic collision is one in which the two colliding objects stick together after the collision (eg: alien slime and a spaceship). Kinetic energy is lost in this collision. Physics 1D03 - Lecture 25

18. v1,i v2,i m1 m2 v1,f v2,f If there are no external forces, then the total momentum is conserved: p1,i + p2,i = p1,f + p2,f This is a vector equation. It applies to each component of p separately. Physics 1D03 - Lecture 25

19. Elastic Collisions In one dimension (all motion along the x-axis): 1) Momentum is conserved: In one dimension, the velocities are represented by positive or negative numbers to indicate direction. 2) Kinetic Energy is conserved: We can solve for two variables if the other four are known. Physics 1D03 - Lecture 25

20. One useful result: for elastic collisions, the magnitude of the relative velocity is the same before and after the collision: |v1,i – v2,i | = |v1,f – v2,f | (This is true for elastic collisions in 2 and 3 dimensions as well). • An important case is a particle directed at a stationary target (v2,i = 0): • Equal masses: If m1 = m2, then v1,f will be zero (1-D). • If m1 < m2, then the incident particle recoils in the opposite direction. • If m1 > m2, then both particles will move “forward” after the collision. before after Physics 1D03 - Lecture 25

21. v1 -v1 v1 2v1 v1 Elastic collisions, stationary target (v2,i = 0): • Two limiting cases: • If m1 << m2 , the incident particle rebounds with nearly its original speed. 2) If m1 >> m2 , the target particle moves away with (nearly) twice the original speed of the incident particle. Physics 1D03 - Lecture 25

22. Concept Quiz A tennis ball is placed on top of a basketball and both are dropped. The basketball hits the ground at speed v0. What is the maximum speed at which the tennis ball can bounce upward from the basketball? (For “maximum” speed, assume the basketball is much more massive than the tennis ball, and both are elastic). • v0 • 2v0 • 3v0 ? v0 v0 Physics 1D03 - Lecture 25

23. Example An angry 60.0 kg physicist standing on a frozen lake throws a 0.5 kg stone to the east with a speed of 24.0 m/s. Find the recoil velocity of the angry physicist. Physics 1D03 - Lecture 25

24. Example – inelastic collision: A neutron, with mass m = 1 amu (atomic mass unit), travelling at speed v0, strikes a stationary deuterium nucleus (mass 2 amu), and sticks to it, forming a nucleus of tritium. What is the final speed of the tritium nucleus? Physics 1D03 - Lecture 25

25. An elastic collision: Two carts moving toward each other collide and bounce back. If cart 1 bounces back with v=2m/s, what is the final speed of cart 2 ? 5 m/s 6 m/s 4 kg 2 kg Physics 1D03 - Lecture 25

26. Example A bullet of mass m is shot into a block of mass M that is at rest on the edge of a table. If the bullet embeds in the block, determine the velocity of the bullet if they land a distance of x from the base of the table, which has a height of h. Physics 1D03 - Lecture 25

27. 10 min rest Physics 1D03 - Lecture 25

28. Impulse , or dp = F dt Newton #2: For a constant force, Dp = FDt . The vector quantity FDt is called the Impulse:J = FΔt = Δp (change in p) = (total impulse from external forces) (Newton’s Second Law again) (Extra) In general (force not constant), we integrate: where the integral gives the area under a curve… Physics 1D03 - Lecture 25

29. Area = t t Impulse is the area under the curve. The average force is the constant force which would give the same impulse. The impulse – momentum theorem, J=FΔt=Δp, tells us thatwe do not need to know the details of the force/interaction,we only need to know the area under the curve=integralof the force. Compare with work: W = FDx ; so the work-energy theorem (derived from Newton #2) is DK = FDx. Physics 1D03 - Lecture 25

30. Quiz A 100g rubber and a 100g clay ball are thrown at a wall with equal speed. The rubber ball bounces back while the clay ball sticks to the wall.Which ball exerts a larger impulse on the wall ? • the clay bass b/c it sticks • the rubber ball b/c it bounces • they are equal because the have equal momenta • neither exerts an impulse b/c the wall doesn’t move Physics 1D03 - Lecture 25

31. Example 1 A golf ball is launched with a velocity of 44 m/s. The ball has a mass of 50g. Determine the average force on the ball during the collision with the club, if the collision lasted 0.01 s. Physics 1D03 - Lecture 25

32. Example 2 Use the impulse-momentum theorem to find how long a falling object takes to increase its speed from 5.5m/s to 10.4m/s. Physics 1D03 - Lecture 25

33. Example 3 • A 150 g baseball is thrown with a speed of 20m/s. It is hit straight back toward the pitcher at a speed of 40m/s. The interaction force is shown by the graph:What is the maximum force Fmax that the bat exerts on the ball ? F Fmax t 6ms Physics 1D03 - Lecture 25

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