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Linear momentum and Collisions

Linear momentum and Collisions. Chapter 9. The center of mass - System of particles / - Solid body Newton’s Second law for a system of particles Linear Momentum - System of particles / - Conservation IV. Collision and impulse

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Linear momentum and Collisions

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  1. Linear momentum and Collisions Chapter 9

  2. The center of mass • - System of particles / - Solid body • Newton’s Second law for a system of particles • Linear Momentum • - System of particles / - Conservation • IV. Collision and impulse • - Single collision / - Series of collisions • Momentum and kinetic energy in collisions • VI. Inelastic collisions in 1D -Completely inelastic collision/ Velocity of COM Center of mass and linear momentum

  3. VII.Elastic collisions in 1D • VIII. Collisions in 2D • IX. Systems with varying mass • X. External forces and internal energy changes

  4. I. Center of mass The center of mass of a body or a system of bodies is the point that moves as though all the mass were concentrated there and all external forces were applied there.

  5. System of particles: Two particles of massesm1 andm2separated by a distanced Origin of reference system coincides withm1

  6. System of particles: Choice of the reference origin is arbitrary Shift of the coordinate system but center of mass is still at the same distance from each particle. The center of mass lies somewhere between the two particles. General: M= total mass of the system

  7. System of particles: We can extend this equation to a general situation fornparticles that strung along x-axis. The total mass of the systemM=m1+m2+m3+……+mnThe location of center of the mass: 3D:

  8. System of particles: 3D: The vector form Position of the particle: Position COM: M= mass of the object

  9. Solid bodies: Continuous distribution of matter. Particles = dm (differential mass elements). 3D: M= mass of the object Assumption: Uniform objects  uniform density

  10. The center of mass of an object with a point, line or plane of symmetry lies on that point, line or plane. The center of mass of an object does not need to lie within the object (Examples: doughnut, horseshoe )

  11. Problem solving tactics: • (1) Use object’s symmetry. • (2) If possible, divide object in several parts. Treat each of these parts as a particle located at its own center of mass. • Chose your axes wisely. Use one particle of the system as origin of your reference system or let the symmetry lines be your axis.

  12. A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106. If the bonds are 0.100 nm long, where is the center of mass of the molecule?

  13. y A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106. If the bonds are 0.100 nm long, where is the center of mass of the molecule? x

  14. A uniform piece of sheet steel is shaped as in Figure. Compute the xand ycoordinates of the center of mass of the piece.

  15. A3 A uniform piece of sheet steel is shaped as in Figure. Compute the xand ycoordinates of the center of mass of the piece. A2 A1

  16. A3 A uniform piece of sheet steel is shaped as in Figure. Compute the xand ycoordinates of the center of mass of the piece. A2 A1

  17. II. Newton’s second law for a system of particles Motion of the center of mass: Center of the mass of the system moves as a particle whose mass is equal to the total mass of the system. Fnetis the net of all external forces that act on the system. Internal forces (from one part of the system to another) are not included. The system is closed: no mass enters or leaves the system during the movement. (M=total system mass). acomis the acceleration of the system’s center of mass.

  18. Prove: (*) includes forces that the particles of the system exert on each other (internal forces) and forces exerted on the particles from outside the system (external). Newton’s third law internal forces from third-law force pairs cancel out in the sum (*) Only external forces.

  19. Linear Momentum The linear momentum of a particle or an object that can be modeled as a particle of mass mmoving with a velocityv is defined to be the product of the mass and velocity: p = mv The terms momentum and linear momentum will be used interchangeably in the text

  20. III. Linear momentum The linear momentum of a particle is a vectorp defined as: Momentum is a vector with magnitude equalmvand have direction of v. SI unitof the momentum iskg-meter/second • Momentum can be expressed in component form: • px = m vxpy= m vypz = m vz

  21. Newton called the product mv the quantity of motion of the particle Newton II law in terms of momentum: The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of the force.

  22. System of particles: The total linear momentPis the vector sum of the individual particle’s linear momentum. The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. Net external force acting on the system.

  23. Conservation: If no external force acts on a closed, isolated system of particles, the total linear momentumP of the system cannot change. Closed:no matter passes through the system boundary in any direction.

  24. Conservation of Linear Momentum If no net external force acts on the system of particles the total linear momentumPof the system cannot change. Each component of the linear momentum is conserved separately if the corresponding component of the net external force is zero. If the component of the net external force on a closed system is zero along an axis  component of the linear momentum along that axis cannot change. The momentum is constant if no external forces act on a closed particle system. internal forces can change the linear momentum of portions of the system, but they cannot change the total linear momentum of the entire system.

  25. IV. Collision and impulse Collision:isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time. Impulse: Measures the strength and duration of the collision force Third law force pair FR = - FL Single collision

  26. Impulse-linear momentum theorem The change in the linear momentum of a body in a collision is equal to the impulse that acts on that body. Units: kg m/s Favgsuch that: Area underF(t)vsΔtcurve = Area under Favgvs t

  27. An estimated force-time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.

  28. (a) An estimated force-time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball. area under curve (b) (c)

  29. V. Momentum and kinetic energy in collisions Assumptions:Closed systems (no mass enters or leaves them) Isolated systems (no external forces act on the bodies within the system) Elastic collision: If the total kinetic energy of the system of two colliding bodies is unchanged (conserved) by the collision. Example:Ball into hard floor. Inelastic collision: The kinetic energy of the system is not conserved  some goes into thermal energy, sound, etc. After the collision the bodies lose energy and stick together. Completely inelastic collision: Example:Ball of wet putty into floor

  30. Velocity of the center of mass: In a closed, isolated system, the velocity of COM of the system cannot be changed by a collision. (No net external force). Completely inelastic collision V = vcom

  31. VII. Elastic collisions in 1D In an elastic collision, the kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change. Stationary target: Closed, isolated system  Linear momentum Kinetic energy

  32. Stationary target: v2f>0 always v1f>0 if m1>m2  forward mov. v1f<0 if m1<m2  rebounds Equal masses:m1=m2  v1f=0 and v2f = v1i In head-on collisions bodies of equal masses simply exchange velocities.

  33. Massive target:m2>>m1 v1f ≈ -v1iand v2f ≈ (2m1/m2)v1i Body 1 bounces back with approximately same speed. Body 2 moves forward at low speed. Massive projectile:m1>>m2 v1f ≈ v1iand v2f ≈ 2v1i Body 1 keeps on going scarcely lowed by the collision. Body 2 charges ahead at twice the initial speed of the projectile.

  34. VII. Elastic collisions in 1D Moving target: Closed, isolated system  Linear momentum Kinetic energy

  35. VIII. Collisions in 2D Closed, isolated system  Linear momentum conserved Elastic collision  Kinetic energy conserved Example: If the collision is elastic 

  36. Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3Mmoves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg.

  37. Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3Mmoves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg. (a) (motion toward the left).

  38. Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3Mmoves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg. (b)

  39. A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball?

  40. A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball? Assume the initial direction of the ball in the –x direction. (a) (b) W

  41. Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

  42. Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision. - speed of at B before collision.

  43. Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision. , speed of at B just after collision.

  44. As shown in Figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of lengthℓand negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

  45. As shown in Figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of lengthℓand negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest.

  46. As shown in Figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of lengthℓand negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Momentum of the bob-bullet system is conserved in the collision:

  47. A small block of mass m1 = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure (a). When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height hof the wedge?

  48. A small block of mass m1 = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure (a). When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height hof the wedge? (a) The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have:

  49. A small block of mass m1 = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure (a). When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height hof the wedge? (b) Using conservation of energy for the block-wedge-Earth system as the block slides down the smooth (frictionless) wedge, we have:

  50. IV. Systems with varying mass Rocket Propulsion The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel. The initial mass of the rocket plus all its fuel is M + Δm at time tiand velocity v The initial momentum of the system is pi = (M + Δm) v

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