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Semester 1 2014 / 2015

CHAPTER 3: LINEAR MOMENTUM & COLLISIONS. Semester 1 2014 / 2015. Nor Ashbahani Bt Mohamad Kajaan ashbahani@unimap.edu.my 012-2746398. Linear Momentum and Collisions. OBJECTIVES. Ability to compute linear momentum and components of momentum

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Semester 1 2014 / 2015

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  1. CHAPTER 3: LINEAR MOMENTUM & COLLISIONS Semester 1 2014 / 2015 Nor Ashbahani Bt Mohamad Kajaan ashbahani@unimap.edu.my 012-2746398

  2. Linear Momentum and Collisions

  3. OBJECTIVES • Ability to compute linear momentum and components of momentum • Ability to relate impulse and momentum & kinetic energy and momentum • Ability to describe and solve the conditions on kinetic energy and momentum in elastic and inelastic collisions.

  4. TOPICS: • Linear Momentum • Impulse • Conservation of Linear Momentum • Elastic and Inelastic Collisions

  5. Linear Momentum • Definition The linear momentum of an object is the product of its mass and velocity • Note that momentum is a vector – it has both a magnitude and a direction. • SI unit of momentum: kg•m/s. This unit has no special name.

  6. Linear Momentum • For a system of objects, the total momentum is the vector sum of each.

  7. Example 1: A 100 kg football player run with a velocity of 4.0 m/s straight down the field. A 1.0 kg artillery shell leaves the barrel of a gun with a muzzle velocity of 500 m/s. Which has the greater momentum (magnitude), the football player or the shell?

  8. Example 2: What is the total momentum for each of the systems of particles illustrated in Fig (a)?

  9. What is the total momentum for each of the systems of particles illustrated in Fig (b)?

  10. Linear Momentum • The change in momentum is the difference between the momentum vectors. • Here, the vector sum is zero, but the vector difference, or change in momentum, is not. (The particles are displaced for convenience.)

  11. Linear Momentum • The change in momentum is found by computing the change in the components.

  12. Linear Momentum • If an object’s momentum changes, a force must have acted on it. • The net force is equal to the rate of change of the momentum.

  13. Impulse • Impulse is the change in momentum: • SI units of impulse : newton-second (N.s)/ 1kgm/s= 1N.s • Dimension of impulse MLT-1 • Impulse is not property of the particle, but is a measure of the change in momentum of the particle

  14. Impulse The area under the average force curve (FavΔt, within the dashed red lines) is the same as the area under the F-versus-t curve, which is usually difficult to evaluate.

  15. Impulse When a moving object stops, its impulse depends only on its change in momentum. This can be accomplished by a large force acting for a short time, or a smaller force acting for a longer time.

  16. Example 3: A golfer drives a 0.1 kg ball from an elevated tee, the ball an initial horizontal speed of 40m/s. The club and the ball are in contact for 1ms. Calculate average force exerted by the club on the ball during his time.

  17. Conservation of Linear Momentum • If there is no net force acting on a system, its total momentum cannot change. • This is the law of conservation of momentum. • If there are internal forces, the momentum of individual parts of the system can change, but the overall momentum stays the same.

  18. Conservation of Linear Momentum • For the linear momentum of an object to be conserved, its follow Newton’s Second Law • If the net force acting on a particle is zero, hence Fnet = ∆p / ∆t = 0 ∆p = 0 = p - p0 • Therefore conservation of linear momentum:

  19. Conservation of Linear Momentum • In this example, there is no external force, but the individual components of the system do change their momentum. • The spring force is an internal force, so the momentum of the system is conserved.

  20. Example 4: Two masses, m1=1kg and m2=2kg, are held on either side of a light compresses spring by a light string joining them (fig in slide 19). The string is burned (neglect external force), and the masses move apart on the frictionless surface, with m1 having a velocity of 1.8m/s to the left. Velocity of m2?

  21. Types of Collisions: Elastic Collisions: • Total kinetic energy is conversed • the total kinetic energy of all the objects of the system after the collision is the same as the total kinetic energy before the collision

  22. Types of Collisions: Inelastic collision: • total kinetic energy is not conversed

  23. Energy and Momentum in Inelastic Collisions • Figure above, m1=m2 and V10=-V20 • Total momentum before collision is zero. • After collisions, the balls are stuck together and stationary, so total momentum is unchanged, still zero.

  24. One ball is initially at rest, and the ball stick together after collisions. • Both balls have same velocity • Assume balls have different mass; • Since the momentum is conserved even in inelastic collisions,

  25. before Lets us consider how much kinetic energy has been lost; after

  26. Substitute v to equation Kf and simplify the results;

  27. (m2 initially at rest, completely inelastic collision only) • In a completely inelastic collision, the maximum amount of kinetic energy is lost, consistent with the conservation of momentum.

  28. Example 5: A 30 g bullet with speed 400 m/s strikes a glancing blow to a target brick of mass 1.0 kg. The brick breaks into two fragments. The bullet deflect at an angle of 30 degree and has a reduced speed of 100m/s. One piece of the brick with mass 0.75 kg goes off to the right or in the initial direction of the bullet with speed 5m/s. • Sketches the situations • Calculate the speed and direction of 2nd piece

  29. Energy and Momentum in Inelastic Collisions • For general elastic collision of two objects, Conservation of kinetic energy Conservation of momentum

  30. Consider this situation, if the ball m2 is stationary, the equation for elastic collision; • Rearrange these equations gives; m1(V1o2 - V12) = m2V22 ……(1) and m1(V1o-V1)=m2V2 …………(2)

  31. Using algebraic relationship x2 + y2 = (x+y) (x-y) and dividing equation (1) by equation (2), we get; m1(V1o +V1) (V1o-V1)/m1(V1o-V1) = m2V22/m2V2 • V1o + V1 = V2 …….(3) • Equation 3 can be used to eliminate V1 or V2 from Equation total momentum

  32. Final velocities for elastic, head on collision with m2 initially stationary

  33. Example 6: A 3.0 kg object with speed of 2.0 m/s in the positive x-direction has head on elastic collision with a stationary 0.70 kg object located at x=0. Calculate the distance separating the objects 2.5 after the collision.

  34. Example 7: A 7.1 kg bowling ball with a speed of 6.0 m/s has head on elastic collision with a stationary 1.6 kg pin. (i) Calculate velocity of each object after collision (ii) Total momentum after collision.

  35. SUMMARY • The linear momentum of an object is the product of its mass and velocity • For a system of objects, the total momentum is the vector sum of each

  36. SUMMARY • Impulse is the change in momentum • conservation of linear momentum • Energy and Momentum in Inelastic Collisions

  37. SUMMARY • Energy and Momentum in Inelastic Collisions

  38. SUMMARY • Final velocities for elastic, head on collision with m2 initially stationary

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