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Chapter 9: Impulse, Momentum, and Collisions  Up to now we have considered forces which have a constant value (except the spring) throughout the motion.

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Presentation on theme: "Chapter 9: Impulse, Momentum, and Collisions  Up to now we have considered forces which have a constant value (except the spring) throughout the motion."— Presentation transcript:

1 Chapter 9: Impulse, Momentum, and Collisions  Up to now we have considered forces which have a constant value (except the spring) throughout the motion and no explicit time duration  Now, lets consider a force which has a time duration (usually short) and with a magnitude that may vary with time – examples: a bat hitting a baseball, a car crash, a asteroid or comet striking the Earth, etc.  It is difficult to deal with a time-varying force, so we might take the mean value F t t0t0 tftf tt

2  Define a new quantity by multiplying the force by the time duration - a vector, points in the same direction as the force - has units of N s  Define another quantity, but which gives a measure of the motion (similar to K) - a vector, points in same direction as the velocity - units of kg m/s = N s  Linear momentum and K are related

3 Example A car of mass 760 kg is traveling east at a speed of 10.0 m/s. The car hits a wall and rebounds (moving west) with a speed of 0.100 m/s. Determine its momentum and K before and after the impact. Determine the impulse. Solution: Given: m = 750 kg,

4  Now, from the definition of acceleration and Newton’s 2 nd Law:

5 Impulse-Momentum Theorem  The Impulse-Momentum Theorem says that if an impulse (force*time duration) is applied to an object, its momentum changes  In this example, the impact of the car with the wall applies an impulse to the car  car’s p changes

6 Example Problem A 0.500-kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor? 0 12 3 y Solution: Given: m = 0.500 kg, h 0 =1.20 m, h 3 =0.7 m, h 1 =h 2 =0

7 Method: need momentum before and after impact  need velocities  use conservation of energy Conservation of mechanical energy is satisfied between 0 and 1 and between 2 and 3, but not between 1 and 2

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9 Collisions  Involves two (or more) objects which may have their motion (velocity, momentum) altered by collisions  These concepts are applicable to the collisions of atoms, billiard balls, cars, planetary objects, galaxies, etc.  Say, we have a collection of interacting particles numbered 1, 2, 3, … We can define the Total Momentum of the system (all the particles) as just the sum of all the individual momenta

10  Imagine that these particles interact in some way – collide and scatter  As long as there are no net external forces acting on the system (collection of objects), the Total Linear Momentum does not change  Which means the Total Linear Momentum is the same before the collision, during the collision, and after the collision  Conservation of Linear Momentum


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