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## 5.1 Impulse and Momentum

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**Chapter 5 Momentum**Ewen et al. 2005) 5.1 Impulse and Momentum • Objective: • Use impulse and momentum in describing motion.**Momentum**• Is a measure of the amount of inertia and motion and object has. • Another way to think about momentum is how difficult it is to bring a moving object to rest. • It is the product of mass and velocity. • Units? • kgm/s and slugft/s • Momentum is a vector quantity.**Example 1**• Find the momentum of an auto with mass 105 slugs traveling 60.0 mi/h.**Example 2**• Find the momentum of an auto with mass 1350 kg traveling 75.0 km/h.**Example 3**• Find the velocity a bullet of mass 1.00 x 10-2 kg would have to have so that it has the same momentum as a lighter bullet of mass 1.80 x 10-3 kg and velocity of 325 m/s. Problems 5.1: Do problems 1-11, odd.**Impulse**• Impulse on an object is the product of the force and the time interval during which the force acts on the object. • How are impulse and momentum related? • Recall that and**Combining Eq. 1 and Eq. 2**• Then, distributing m • Multiply both sides by t • Thus impulse (Ft) is equal to final minus initial momentums (mv), i.e. change in momentum.**After this analysis, we see that impulse is the measure of**the change in momentum of an object in response to an exerted force. • Some interesting outcomes • Given enough time and distance, any force no matter how small can bring the largest object to p = 0 when given enough time. • The importance of “followthough” in sports.**Example 4**• A 17.5-g bullet is fired at a muzzle velocity of 582 m/s from a gun with a mass of 8.00 kg and a barrel length of 75.0 cm. • How long is the bullet in the barrel? • What is the force on the bullet while it is in the barrel? • Find the impulse exerted on the bullet while it is in the barrel. • Find the bullet’s momentum as it leaves the barrel.**Law of Conservation of Momentum**• When no outside forces are acting on a system of moving objects, the total momentum of a system remains constant.**Example of Conservation of Momentum**Both at Rest After Push vboy = -0.40 m/s vman = ? mman = 75 kg mboy = 35 kg**Example 5**• What force is required to slow a 1450 kg care traveling 115 km/h to 45.0 km/h within 3.00s? How far does the car travel during its deceleration?