Presentation on theme: "Collisions and Conservation of Energy Consider the above system. A) What is the momentum for each object? B) What is the total momentum for the system?"— Presentation transcript:
Collisions and Conservation of Energy Consider the above system. A) What is the momentum for each object? B) What is the total momentum for the system?
Bellwork Answer the 3 kg mass has a positive momentum of 6 kg m/sec the 2 kg mass has a negative momentum of 6 kg m/sec each mass is moving, but the total momentum of the system equals 0.
Warm-Up 12/09/09 A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s momentum compare to the rate of change of the pebble’s momentum? A. greater than B. less than C. equal to D. Cannot be determined
Solution to Warm-Up C – equal to The rate of change of momentum is, in fact, the force. Remember that F = p/ t. Since the force exerted on the boulder and the pebble is the same, then the rate of change of momentum is the same.
Follow-up A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble. How does the rate of change of the boulder’s velocity compare to the rate of change of the pebble’s velocity? A. greater than B. less than C. equal to D. Cannot be determined
To understand the properties of momentum, we must first re-examine Newton's Laws Newton's 3rd Law F AB = - F BA F AB t = - F BA t m B v fB - m B v oB = - (m A v fA - m A v oA ) m B v fB + m A v fA = m A v oA + m B v oB Σp f = Σp o The Law of Conservation of Momentum states the sum of the momenta before a collision equals the sum of the momenta after a collision.
Conservation of Momentum Σp o = Σp f The total linear momentum of a system is conserved if the net external force on the system is zero. Consider the cue ball that makes a head on collision with an 8 ball at rest. Define the system. – Before the collision – After the collision 8 8
Finding the Sum of Momentum Remember: momentum is a vector quantity. You must use vector addition to determine the total momentum of a system. In a collision, conservation of momentum can be written as m 1 v 1 +m 2 v 2 = m 1 v 1o +m 2 v 2o Practice with Conservation of Momentum: Solve problems A-F ctype=5&filename=Compilations_CPworkbook_ ConservationMomentum.xml
Example 1: A 50 kg pitching machine (excluding the baseball) is placed on a frozen pond. The machine launches a 0.40 kg baseball with a speed of 35 m/s in the eastward direction. Describe the velocity of the machine after the ball is launched. m 1 v 1 +m 2 v 2 = m 1 v 1o +m 2 v 2o
Conceptual Reasoning for Example 1 Before the ball is launched, neither the ball nor the machine is in motion so the total initial momentum is zero. When the ball is launched in the eastward direction, it has a final momentum in the eastward direction. Therefore, the machine must have a final momentum in the westward direction to conserve the total momentum of the system. Therefore, the velocity of the machine after the ball is launched is in the westward direction.
Solution to Example 1 Given: m 1 =50 kg (machine) m 2 =0.40 kg (ball) v 1o = 0 v 2o = 0 v 1 = ? v 2 = 35 m/s east Here the “collision” is the launching of the ball. m 1 v 1 +m 2 v 2 = m 1 v 1o +m 2 v 2o solve for v 1
Example 2 A major league catcher catches a fastball moving at 95 mi/h and his hand and glove recoil 10.0 cm in bringing the ball to rest. If it took seconds to bring the ball (with a mass of 250 g) to rest in the glove, (a) what is the magnitude and direction of the change in momentum of the ball? (b) Find the average force the ball exerts on the hand and glove.
Solution to Example 2 Given a) The change in momentum is: b) The average force is:
Example 3: A 10-gram bullet moving at 250. m/sec bores through a piece of lucite 2.0 cm thick and emerges out the other side at 200. m/sec. How much did the bullet’s momentum change as it moved through the lucite? If the bullet takes s to travel through the lucite, what is the average force exerted on the bullet by the lucite? What acceleration did the bullet experience while traveling through the lucite?
Solution to Example 3 Given: m= 10 g = kg, v f = 200 m/s, v 0 = 250 m/s, t = s
Quick Quiz Amy (150 lbs) and Gwen (50 lbs) are standing on slippery ice and push off each other. If Amy slides at 6 m/s, what speed does Gwen have? 1) 2 m/s 2) 6 m/s 3) 9 m/s 4) 12 m/s 5) 18 m/s 150 lbs 50 lbs
Elastic and Inelastic Collisions There are two general categories of collisions: elastic and inelastic. Elastic collisions occur when objects bounce off each other AND kinetic energy is conserved during the collision. These types of collisions are more "ideal" in nature and are difficult to observe. Inelastic collisions occur when objects collide and energy is lost during the collision. That is, ΣKE before > ΣKE after. During perfectly inelastic collisions the objects stick together during the collision and leave as one single mass. Now solve problems G-K
Air Track Collisions The degree to which a collision is considered to be elastic or inelastic is measured by a quantity called the coefficient of restitution, e, which is defined as the ratio of the (relative velocities of recession) /(relative velocities of approach) for the two objects involved in the collision. e = (v f2 - v f1 ) / (v o1 - v o2 ) When e = 1, the collision is perfectly elastic; when e = 0 is it perfectly inelastic. /Flash/ClassMechanics/AirTrack/AirTrack.html /Flash/ClassMechanics/AirTrack/AirTrack.html
Example 4: Suppose that a 3-kg mass moving at 5 m/sec strikes a stationary 1-kg mass whereupon they stick other. What will be the final velocity of the combined mass, v c, after this inelastic collision? Example 5: Suppose that a 3-kg mass moving at 2 m/sec strikes a 1-kg mass moving towards it at 10 m/sec whereupon they stick together. What will be the final velocity of the combined mass, v c, after this inelastic collision? Example 6: Suppose a 1-kg pistol containing a 10-gram bullet is resting on a table when it is accidentally discharged. If the bullet has a muzzle velocity of 150 m/sec, how fast will the pistol recoil? Example 7: How do the impulses an object receives compare when (1) it strikes a wall and sticks to it, versus (2) it rebounds elastically off of the wall?
Solution to Example 4 Given: m 1 = 3 kgm 2 = 1 kg v 1o = 5 m/s v 2o = 0 m/s v combined = ? Σmv before = Σmv after (3kg)(5m/s) + (1kg)(0m/s) = (3kg+1kg)v c 15kgm/s = 4kg v c v c = 3.75 m/sec
Solution to Example 5 Given: m 1 = 3 kgm 2 = 1 kg v 1o = +2 m/s v 2o = -10 m/s Σmv before = Σmv after (3kg)(2m/s) + (1kg)(-10m/s) = (3kg+1kg)v c -4kgm/s = 4kg v c v c = - 1 m/sec Note that the original velocity of the 3-kg mass is +2 m/sec while the original velocity of the 1-kg mass is - 1 m/sec signifying that they are traveling towards each other in opposite directions. Finally, v c being negative tells us that the combined mass is moving in the original direction of the 1-kg mass.
Solution to Example 6 Given: m 1 = 1 kgm 2 = 10 g = 0.01 kg v 1o = 0 m/sv 2o = 0 m/s v 1’ = ? m/s v 2’ = 150 m/s Σmv before = Σmv after (1kg +.01kg)(0m/s) = (0.01kg)(150m/s) + (1kg)v 1’ 0 kgm/s = v f v f = -1.5 m/sec Note that in order to compare their respective momenta, both the pistol and the bullet have to have their masses stated in kilograms (10 grams = 0.01 kg). Also note that the negative sign on the pistol's final velocity signifies that it is traveling in a direction opposite to the direction of the bullet. We usually say that the pistol "recoils" with a speed of 1.5 m/sec.
Solution to Example 7: the ball that bounces receives an impulse that is twice as large as the ball that sticks to the wall When the ball sticks to the wall, its final velocity equals zero. When the ball rebounds off of the wall, its final velocity equals - v o. The impulse delivered to the ball by the wall equals: The impulse delivered to the ball by the wall equals (net F)t = mv f - mv o (net F)t = m(0) - mv o (net F)t = - mv o (net F)t = mv f - mv o (net F)t = m(-v o ) - mv o (net F) t = - 2mv o
Homework Pass Challenge: When objects travel along diagonal paths, they have momentum in each of the x and y dimensions. Consequently momentum vectors must be resolved into their x- and y-components when working two-dimensional collisions. Consider the following scenario A 2-kg mass traveling left at 3 m/sec along the positive x-axis and a 3-kg mass traveling upwards along the positive y-axis at 2 m/sec collide with each other. If the two masses stick together during the collision, how fast and in what direction will they leave the collision?
x-componentsy-components BEFORE COLLISION Green ball 2(-3) = -6 kgm/s Purlple ball 0 0 3(2) = +6 kgm/s AFTER COLLISION Green and purple balls stick together -(2kg+3kg)v c cos +(2kg + 3kg)v c sin ANALYSISConservation of Momentum -6kgm/s + 0 = -(5kg)v c cos v c cos = 6/ kgm/s = +(5kg)v c cos v c sin = 6/5 tan( ) = 1 = 45 o v c cos(45 o ) = 6/ v c = 1.2 v c = 1.2/0.707 v c = 1.70 m/s