# Collisions and Conservation of Energy

## Presentation on theme: "Collisions and Conservation of Energy"— 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? greater than less than equal to Cannot be determined

Solution to Warm-Up C – equal to
The rate of change of momentum is, in fact, the force. Remember that F = Dp/Dt. 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? greater than less than equal to Cannot be determined

To understand the properties of momentum, we must first re-examine Newton's Laws
Newton's 3rd Law FAB = - FBA FAB t = - FBA t mBvfB - mBvoB = - (mAvfA- mAvoA) mBvfB + mAvfA = mAvoA + mBvoB Σpf = Σpo 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
Σpo = Σpf 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 If the system is taken to be only the 8-ball at rest, then we isolate it with a dotted borer around it. So long as not outside force acts on it, there will be no impulse on it and no change in its momentum. But when the cue ball strikes it, there is an outside force and an impulse on it. Its momentum changes as it speeds away with the speed of the incident cue ball. Or take the system to be the cue ball. Initially it has a momentum of mv. Then it strikes the 8-ball and its momentum undergoes a change. The reaction force by the 8-ball brings it to a halt. Now consider the system of both balls. Before the collision, the momentum is that of the moving cue ball. When the balls strike, no outside force acts, for the interaction is between the two balls, both parts of the same system. So no impulse acts on the system and no change in momentum of the system occurs. In this case the momentum is conserved. It is the same before and after the collision. Momentum of a system is conserved only when no external impulse is exerted on the system.

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 m1v1 +m2v2 = m1v1o +m2v2o Practice with Conservation of Momentum: Solve problems A-F Use “Numbered Heads” Structure Students number off 1-4. Teacher poses a question. Team members lean toward the center of the team so they can quietly discuss the answer. Anyone might be called to give the team answer so they should all be involved in the discussion. When everyone knows the answer, they sit back in their seats and stop talking. Use the student selector spinner to choose a number from each team to respond. The student whose number is called must give an answer outloud or displayed on a whiteboard.

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. m1v1 +m2v2 = m1v1o +m2v2o

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.

Here the “collision” is the launching of the ball.
Solution to Example 1 Given: m1=50 kg (machine) m2=0.40 kg (ball) v1o = v2o= 0 v1 = ? v2 = 35 m/s east Here the “collision” is the launching of the ball. m1v1 +m2v2 = m1v1o +m2v2o solve for v1

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
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, vf = 200 m/s, v0 = 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, ΣKEbefore > ΣKEafter. 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 = (vf2 - vf1) / (vo1 - vo2) When e = 1, the collision is perfectly elastic; when e = 0 is it perfectly inelastic.

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, vc, 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, vc, 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: m1 = 3 kg m2 = 1 kg
v1o = 5 m/s v2o = 0 m/s vcombined = ? Σmvbefore = Σmvafter (3kg)(5m/s) + (1kg)(0m/s) = (3kg+1kg)vc 15kgm/s = 4kg vc vc = 3.75 m/sec

Solution to Example 5 Given: m1 = 3 kg m2 = 1 kg
v1o = +2 m/s v2o = -10 m/s Σmvbefore = Σmvafter (3kg)(2m/s) + (1kg)(-10m/s) = (3kg+1kg)vc -4kgm/s = 4kg vc vc = - 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, vc being negative tells us that the combined mass is moving in the original direction of the 1-kg mass.

Solution to Example 6 Given: m1 = 1 kg m2 = 10 g = 0.01 kg
v1o = 0 m/s v2o = 0 m/s v1’ = ? m/s v2’ = 150 m/s Σmvbefore = Σmvafter (1kg +.01kg)(0m/s) = (0.01kg)(150m/s) + (1kg)v1’ 0 kgm/s = vf vf = -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 -vo. The impulse delivered to the ball by the wall equals: The impulse delivered to the ball by the wall equals (net F)t = mvf - mvo (net F)t = m(0) - mvo (net F)t =  - mvo (net F)t = mvf - mvo (net F)t = m(-vo) - mvo (net F) t =  - 2mvo

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?

Conservation of Momentum
x-components y-components BEFORE COLLISION Green ball 2(-3) = -6 kgm/s Purlple ball 0 3(2) = +6 kgm/s AFTER COLLISION Green and purple balls stick together -(2kg+3kg)vc cosq +(2kg + 3kg)vc sinq ANALYSIS Conservation of Momentum -6kgm/s + 0 = -(5kg)vc cosq vc cosq = 6/5 0 + 6kgm/s = +(5kg)vc cosq vc sinq = 6/5 tan(q) = 1 q = 45o vc cos(45o) = 6/5 0.707 vc = 1.2 vc = 1.2/0.707 vc = 1.70 m/s