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 12/09/09A 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 thanB. less thanC. 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 = 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?

A. greater thanB. less thanC. equal to D. 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 MomentumΣpo = Σpf

• The total linear momentum of a system is conserved if the net external force on the system is zero.

• Consider the cue ballthat makes a head oncollision with an 8 ball at rest.

• Define the system. – Before the collision– After the collision

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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

http://dev.physicslab.org/DocumentPrint.aspx?doctype=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.

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.

Solution to Example 1

• Given: m1=50 kg (machine) m2=0.40 kg (ball) v1o = 0 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 0.00470 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 0.00025 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 3Given: m= 10 g = 0.010 kg, vf = 200 m/s, v0 = 250 m/s, t = 0.00025 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/s2) 6 m/s3) 9 m/s4) 12 m/s5) 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. • http://faraday.physics.utoronto.ca/PVB/Harrison

/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, 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/sΣmvbefore = 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 = 1.5 + 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 ballby the wall equals:

The impulse delivered to the ballby 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?

x-components y-components BEFORECOLLISION

Green ball 2(-3) = -6 kgm/sPurlple ball 0

03(2) = +6 kgm/s

AFTER COLLISIONGreen and purple balls stick together

-(2kg+3kg)vc cosq +(2kg + 3kg)vc sinq

ANALYSIS Conservation of Momentum-6kgm/s + 0 = -(5kg)vc cosqvc cosq = 6/5

0 + 6kgm/s = +(5kg)vc cosqvc sinq = 6/5

tan(q) = 1q = 45o

vc cos(45o) = 6/50.707 vc = 1.2vc = 1.2/0.707vc = 1.70 m/s