Momentum Momentum is defined as the product of the mass × velocity of an object.  p = m × v where m is the mass in kg, v is the velocity in m/s, and the momentum, p, in kgm/s. Momentum is a vector quantity meaning that it has magnitude and direction.

The direction of momentum is that of the velocity of the object. From Newton’s 2 nd Law:  F net = ma  a = ΔvΔv ΔtΔt Therefore, F net = mΔvmΔv ΔtΔt

F net Δt = mΔv F net Δt is called impulse where F net is measured in N and Δt measured in s.  F net Δt = Ns  Impulse is a vector quantity and its direction is that of F net.

Impulse and Momentum Momentum can be thought of as “inertia in motion”. When a car on Thurber’s Avenue crashes into a guard rail, a large force is exerted on the guard rail.  The force of impact comes from a change in velocity or deceleration of the car.

 The force of impact comes from a change in velocity or deceleration of the car.  The force of impact is directly proportional to the change in velocity of the moving car and to its mass. F net α m × v To change the momentum of an object, you must consider the impulse.

 Impulse depends on the magnitude of the force and the time of impact (time the objects are in physical contact). Hitting a golf ball for distance (teeing off), throwing a baseball or football for distance requires a “follow through”.  These situations call for the largest change in momentum (mΔv) which is achieved by using a large force for as long as possible (FΔt).

When receiving a punch, it is best to “roll with the punch”.  By rolling with the punch, you extend the time of impact which minimizes F net.  F net = mΔv/Δt  The same change in momentum will occur in either case but it is best to maximize the value of Δt rather then the F net.

When throwing a punch, you try to pull your fist back as quickly as possible.  By pulling back your fist as quickly as possible, it minimizes the Δt which maximizes the F net. Didn’t you use this idea as a youngster jumping from a high height?

Momentum Problems A kg baseball with a velocity of 7.5 m/s strikes a wall perpendicularly and rebounds at 8.9 m/s. m = kgv 1 = 7.5 m/sv 2 = -8.9 m/s (a)What is the change in momentum of the baseball? Δmv = mΔv = kg × (7.5 m/s – -8.9 m/s) = 2.4 kgm/s 2

(b) If the force of impact acted s, what was the average force on the wall? (c) What average force is exerted on the baseball? By Newton’s 3 rd Law, 53 N. F net = mΔvmΔv ΔtΔt 2.4 kgm/s 2 F net = s = 53 N

A 26 kg boy is riding on a wagon going at 3.0 m/s N. The boy jumps off the back of a wagon in such a way that he has no velocity when he lands on the ground. The mass of the wagon is 10. kg. m b = 26 kgm w = 10. kg v b = 3.0 m/s N v b ’ = 0 m/s

(a)What was the boy's change in momentum? Δp b = mΔv = 26 kg × -3.0 m/s = -78 kgm/s S (b)What was the wagon's change in momentum? Δp system = 0, therefore the change in momentum of the wagon is 78 kgm/s N.

(c) How fast does the wagon go after the boy jumps off? Δp w = mΔv Δv = Δp w /m = 78 kgm/s/10. kg = 7.8 m/s N v f = v i + Δv = 3.0 m/s m/s = 10.8 m/s N

Conservation of Momentum When studying the momentum changes in collisions, you must focus on a closed and isolated system.  A system is defined as a collection of objects.  A system is closed if no objects enter or leave the system.

 A system is isolated if no external forces are exerted on it. Consider a game of pool when there is only the 8-ball and the cue ball.  The system is closed because both balls only interact with each other.  The system is isolated because rolling friction (µ k = 0) can be ignored.

 8 A B pApA pBpB where m A = m B, v A > v B, and p A > p B Assume neither ball is spinning (no english or angular momentum) because that would complicate matters.

 At the instant of impact. 8 FAFA FBFB 8 After impact. v A = 0v B > 0

The Conservation of Momentum states that in a closed and isolated system, the momentum before the collision equals the momentum after the collision.  Δp = 0 p i = p f  Momentum may be redistributed in a collision but Δp system = 0.  Internal forces are forces between objects within the system.

 External forces are exerted by objects outside a system.  Momentum is conserved only when internal forces are present.

Conservation of Momentum Problems A 7450 kg truck traveling at 4.9 m/s 090° collides with a 1575 kg car traveling at 21 m/s 240°. After colliding, the truck and car move together. What is their velocity after the collision? m 1 = 7450 kgm 2 = 1575 kg v 1 = 4.9 m/s 090°v 2 = 21 m/s 240°

. m1v1m1v1 m2v2m2v2 θ1θ1 θ2θ2 mRvRmRvR θ 1 = 30° θ 2 = ? m R v R = ((m 1 v 1 ) 2 + (m 2 v 2 ) 2 - 2×m 1 v 1 ×m 2 v 2 cosθ) ½ m R v R = ((7450 kg × 4.9 m/s) 2 + (1575 kg × 21 m/s) × 7450 kg × 4.9 m/s × 1575 kg × 21 m/s × cos30°) ½

. m R v R = 1.8 x 10 4 kgm/s v R = 2.0 m/s sinθ 1 mRvRmRvR = sinθ 2 m2v2m2v2 θ2θ2 = 67° sin30° = sinθ kg × 21 m/s 1.8 x 10 4 kgm/s v R = 2.0 m/s, 157°

Two angry hockey players push each other. The mass of the first hockey player is 70. kg and the second has a mass of 50. kg. If the first hockey player has a velocity of 6.2 m/s N, how far apart are they after 3.0 s? Assume friction is negligible. m 1 = 70.kgm 2 = 50. kg v 1 = 6.2 m/s Nv 2 = ? Δt = 3.0 s µ = 0

Δp = 0 p i = p f m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ 0 = m 1 v 1 ’ + m 2 v 2 ’ 0 = 70. kg × 6.2 m/s kg × v 2 ’ V 2 ’ = -8.7 m/s = 8.7 m/s S

Δx T = Δx 1 + Δx 2 = v 1 × Δt 1 + v 2 × Δt 2 Δx T = 6.2 m/s × 3.0 s m/s × 3.0 s Δx T = 45 m

Conservation of Kinetic Energy Kinetic energy is the energy due to the motion of an object.  KE = ½mv 2 where m is the mass in kg, v is the velocity in m/s, and KE is the kinetic energy in J (joules).  Kinetic energy is a scalar quantity.

The Law of Conservation of Energy states that within a closed and isolated system, energy can change form but the total amount of energy is constant.  ΔE = 0 When two objects collide, the force each object exerts on the other slightly change the shape of the object.  KE PE

Momentum And Energy Together These two conservation laws, momentum and energy, are both needed to give a “complete picture” of what happens in any collision.  The conservation of momentum gives the final momentum after the collision. What it does not tell us is the distribution between mass and velocity.

The conservation of kinetic energy tells us how the distribution between mass and velocity by considering the elastic properties of the colliding objects.  The conservation of energy puts an upper limit of the velocity of the impacted object and how close you get to this upper limit depends on the elasticity.

Remember the three types of collisions:  Elastic Collisions After the collision, the objects separate. Both momentum and kinetic energy are conserved.

 Perfectly Inelastic Collisions After the collision, the objects stick together and share a common velocity. Only momentum is conserved.

 Inelastic Collisions The objects deform during the collision so the total kinetic energy decreases but the objects separate after the collision. Only momentum is conserved.

A 600. mg popcorn kernel feeling the heat is moving around at 7.4 cm/s before it pops and breaks into two pieces of equal mass. If one piece comes to an abrupt halt, determine the change in kinetic energy. m 1 = 600. mgm 1 ’ = 300. mgv 1 ’ = v 1 ’ v 1 = 7.4 cm/sm 2 ’ = 300. mgv 2 ’ = 0

Δp = 0 p i = p f m 1 v 1 = m 1 v 1 ’ + m 2 v 2 ’ 600. mg × 7.4 m/s = 300. mg × v 1 ’ v 1 ’ = 14.8 m/s

ΔKE = 0 KE i = KE f ½m 1 v 1 2 = ½m 1 v 1 ’ 2 + ½m 2 v 2 ’ 2 ½ × (600. mg× 1 g 10 3 mg ×× 1 kg 10 3 g (7.4 cm s × 1 m 10 2 cm )2)2 =1.6 x J

(14.8 cm s × 1 m 10 2 cm )2)2 =3.3 x J ½ × (300. mg× 1 g 10 3 mg ×× 1 kg 10 3 g ΔKE = KE f – KE i = 3.3 x J x J ΔKE = KE f – KE i = 1.6 x J

A Toyota with a mass of 575 kg moving at 15 m/s collides into the rear of a Ford pick-up which has a mass of 1575 kg and is moving at 5.0 m/s. The two vehicles lock together and continue sliding forward. (a)What is their final velocity? m 1 = 575 kgm 2 = 1575 kg v 1 = 15 m/s v 2 = 5.0 m/s v 1 ’ = ? v 2 ’ = ?

Δp = 0 p i = p f m 1 v 1 + m 2 v 2 = (m 1 +m 2 )v’ 575 kg × 15 m/s kg × 5.0 m/s = (575 kg kg) × v’ v’ = 7.7 m/s

(b) How much kinetic energy was “lost” in the collision? ΔKE = 0 KE i = KE f ½m 1 v ½m 2 v 2 2 = ½(m 1 + m 2 )v’ ½ × 575 kg × (15 m/s) 2 + ½ × 1575 kg × (5.0 m/s) 2 = ½ × (575 kg kg) × (7.7 m/s) 2

84000 J ≠ J ΔKE = J – J = 2.0 x 10 4 J Therefore, 2.0 x 10 4 J were converted into heat and sound which represent a wasted form of energy.

A 15 g bullet is shot into a wooden block which has a mass of 5085 g. The block is at rest on a horizontal surface. After the bullet becomes imbedded in the block, the two move at 1.0 m/s. Determine the velocity of the bullet before striking the block. m b = 15 gm w = 5085 g v b = ? v w = 0 v b ’ = 1.0 m/sv w ’ 1.0 m/s

Wrap Up Questions If two particles have equal kinetic energies, are their momenta always equal? No, the momenta would only be equal if their masses were equal.

Does a large force always produce a larger impulse than a smaller force? Justify your answer. The impulse, F net Δt, is dependent on both the net force and the time interval the net force acts. The statement would be true only if the time intervals were equal.

The cue ball collides with the 8-ball which is initially at rest. Is it possible for both the cue ball and 8-ball to be at rest immediately after the collision? The conservation of momentum (Δp = 0) prohibits this from happening. If the system (the cue ball and the 8-ball) had momentum before the collision, then there has to be the same momentum before the collision.

Would it be possible for one ball to be at rest after the collision? Yes, whenever two equal masses experience a head-on elastic collision, the stationary ball will have all of the momentum after the collision.

Is momentum conserved when a basketball bounces off the floor? Yes, if the system is considered to be both the basketball and the earth. However, due to the mass of the earth, its momentum would be undetectable. Too understand why the earth must be included, take a look at the next slide.

Assume the ball moving downward to have a positive velocity, then the ball moving upward would have a negative velocity. m1v1m1v1 -m 1 v 1 ’ Δp = 0 p i = p f m 1 v 1 = - m 1 v 2 How can v 1 = - v 2 ?

Can a collision occur such that all of the kinetic energy is lost? Yes, provided that equal masses are approaching each other with the same speeds, such that the momentum before the collision is equal to zero. If the collision is perfectly inelastic, then the kinetic energy is converted into elastic potential energy which deforms the objects.