MOMENTUM and IMPULSE

• LINEAR MOMENTUM

• Momentum describes an object’s Momentum describes an object’s motionmotion

• Momentum Momentum a vector quantity a vector quantity defines as the product of an defines as the product of an objects mass and velocity.objects mass and velocity.

• The symbol for momentum is: The symbol for momentum is:

Ch. 6-1 in textbook.

OBJECTIVES: •compare the momentum of different moving objects• Compare the momentum of the same objects moving with different velocities•Identify examples of change in the momentum of an object•Describe changes in momentum in terms of force and time

Linear Momentum…

• The equationfor momentum is: = mv

Where is momentummomentum, m is mass, and v is velocity.

Linear Momentum

• Momentum is a VECTOR QUANTITY.VECTOR QUANTITY.

• It’s direction matches that of its velocity.

• The SI units of momentum are Kg x m/s

Linear Momentum

• Think about coasting down a steep hill on your bike…– The faster you move, the more momentum you

have and the more difficult it is to come to a stop.

• At the same speed…the larger the mass of the object the more momentum that object has.

50 mph 50 mph

Sample Problem

• A 22502250 kg truck has a velocity of 2525 m/s to the east. What is the momentum of the truck?

GIVEN FORMULA SUBSTITUTION ANSWER and UNITS

M = 2250 kgM = 2250 kg = m x v = 2250 (25) 56250 =

V = 25 m/sV = 25 m/s 5.6 x 105.6 x 104 4 Kg x m/s Kg x m/s to the east.to the east.

=

Practice Problems…

• A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.– A. What is the total momentum of the child and

bike together?– B. What is the momentum of the child?– C. What is the momentum of the bike?

A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.A. What is the total momentum of the child and bike together?A. What is the total momentum of the child and bike together?B. What is the momentum of the child?C. What is the momentum of the bike?

GIVEN FORMULA SUB ANS/UNITS

M(child) = 21 kg = mv = (21 +5.9)(4.5) = 121.05 Kg x m/s = 121.05 Kg x m/s to the northwestto the northwest

M (bike) = 5.9 kg = (26.9)(4.5)

V = 4/5 m/s

A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.A. What is the total momentum of the child and bike together?B. What is the momentum of the child?B. What is the momentum of the child?C. What is the momentum of the bike?

GIVEN FORMULA SUB ANS/UNITS

M(child) = 21 kg = mv = (21)(4.5) = 94 Kg x m/s to the northwest

M (bike) = 5.9 kg

V = 4/5 m/s

A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.A. What is the total momentum of the child and bike together?B. What is the momentum of the child?C. What is the momentum of the bike?C. What is the momentum of the bike?

GIVEN FORMULA SUB ANS/UNITS

M(child) = 21 kg = mv = (5.9)(4.5) = 27 Kg x m/s to = 27 Kg x m/s to the northwestthe northwest

M (bike) = 5.9 kg

V = 4/5 m/s

Change in Momentum

• A change in momentum takes force and time.

F = t

Force = Change in momentum time interval

IMPULSE MOMENTUM THEOREM

• (Page 210 in textbook)• Impulse for a constant external force, the product of the

force and the time over which it acts on an object.

• Impulse momentum theorem:

F t = or

F t = = mv(final) – mv(initial)

IMPULSE MOMENTUM THEOREM

• A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.

A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.

GIVEN FORMULA SUB ANS/UNIT

M = 1400 kg F t = = mv(final) – mv(initial) F = (1400)(0) – (1400)(15) 0.30

Vi = 15 m/s F = mv(final) – mv(initial)

tF = 21000 kg x m/s 0.30

F = 70000 N to the east

Vf = 0

t = .30 s

Figure 6-3 in text…

• Stopping times and distance depend on the impulse-momentum theorem.

• Two trucks with the same mass traveling at the same speed, but one is loaded down…

• The truck with a load in the bed, MUST undergo a GREATER change in momentum than the truck without a load when arriving upon the same stop sign.

Figure 6-3 in text…

• The loaded down truck’s momentum is twice as great, so its change in momentum is also twice as great. The applied forces are the same, so the time period must be twice as great as well.

= F t

Figure 6-3 in text…

• How do the stopping distances compare?– The loaded truck’s time period is twice as great while its

acceleration is half as much (F-ma). – Because x = Vi t + ½ a t 2

– the loaded truck’s stopping distance is 2 times as great as the empty trucks.

A 2250 kg car traveling to the west A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to slows down uniformly from 20.0 m/s to 5.0 m/s. How long does it take the car 5.0 m/s. How long does it take the car to decelerate if the force on the car is to decelerate if the force on the car is 8450 N to the east? How far does the 8450 N to the east? How far does the car travel during deceleration?car travel during deceleration?

A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.0 m/s. How long does it take the car to decelerate if the How long does it take the car to decelerate if the force on the car is 8450 N to the east? force on the car is 8450 N to the east? How far does the car travel during deceleration?

GIVEN FORMULA SUB ANS/UNIT

M = 2250 kg F t = = mv(final) – mv(initial) t= (2250)(-5) – (2250)(-20) 8450

Vi = -20 m/s(neg b/c going west)

t = mv(final) – mv(initial)

F t = 33750 kg x m/s 8450 kg x m2/s2 t= 3.99s

Vf = -5 m/s

F = 8450 N to the east (+)

A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.0 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel How far does the car travel during deceleration?during deceleration?

GIVEN FORMULA SUB ANS/UNIT

M = 2250 kg x = ½ (vi(final) + v(initial))t x= ½ (-20 - 5) (3.99)

Vi = -20 m/s(neg b/c going west)

X= -50 m or 50 m to the West

Vf = -5 m/s

T = 4

A change in momentum over a longer time requires less force.

• If the speed of a particle is doubled, then the momentum increases by a factor of 2.

• The thespeed of a particle is doubled, then the KE increases by a factor of 4.

Impulse and Momentum

• What is the relationship???Impulse is equivalent to a change in momentum.

When comparing the momentum of 2 moving objects…The less massive object will have less momentum if the velocities are the same.

Which has the GREATEST momentum?

A: Turtle with a mass of 91 kg moving at 1.4 m/s

B: 1.8 kg Roadrunner moving at a 6.7 m/s

C: Hare with a mass of 2.7 kg moving at 7 m/s

D: 270 kg Tortoise moving at 0.5 m/s

Which has the GREATEST momentum?

A: Turtle with a mass of 91 kg moving at 1.4 m/s127.4 kg*m/s

B: 1.8 kg Roadrunner moving at a 6.7 m/s12.06 kg*m/s

C: Hare with a mass of 2.7 kg moving at 7 m/s18.9 kg*m/s

D: 270 kg Tortoise moving at 0.5 m/s135 kg*m/s

Review…

• If the speed of a particle is doubled…– By what factor is its momentum changed?– What happens to its KE?

Review

• 1- Momentum increases by a factor of 2

• KE increases by a factor of 4

• Why the difference?

HOMEWORK

• Page 209 # 1 and 3• Page 211 # 1 and 3

CONSERVATION of MOMENTUM

• ρ1i + ρ2i = ρ1f + ρ2f

• m 1 v1i + m 2 v2i = m 1 v1f + m 2 v2f

TOTAL INITIAL MOMENTUM = TOTAL FINAL MOMENTUM

COPY DOWN IN NOTES…

• The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects.

READING

• Read Page 216-217 on momentum conservation

• Understand that momentum is conserved during collisions AND for objects that are pushing away from each other.

EXAMPLE

• 2 Skaters pushing away from each other (fig 6-7)Both initially at rest with a momentum of ρ1i = ρ2i =0

When they push away from each other, they move in opposite directions with EQUAL but OPPOSITE momentum so that the final momentum is also 0.

ρ1f + ρ2f

SAMPLE

• A 76 kg boater, initially at rest in a stationary 45 kg boat, steeps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

• m1i v1i + m2i v2i =m1f v1f +m2f v2f

Solution

• m1i v1i + m2i v2i =m1f v1f +m2f v2f

• 190 kg m/s + 45 kg ( v2f) = 0

• 45 kg ( v2f) = -190 kg m/s • ( v2f) = -190 kg m/s

45 kg ( v2f) = -4.2 m/s

The negative sign indicates that the boat is moving left. (opposite direction of the boater)

HOMEWORK

• Page 219: 1-2

• Page 221: 1 and 3