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Algebra Based Basic Momentum & Impulse Notes
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MomentumMomentum
Chapter 6Chapter 6
Momentum (Symbol = p) [Unit = kg m/s] Inertia in motion The Quantity of MotionMomentum = Mass • Velocity
p = m • vNewton’s 2nd Law
Impulse = change in momentum
( )
f i
f i
f i
v vvF ma a t tv v
t F m ttFt mv mv
Impulse (Symbol = I)[Unit = Nsec] A force in a certain time
Concept: In order to change momentum, we need a force in a certain amount of time!
f ipI Ft mv mv
f ipI Ft mv mv
Changing MomentumDifference between hitting concrete bridge vs. yellow barrels full of pea gravelSeat belt vs. Air bag
Δp = Ft Δp = Ft
0 - m v = FT 0 - m v = Fthttp://www.regentsprep.org/Regents/physics/phys01/impulse/default.htm
Momentum is a vector, so direction is important!
A 4.0 x 104 N rocket is at rest when it fires its thrusters. If the thrusters put out an average force of 5.0 x 103 N over 20 sec, then (a) What is the impulse the rocket experiences?(b) What is the rocket’s final velocity?
(a) Impulse I = F tI = 5000 (20)I = 100,000 Nsec
(b) v I = p = mvf – mvi
100,000 = 4080 vf – 4080(0) vf = 24.5 m/s
Conservation of MomentumConservation of Momentum
Law of Conservation of MomentumLaw of Conservation of Momentum Momentum cannot be created or destroyed Momentum cannot be created or destroyed
Momentum In = Momentum outMomentum In = Momentum out
ppinin = p = poutout
2 Types of CollisionsElastic Energy is conservedInelastic Energy is lost to heat, sound, etc.
Since we work in a happy, ideal world, we will deal with all elastic collisions.
This is for Collisions!
1.1. They hit and all the momentum is transferredThey hit and all the momentum is transferred
mA = 4 kg
vA = 3 m/s
mB = 2 kg
vB = 0 m/s
Before Collision
mA = 4 kg
vA = 0 m/s
mB = 2 kg
vB = ? m/s
After Collision
pin = pout
mAvA + mBvB = mAvA + mBvB
(4)(3) + (2)(0) = 4(0) + 2(vB)
vB = 6.00 m/s
2.2. They hit stick togetherThey hit stick together
4 kg
3 m/s
2 kg
0 m/s
Before Collision
4 kg
? m/s
2 kg
? m/s
After Collision
pin = pout
mAvA + mBvB = (mA+mB)vAB
(4)(3) + (2)(0) = (4+2)vB
vAB = 2.00 m/s
3.3. They hit and bounce awayThey hit and bounce away
4 kg
2 m/s
2 kg
-3 m/s
Before Collision
4 kg
-1 m/s
2 kg
? m/s
After Collision
pin = pout
mAvA + mBvB = mAvA + mBvB
(4)(2) + (2)(-3) = (4)(-1) + (2)vB
vAB = 3.00 m/s
? ?
?
3.3. They hit and bounce awayThey hit and bounce away
Before Collision After Collision
pin = pout
mAvA + mBvB = mAvA + mBvB
(4)(2) + (2)(-3) = (4)(-1) + (2)vB
vAB = 3.00 m/s
4 kg
2 m/s
2 kg
-3 m/s
4 kg
-1 m/s
2 kg
? m/s
? ?
4.4. Both start with zero velocityBoth start with zero velocity
4 kg
0 m/s
2 kg
0 m/s
Before Collision
4 kg
? m/s
2 kg
20 m/s
After Collision
pin = pout
mAvA + mBvB = mAvA + mBvB
(4)(0) + (2)(0) = (4)(vA) + (2)(20)
vA = -10.0m/s
1.1. They hit and all the momentum is transferredThey hit and all the momentum is transferred
mA = 4 kg
vA = 3 m/s
mB = 2 kg
vB = 0 m/s
Before Collision
mA = 4 kg
vA = 0 m/s
mB = 2 kg
vB = ? m/s
After Collision
2.2. They hit stick togetherThey hit stick together
4 kg
3 m/s
2 kg
0 m/s
Before Collision
4 kg
? m/s
2 kg
? m/s
After Collision
3.3. They hit and bounce awayThey hit and bounce away
4 kg
2 m/s
2 kg
-3 m/s
Before Collision
4 kg
-1 m/s
2 kg
? m/s
After Collision? ?
?
3.3. They hit and bounce awayThey hit and bounce away
Before Collision After Collision
pin = pout
mAvA + mBvB = mAvA + mBvB
(4)(2) + (2)(-3) = (4)(-1) + (2)vB
vAB = 3.00 m/s
4 kg
2 m/s
2 kg
-3 m/s
4 kg
-1 m/s
2 kg
? m/s
? ?
4.4. Both start with zero velocityBoth start with zero velocity
4 kg
0 m/s
2 kg
0 m/s
Before Collision
4 kg
? m/s
2 kg
20 m/s
After Collision
