Collisions and Conservation of Momentum
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Transcript of Collisions and Conservation of Momentum
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Collisions and Conservation of Collisions and Conservation of MomentumMomentum
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A Collision of Two MassesA Collision of Two MassesWhen two masses m1 and m2 collide, we will
use the symbol u to describe velocities before collision.
The symbol v will describe velocities after collision.
BeforeBefore m1
u1m2
u2
m1
v1 m2
v2AfterAfter
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A Collision of Two BlocksA Collision of Two Blocks
m1 Bm2
“u”= Before “v” = After
m1
u1m2
u2BeforeBefore
m2v2m1
v1AfterAfter
Collision
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Conservation of EnergyConservation of Energy
m1 m2
u1u2
The kinetic energy beforebefore colliding is equal to the kinetic energy afterafter colliding plus the energy lostlost in the collision.
2 2 2 21 1 1 11 1 2 2 1 1 2 22 2 2 2m u m u m v m v Loss 2 2 2 21 1 1 1
1 1 2 2 1 1 2 22 2 2 2m u m u m v m v Loss
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Example 1.Example 1. A A 2-kg2-kg mass moving at mass moving at 4 m/s4 m/s collides with a collides with a 1-kg1-kg mass initially at rest. mass initially at rest. After the collision, the 2-kg mass moves After the collision, the 2-kg mass moves at at 1 m/s1 m/s and the 1-kg mass moves at and the 1-kg mass moves at 2 2 m/sm/s. What energy was lost in the . What energy was lost in the collision?collision?
It’s important to draw and label a sketch It’s important to draw and label a sketch with appropriate symbols and given with appropriate symbols and given
information.information.
m2
u2 = 0
m1
u1 = 4 m/s
m1 = 2 kg m1 = 1 kg
BEFOREBEFORE
m2
v2 = 2 m/s
m1
v1 = 1 m/s
m1 = 2 kg m1 = 1 kg
AFTERAFTER
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Example 1 (Continued).Example 1 (Continued). What What energy was lost in the collision? energy was lost in the collision? Energy is conserved.Energy is conserved.
m2
uu22 = 0= 0
m1
uu1 1 = = 4 m/s4 m/s
mm1 1 = = 2 kg2 kg mm2 2 = = 1 kg1 kg
m2
vv22 = = 2 m/s2 m/s
m1
vv1 1 = = 1 m/s1 m/s
mm1 1 = = 2 kg2 kg mm2 2 = = 1 kg1 kg
BEFORBEFORE:E:
2 2 21 1 11 1 2 22 2 2 (2 kg)(4 m 0 16 J/s)m u m u
2 2 2 21 1 1 11 1 2 22 2 2 2(2 kg)(1 m/s) (1 kg)(2 m/s) 3 Jm v m v AFTERAFTER
Energy Conservation: K(Before) = K(After) + Loss
Loss = 16 J – 3 J Energy Loss = 13 J
Energy Loss = 13 J
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Impulse and MomentumImpulse and Momentum
A BuA
uB
A BvA vB
B--FFAA tt FFB B tt
Opposite but Equal F t
Ft = mvf– mvo
FBt = -FAt
Impulse = p
mBvB - mBuB = -(mAvA - mAuA)
mAvA + mBvB = mAuA + mBuBmAvA + mBvB = mAuA + mBuBSimplifying:
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Conservation of Conservation of MomentumMomentum
A BuA
uB
A BvA vB
B--FFAAtt FFB B tt
The total momentum AFTER a collision is equal to the total momentum BEFORE.
Recall that the total energy is also conserved:
KKA0A0 + K + KB0B0 = K = KAfAf + K + KBfBf + Loss + LossKKA0A0 + K + KB0B0 = K = KAfAf + K + KBfBf + Loss + Loss
Kinetic Energy: K = ½mvKinetic Energy: K = ½mv22
mAvA + mBvB = mAuA + mBuBmAvA + mBvB = mAuA + mBuB
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Example 2:Example 2: A A 2-kg2-kg block block A A and a and a 1-kg1-kg block block BB are pushed together against a are pushed together against a spring and tied with a cord. When the spring and tied with a cord. When the cord breaks, the cord breaks, the 1-kg1-kg block moves to block moves to the right at the right at 8 m/s8 m/s. What is the . What is the velocity of the velocity of the 2 kg2 kg block? block?
A B
The initial velocities are The initial velocities are zerozero, so that the total , so that the total
momentum momentum beforebefore release is zero.release is zero.
mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBB0 0
mAvA = - mBvB vA = - mBvB
mA
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Example 2 (Continued)Example 2 (Continued)
mmAAvvAA+ m+ mBBvvBB = m = mAAuuAA + m + mBBuuBB0 0
mAvA = - mBvB vA = - mBvB
mA
A B
2 kg1 kg A B
8 m/svA2
vA = - (1 kg)(8 m/s)
(2 kg)vA = - 4 m/svA = - 4 m/s
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Elastic or Inelastic?Elastic or Inelastic?
An elastic collision loses no energy. The deform-ation on collision is fully restored.
In an inelastic collision, energy is lost and the deformation may be permanent. (Click it.)
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Perfectly Inelastic Perfectly Inelastic CollisionsCollisions
Collisions where two objects stick together and have a common velocity
after impact.
Collisions where two objects stick together and have a common velocity
after impact.
Before After
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Example 3:Example 3: A A 60-kg60-kg football player football player stands on a frictionless lake of ice. He stands on a frictionless lake of ice. He catches a catches a 2-kg2-kg football and then football and then moves at moves at 40 cm/s40 cm/s. What was the . What was the initial velocity of the football?initial velocity of the football?
Given: uB= 0; mA= 2 kg; mB= 60 kg; vA= vB= vC vC = 0.4 m/s
AA
BB
mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBBMomentum:0
(m(mAA + m + mBB)v)vCC = m = mAAuuAA
(2 kg + 60 kg)(0.4 m/s) = (2 kg)uA
P. Inelastic collision:
uuAA= 12.4 m/s= 12.4 m/s uuAA= 12.4 m/s= 12.4 m/s
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Example 3 (Cont.):Example 3 (Cont.): How much How much energy was lost in catching the energy was lost in catching the football?football?
0
½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss
154 J = 4.96 J + Loss Loss = 149 JLoss = 149 JLoss = 149 JLoss = 149 J
97% of the energy is lost in the collision!!
2 2 21 1 12 2 2 ( ) LossA A B B A B Cm u m u m m v
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General: Perfectly General: Perfectly InelasticInelastic
Collisions where two objects stick together and have a common velocity vC
after impact.Conservation of Momentum:Conservation of Momentum:
Conservation of Conservation of Energy:Energy:
( )A B c A A B Bm m v m u m u ( )A B c A A B Bm m v m u m u
2 2 21 1 12 2 2 ( )A A B B A B cm u m u m m v Loss 2 2 21 1 1
2 2 2 ( )A A B B A B cm u m u m m v Loss
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Example 4.Example 4. An An 87-kg87-kg skater skater BB collides with a collides with a 22-kg22-kg skater skater AA initially at rest on ice. They initially at rest on ice. They move together after the collision at move together after the collision at 2.4 m/s2.4 m/s. . Find the velocity of the skater Find the velocity of the skater BB before the before the collision.collision.
AABB
uuBB = ?= ?uuAA = 0= 0
Common speed Common speed after colliding: after colliding: 2.4 2.4
m/s.m/s.
22 kg22 kg
87 kg87 kg( )A A B B A B Cm u m u m m v ( )A A B B A B Cm u m u m m v
vvBB= v= vA A = v= vCC = = 2.4 m/s2.4 m/s
(87(87 kg)kg)uuBB = (87 kg + 22 kg)(2.4 = (87 kg + 22 kg)(2.4
m/s)m/s)(87 kg)(87 kg)uuBB =262 kg =262 kg
m/sm/s
uB = 3.01 m/s
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Example 6. Nonperfect inelastic:Example 6. Nonperfect inelastic: A A 0.150 kg0.150 kg bullet bullet is fired at is fired at 715 m/s715 m/s into a into a 2-kg2-kg wooden block at rest. wooden block at rest. The velocity of block afterward is The velocity of block afterward is 40 m/s40 m/s. The bullet . The bullet passes through the block and emerges with what passes through the block and emerges with what velocity?velocity?
A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u BB
AA
uuB B = 0= 0
(0.150 kg)(0.150 kg)vvAA+ + (2 kg)(40 m/s) =(2 kg)(40 m/s) = (0.150 kg)(715 (0.150 kg)(715
m/s)m/s)0.1500.150vvAA+ + (80 m/s) =(80 m/s) = (107 (107
m/s)m/s)0.1500.150vvAA = = 27.2 27.2
m/s)m/s)
27.2 m/s
0.150Av
vA = 181 m/s
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Completely Elastic Completely Elastic CollisionsCollisions
Collisions where two objects collide in such a way that zero energy is lost in the
process.
APPROXIMATIONS!APPROXIMATIONS!
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Velocity in Elastic Velocity in Elastic CollisionsCollisions
A B
A B
uBuA
vA vB
1. Zero energy lost.
2. Masses do not change.
3. Momentum conserved.
(Relative v After) = - (Relative v Before)
Equal but opposite impulses (F t) means that:
For elastic collisions: vA - vB = - (uA - uB)vA - vB = - (uA - uB)
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Example 6:Example 6: A A 2-kg2-kg ball moving to ball moving to the right at the right at 1 m/s1 m/s strikes a strikes a 4-kg4-kg ball ball moving left at moving left at 3 m/s3 m/s. What are the . What are the velocities after impact, assuming velocities after impact, assuming complete elasticity?complete elasticity?
A B
A B
3 3 m/sm/s
1 m/s1 m/s
vvAA vvBB1 kg1 kg 2 kg2 kg
vvAA - v - vBB = - (u = - (uAA - u - uBB))
vvA A - v- vBB = u = uBB - u - uAA
vvAA - v - vBB = (-3 m/s) - (1 m/s)
From conservation of energy (relative v):
vA - vB = - 4 m/s vA - vB = - 4 m/s
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Example 6 (Continued)Example 6 (Continued)
A B
A B
3 3 m/sm/s1 m/s1 m/s
vvAA vvBB1 kg1 kg 2 kg2 kg
mmAAvvAA + m + mBBvvBB = m = mAAuuAA + m + mBBuuBB
Energy: Energy: vvAA - v - vBB = = - 4 m/s- 4 m/s
(1 kg)vvAA+(2 kg)vvBB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)
vvAA + 2v + 2vBB = -5 m/s
Momentum also conserved:
vvAA - v - vBB = = - 4 m/s- 4 m/s
Two independent equations to
solve:
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Example 6 (Continued)Example 6 (Continued)
A B
A B
3 3 m/sm/s1 m/s1 m/s
vvAA vvBB1 kg1 kg 2 kg2 kg
vA + 2vB = -5 m/s
vvAA - v - vBB = = - 4 m/s- 4 m/s
Subtract:0 + 3vvB2B2 = - = - 1 m/s1 m/s
vB = - 0.333 m/svB = - 0.333 m/s
Substitution:
vvAA - v - vBB = = - 4 m/s- 4 m/s
vvA2A2 - - (-0.333 m/s)(-0.333 m/s) = = - 4 m/s- 4 m/s
vA= -3.67 m/svA= -3.67 m/s
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Example 7.Example 7. A A 0.150 kg0.150 kg bullet is fired at bullet is fired at 715 m/s715 m/s into a into a 2-kg2-kg wooden block at rest. The velocity of wooden block at rest. The velocity of block afterward is block afterward is 40 m/s40 m/s. The bullet passes through . The bullet passes through the block and emerges with what velocity?the block and emerges with what velocity?
A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u BB
AA
uuB B = 0= 0
(0.150 kg)(0.150 kg)vvAA+ + (2 kg)(40 m/s) =(2 kg)(40 m/s) = (0.150 kg)(715 (0.150 kg)(715
m/s)m/s)0.1500.150vvAA+ + (80 m/s) =(80 m/s) = (107 (107
m/s)m/s)0.1500.150vvAA = = 27.2 27.2
m/s)m/s)
27.2 m/s
0.150Av
vA = 181 m/s
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Example 8a: Example 8a: P. inelastic collision: Find P. inelastic collision: Find vvCC..
AA BB
5 kg5 kg 7.5 kg7.5 kg
uuBB=0=02 m/s2 m/s
AA BB
CommoCommon n vvCC afterafter
vvCC
( )A A B B A B Cm u m u m m v ( )A A B B A B Cm u m u m m v
After hit: After hit: vvBB= v= vAA= v= vCC
(5(5 kg)(2 m/s) = (5 kg + 7.5 kg)kg)(2 m/s) = (5 kg + 7.5 kg)vvCC
12.5 12.5 vvCC =10 m/s =10 m/s
vC = 0.800 m/svC = 0.800 m/s
In an completely inelastic collision, the two balls stick together and move as one
after colliding.
In an completely inelastic collision, the two balls stick together and move as one
after colliding.
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Example 8.Example 8. (b) Elastic collision: Find (b) Elastic collision: Find vvA2A2 and and vvB2B2
AA BB
5 kg5 kg 7.5 kg7.5 kg
vvB1B1=0=02 m/s2 m/s
A A A A B Bm v m v m v A A A A B Bm v m v m v
Conservation of Conservation of Momentum:Momentum:
(5(5 kg)(2 m/s) = (5 kg)kg)(2 m/s) = (5 kg)vvA2A2 + (7.5 kg) + (7.5 kg)
vvBB
AA BB
vvAAvvBB
5 vA + 7.5 vB = 10 m/s
( )A B A Bv v u u
For Elastic For Elastic Collisions:Collisions:
2 m/sA Bv v 2 m/sA Bv v
Continued . . . Continued . . .
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Example 8b (Cont).Example 8b (Cont). Elastic collision: Find Elastic collision: Find vvAA & & vvBB
AA BB
5 kg5 kg 7.5 7.5 kgkg
vvBB =0=02 m/s2 m/s
AABB
vvAAvvBB
Solve Solve simultaneously:simultaneously:
5 vA + 7.5 v B = 10 m/s
2 m/sA Bv v 2 m/sA Bv v
5 5 vvAA + 7.5 + 7.5 vvBB = 10 m/s= 10 m/s
-5 -5 vvAA + 5 + 5 vvBB = +10 m/s= +10 m/s
x (-5)x (-5)
12.5 12.5 vvBB = 20 m/s = 20 m/s
20 m/s1.60 m/s
12.5 Bv
vvAA - 1.60 m/s = -2 m/s- 1.60 m/s = -2 m/s
vA = -0.400 m/svA = -0.400 m/s
vB = 1.60 m/svB = 1.60 m/s
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General: Completely General: Completely ElasticElastic
Collisions where zero energy is lost during a collision (an ideal case).
Conservation of Momentum:Conservation of Momentum:
Conservation of Conservation of Energy:Energy:
2 2 2 21 1 1 12 2 2 2
A A B B A A B B
A B B A
m u m u m v m v Loss
v v u u
2 2 2 21 1 1 12 2 2 2
A A B B A A B B
A B B A
m u m u m v m v Loss
v v u u
A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u
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Summary of Formulas:Summary of Formulas:Conservation of Momentum:Conservation of Momentum:
Conservation of Conservation of Energy:Energy:
2 2 2 21 1 1 12 2 2 2A A B B A A B Bm u m u m v m v Loss 2 2 2 21 1 1 1
2 2 2 2A A B B A A B Bm u m u m v m v Loss
A A B B A A B Bm v m v m u m u A A B B A A B Bm v m v m u m u
For elastic only:For elastic only: A B B Av v u u A B B Av v u u
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The EndThe End