Collisions and Conservation of Momentum

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Collisions and Conservation of Momentum
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A Collision of Two Masses
When two masses m1 and m2 collide, we will use
the symbol u to describe velocities before
collision.
Before
The symbol v will describe velocities after
collision.
After
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A Collision of Two Blocks
Collision
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Conservation of Energy
The kinetic energy before colliding is equal to
the kinetic energy after colliding plus the
energy lost in the collision.
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Example 1. A 2-kg mass moving at 4 m/s collides
with a 1-kg mass initially at rest. After the
collision, the 2-kg mass moves at 1 m/s and the
1-kg mass moves at 2 m/s. What energy was lost in
the collision?
Its important to draw and label a sketch with
appropriate symbols and given information.
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Example 1 (Continued). What energy was lost in
the collision? Energy is conserved.
BEFORE
AFTER
Energy Conservation K(Before) K(After) Loss
Loss 16 J 3 J
Energy Loss 13 J
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Impulse and Momentum
Impulse Dp
FDt mvf mvo
Opposite but Equal F Dt
FBDt -FADt
mBvB - mBuB -(mAvA - mAuA)
mAvA mBvB mAuA mBuB
Simplifying
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Conservation of Momentum
The total momentum AFTER a collision is equal to
the total momentum BEFORE.
mAvA mBvB mAuA mBuB
Recall that the total energy is also conserved
Kinetic Energy K ½mv2
KA0 KB0 KAf KBf Loss
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Example 2 A 2-kg block A and a 1-kg block B are
pushed together against a spring and tied with a
cord. When the cord breaks, the 1-kg block moves
to the right at 8 m/s. What is the velocity of
the 2 kg block?
The initial velocities are zero, so that the
total momentum before release is zero.
mAvA mBvB mAuA mBuB
mAvA - mBvB
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Example 2 (Continued)
vA - 4 m/s
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Example 2 (Cont.) Ignoring friction, how much
energy was released by the spring?
½kx2 ½(2 kg)(4 m/s)2 ½(1 kg)(8 m/s)2
½kx2 48 J
½kx2 16 J 32 J 48 J
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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 Collisions
Collisions where two objects stick together and
have a common velocity after impact.
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Example 3 A 60-kg football player stands on a
frictionless lake of ice. He catches a 2-kg
football and then moves at 40 cm/s. What was the
initial velocity of the football?
A
Given uB 0 mA 2 kg mB 60 kg vA vB vC
vC 0.4 m/s
B
mAvA mBvB mAuA mBuB
Momentum
(mA mB)vC mAuA
P. Inelastic collision
(2 kg 60 kg)(0.4 m/s) (2 kg)uA
uA 12.4 m/s
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Example 3 (Cont.) How much energy was lost in
catching the football?
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 J
97 of the energy is lost in the collision!!
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General Perfectly Inelastic
Collisions where two objects stick together and
have a common velocity vC after impact.
Conservation of Momentum
Conservation of Energy
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Example 4. An 87-kg skater B collides with a
22-kg skater A initially at rest on ice. They
move together after the collision at 2.4 m/s.
Find the velocity of the skater B before the
collision.
vB vA vC 2.4 m/s
(87 kg)uB (87 kg 22 kg)(2.4 m/s)
(87 kg)uB 262 kg m/s
uB 3.01 m/s
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Example 6. Nonperfect inelastic A 0.150 kg
bullet is fired at 715 m/s into a 2-kg wooden
block at rest. The velocity of block afterward is
40 m/s. The bullet passes through the block and
emerges with what velocity?
(0.150 kg)vA (2 kg)(40 m/s) (0.150 kg)(715 m/s)
0.150vA (80 m/s) (107 m/s)
0.150vA 27.2 m/s)
vA 181 m/s
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Completely Elastic Collisions
Collisions where two objects collide in such a
way that zero energy is lost in the process.
APPROXIMATIONS!
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Velocity in Elastic Collisions
1. Zero energy lost.
2. Masses do not change.
3. Momentum conserved.
Equal but opposite impulses (F Dt) means that
(Relative Dv After) - (Relative Dv Before)
vA - vB - (uA - uB)
For elastic collisions
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Example 6 A 2-kg ball moving to the right at 1
m/s strikes a 4-kg ball moving left at 3 m/s.
What are the velocities after impact, assuming
complete elasticity?
vA - vB - (uA - uB)
vA - vB uB - uA
vA - vB (-3 m/s) - (1 m/s)
From conservation of energy (relative v)
vA - vB - 4 m/s
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Example 6 (Continued)
Energy vA - vB - 4 m/s
Momentum also conserved
mAvA mBvB mAuA mBuB
(1 kg)vA(2 kg)vB(1 kg)(1 m/s)(2 kg)(-3 m/s)
vA 2vB -5 m/s
Two independent equations to solve
vA - vB - 4 m/s
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Example 6 (Continued)
Subtract
0 3vB2 - 1 m/s
vB - 0.333 m/s
vA2 - (-0.333 m/s) - 4 m/s
Substitution
vA -3.67 m/s
vA - vB - 4 m/s
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Example 7. A 0.150 kg bullet is fired at 715 m/s
into a 2-kg wooden block at rest. The velocity
of block afterward is 40 m/s. The bullet passes
through the block and emerges with what velocity?
(0.150 kg)vA (2 kg)(40 m/s) (0.150 kg)(715 m/s)
0.150vA (80 m/s) (107 m/s)
0.150vA 27.2 m/s)
vA 181 m/s
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Example 8a P. inelastic collision Find vC.
After hit vB vA vC
(5 kg)(2 m/s) (5 kg 7.5 kg)vC
12.5 vC 10 m/s
vC 0.800 m/s
In an completely inelastic collision, the two
balls stick together and move as one after
colliding.
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Example 8. (b) Elastic collision Find vA2 and
vB2
Conservation of Momentum
(5 kg)(2 m/s) (5 kg)vA2 (7.5 kg) vB
5 vA 7.5 vB 10 m/s
For Elastic Collisions
Continued . . .
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Example 8b (Cont). Elastic collision Find vA
vB
Solve simultaneously
5 vA 7.5 vB 10 m/s
-5 vA 5 vB 10 m/s
vA - 1.60 m/s -2 m/s
vA -0.400 m/s
12.5 vB 20 m/s
vB 1.60 m/s
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General Completely Elastic
Collisions where zero energy is lost during a
collision (an ideal case).
Conservation of Momentum
Conservation of Energy
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Summary of Formulas
Conservation of Momentum
Conservation of Energy
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The End