chapter 9 linear momentum and collisions examples
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Chapter 9Linear Momentum and Collisions EXAMPLES
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Example 9.1 The Archer The archer of mass 60kg is standing on a
frictionless surface (ice). He fires a 0.50kg arrow horizontally at 50 m/s. With what velocity does the archer move across the ice after firing the arrow?
Can we use: Newton’s Second Law ? NO
No information about F or a Energy? NO
No information about work or energy Momentum? YES
The System will be: the archer with bow (particle 1) and the arrow (particle 2)
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Example 9.1 The Archer, final
ΣFx = 0, so it is isolated in terms of momentum in the x-direction
Total momentum before releasing the arrow is 0: p1i + p2i = 0
The total momentum after releasing the arrow is p1f + p2f = 0
m1v1f + m2v2f = 0 v1f = – (m2/m1 )v2f
v1f = – (0.50kg/60kg )(50.0î)m/s
v1f = –0.42 î m/s
The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law
Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow
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Example 9.2 Railroad Cars Collide(Perfectly Inelastic Collision)
Initial Momentum = Final Momentum (One-Dimension)
If: m1v1+m2v2 = (m1 + m2)v’ Given: v2 = 0, v’1 = v’2 = v’ = ?
v’ = m1v1/(m1 + m2) = 240,000/20,000m/s v’ = 12 m/s
m1 = 10,000kg m2 = 10,000kg
m1 + m2 = 20,000 kg
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Example 9.3 How Good Are the Bumpers?
Mass of the car is 1500 kg. The collision lasts 0.105s.
Find: p = I and the average force (Favg) exerted on the car.
Nt
pF
smkgsmkgsmkgpI
Ippp
avg
if
000,178
/400,26/500,22/900,3
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Example 9.4 Tennis Ball Hits the Wall
p = I ? Given to the ball.If m = 0.060 kg and v = 8.00 m/sp = I : change in momentum wall.
Momentum ∕∕ to wall doesn’t change.
Impulse will be wall. Take + direction toward wall,
p = I = mv = m (vf – vi) p = I = m(–vsin45 – vsin45)p = I = –2mvsin45 = –2.1 N.s
Impulse on wall is in opposite direction: 2.1 N.s
vf = –vsin45
vi = vsin45
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Example 9.5 Explosion as a Collision
Initial Momentum = Final Momentum (One-Dimension)
m1v1+m2v2 = m1v’1 + m2v’2
Initially: v = 0Explodes!
Finally:
mv = 0 = m2v’2 + m1v’1
Given: m1 , m2, v’2,
you may compute v’1 v’1= – (m2/m1)v’2
m v = 0
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Example 9.6 Rifle Recoil
Momentum Before = Momentum After
m1v1+m2v2 = m1v’1+ m2v’2
Given: mB = 0.02 kg, mR = 5.00 kg, v’B = 620 m/s
0 = mBv’B + mRv’R
v’R = – mBv’B /mR = – (0.02)(620)/5.00 m/s = – 2.48 m/s (to the left, of course!)
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Example 9.7 Ballistic Pendulum
Perfectly inelastic collision – the bullet is embedded in the block of wood
Momentum equation will have two unknowns
Use conservation of energy from the pendulum to find the velocity just after the collision
Then you can find the speed of the bullet
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Example 9.7 Ballistic Pendulum, final
Before: Momentum Conservation:
After:
Conservation of energy:
Solving for vB:
Replacing vB into 1st equation and
solving for v1A:
BA vmmvm )( 2111
0)(0)( 212
2121 ghmmvmm B
ghvB 2
ghm
mmv A 2
)(
1
211
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Example 9.8 Collision at an Intersection
Mass of the car mc = 1500kgMass of the van mv = 2500kg
Find vf if this is a perfectly inelastic collision (they stick together).
Before collision: The car’s momentum is:
Σpxi = mcvc Σpxi = (1500)(25) = 3.75x104 kg·m/s
The van’s momentum is: Σpyi = mvvv
Σpyi = (2500)(20) = 5.00x104 kg·m/s After collision: both have the same x- and
y-components:Σpxf = (mc + mv )vf cos Σpyf = (mc + mv )vf sin
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Example 9.8 Collision at an Intersection, final
Because the total momentum is both directions is conserved:
Σpxf = Σpxi 3.75x104 kg·m/s = (mc + mv )vf cos = 4000 vf cos (1)
Σpyf = Σpyi 5.00x104 kg·m/s = (mc + mv )vf sin = 4000vf sin (2)
Dividing Eqn (2) by (1) 5.00/3.75 =1.33 = tan = 53.1°
Substituting in Eqn (2) or (1) 5.00x104 kg·m/s = 4000vf sin53.1° vf = 5.00x104/(4000sin53.1° ) vf = 15.6m/s
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Example 9.9 Center of Mass (Simple Case) Both masses are on the x-axis The center of mass (CM) is on the x-axis One dimension
xCM = (m1x1 + m2x2)/M
M = m1+m2
xCM ≡ (m1x1 + m2x2)/(m1+m2)
The center of mass is closer to the particle with the larger mass
If: x1 = 0, x2 = d & m2 = 2m1
xCM ≡ (0 + 2m1d)/(m1+2m1) xCM ≡ 2m1d/3m1 xCM = 2d/3
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Example 9.10 Three Guys on a Raft
A group of extended bodies, each with a known CM and equivalent mass m. Find the CM of the group.
xCM = (Σmixi)/Σmi
xCM = (mx1 + mx2+ mx3)/(m+m+m)
xCM = m(x1 + x2+ x3)/3m = (x1 + x2+ x3)/3
xCM = (1.00m + 5.00m + 6.00m)/3 = 4.00m
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Example 9.11 Center of Mass of a Rod
Find the CM position of a rod of mass M and length L. The location is on the x-axis
(A). Assuming the road has a uniform mass per unit length λ = M/L (Linear mass density)
From Eqn 9.32
But: λ = M/L
2
0
2
00CM
22
11
LM
xM
xdxM
dxxM
xdmM
x
L
LL
22
/
222
CM
LL
M
LML
Mx
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Example 9.11 Center of Mass of a Rod, final.
(B). Assuming now that the linear mass density of the road is no uniform: λ = x
The CM will be:
But mass of the rod and are related by:
The CM will be:
2
2
0 00
LxdxdxdmM
L LL
3
0
2
00CM
3
11
LM
x
dxxM
xdxxM
dxxM
xdmM
x
CM
LLL
LL
L
M
Lx
3
2
23
3 2
33
CM
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Examples to Read!!! Example 9.2 (page 239) Example 9.5 (page 247) Example 9.10 (page 256)
Material from the book to Study!!! Objective Questions: 7-8-13 Conceptual Questions: 3-5-6 Problems: 1-9-11-15-25-26-27-37-40-65
Material for the Final Exam