Chapter 9
Linear Momentum and Collisions EXAMPLES
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)
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
Example 9.2 Railroad Cars Collide
(Perfectly Inelastic Collision)
m1 = 10,000kg
m2 = 10,000kg
m1 + m2 = 20,000 kg
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

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.
p  p f  pi  I 
I  p  3,900kg  m / s   22,500kg  m / s   26,400kg  m / s
p
Favg 
 178,000 N
t
Example 9.4 Tennis Ball Hits the Wall
p = I ? Given to the ball.
If m = 0.060 kg and v = 8.00 m/s
p = I : change in momentum  wall.
 Momentum ∕∕ to wall doesn’t change.


Impulse will be  wall.
 Take + direction toward wall,
p = I = mv = 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
Example 9.5 Explosion as a Collision
Initial Momentum = Final
Momentum (One-Dimension)
m1v1+m2v2 = m1v’1 + m2v’2
 Initially: v = 0
Explodes!
 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
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!)
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
Example 9.7 Ballistic Pendulum, final

Before:

Momentum Conservation:
m1v1A  (m1  m2 )vB

After:

1
2
Conservation of energy:
(m1  m2 )vB2  0  (m1  m2 ) gh  0
Solving for vB: vB  2 gh

Replacing vB into 1st equation and
solving for v1A:
(m  m )
v1 A 
1
2
m1
2 gh
Example 9.8 Collision at an Intersection
Mass of the car mc = 1500kg
Mass 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

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
Σpyf = Σpyi 
5.00x104 kg·m/s = (mc + mv )vf sin = 4000vf sin


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
(1)
(2)
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

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
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

xCM
1

M



1
xdm


M
2M
x
2 L
0



2M
L
0
xdx 
M 
L2
But: λ = M/L
xCM 

2M
L2 

M /L 2 L
L 
2M
2
L
0
xdx
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:
xCM
1

M
xCM 


1
xdm


M
3M

L
0
xdx 

M

L
0
xxdx 
L3
But mass of the rod and  are related by:
L
L
L
M   dm   dx   xdx 
0
 The CM will be:
0
0
xCM
L3
L2
2
L3
2



L
L2 3
3M
3
2

M

L
0
x 2 dx 
Material for the Final Exam


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