Chapter 9Linear 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)

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

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

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

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

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:

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

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

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

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

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

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

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