Physics 211: Lecture 15 - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect15.pdfPhysics 211: Lecture 15, Pg 24 Lecture 15, Act 2 Force and Momentum Two men, one heavier than the other,

Physics 211: Lecture 15, Pg 1

Physics 211: Lecture 15

Today’s Agenda

Inelastic Collisions in two dimensions

Explosions

Elastic collisions in one dimension

Center of mass reference frame

Colliding carts problem

Some interesting properties of elastic collisions

Killer bouncing balls


Momentum Conservation: Review

The concept of momentum conservation is one of the most fundamental principles in physics.

This is a component (vector) equation.

We can apply it to any direction in which there is no external force applied.

You will see that we often have momentum conservation even when kinetic energy is not conserved.

FP

EXT

d

dt

d

dt

P 0 FEXT 0


Comment on Energy Conservation

We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.

Energy is lost:

» Heat (bomb)

» Bending of metal (crashing cars)

Kinetic energy is not conserved since work is done during the collision!

Momentum along a certain direction is conserved when there are no external forces acting in this direction.

In general, momentum conservation is easier to satisfy than energy conservation.


Inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).

v1

v2

V

before after

m1

m2

m1 + m2


Inelastic collision in 2-D...

There are no net external forces acting.

Use momentum conservation for both components.

v1

v2

V = (Vx,Vy)

m1

m2

m1 + m2

P Px i x f, , m v m m Vx1 1 1 2

Vm

m mvx

1

1 21

P Py i y f, , m v m m Vy2 2 1 2

Vm

m mvy

2

1 22

X:

y:



So we know all about the motion after the collision!

V = (Vx,Vy)

Vx

Vy

V

m

m mvx

1

1 21

V

m

m mvy

2

1 22

1

2

11

22

x

y

p

p

vm

vm

V

Vtan



We can see the same thing using vectors:

tan p

p2

1

P

p1

p2

P

p1 p2


Explosion (inelastic un-collision)

Before the explosion: M

m1 m2

v1 v2

After the explosion:


Explosion...

No external forces, so P is conserved.

Initially: P = 0

Finally: P = m1v1 + m2v2 = 0

m1v1 = - m2v2 M

m1 m2

v1 v2


Lecture 14, Act 3 Center of Mass

A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible?

(a) 1 (b) 2 (c) both

m m

v V

v

m

m m

v v

v

m

(1) (2)



m m

v v

v

m

(1)

No external forces, so P must be conserved.

Initially: P = 0

In explosion (1) there is nothing to balance the upward momentum of the top piece so Pfinal 0.

mv mv

mv



No external forces, so P must be conserved.

All the momenta cancel out.

Pfinal = 0.

(2)

m m

v v

v

m

mv

mv

mv


Lecture 15, Act 1 Collisions

A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest.

What is the ratio of initial to final kinetic energy of the system?

(a) 1

(b)

(c) 2

2


Lecture 15, Act 1 Solution

v

m m

m m v / 2

x


Lecture 15, Act 1 Solution

Compute kinetic energies:

v

m m

m m v / 2

2

Imv

2

1K

I

2

K2

1

2

vm2

2

1

FK

2K

K

F

I


Lecture 15, Act 1 Another solution

We can write

m m

m m

K1

2mv2

P

2m

2

P is the same before and after the collision.

The mass of the moving object has doubled, hence the kinetic energy must be half.

2K

K

F

I


Lecture 15, Act 1 Another Question:

Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero?


Lecture 15, Act 1 Another Question

Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero?

YES: If the CM is not moving!

CM

CM


Elastic Collisions

Elastic means that kinetic energy is conserved as well as momentum.

This gives us more constraints

We can solve more complicated problems!!

Billiards (2-D collision)

The colliding objects have separate motions after the collision as well as before.

Start with a simpler 1-D problem

Initial Final


Elastic Collision in 1-D

v1,i v2,i

initial

x

m1 m2

v1,f v2,f

final

m1 m2



v1,i v2,i

v1,f v2,f

before

after

x

m1 m2

Conserve PX:

m1v1,i + m2v2,i = m1v1,f + m2v2,f

Conserve Kinetic Energy:

1/2 m1v21,i +

1/2 m2v22,i =

1/2 m1v21,f +

1/2 m2v22,f

Suppose we know v1,i and v2,i

We need to solve for v1,f and v2,f

Should be no problem 2 equations & 2 unknowns!



However, solving this can sometimes get a little bit tedious since it involves a quadratic equation!!

A simpler approach is to introduce the Center of Mass Reference Frame

m1v1,i + m2v2,i = m1v1,f + m2v2,f

1/2 m1v21,i +

1/2 m2v22,i =

1/2 m1v21,f +

1/2 m2v22,f

Airtrack

Collision

balls


CM Reference Frame

We have shown that the total momentum of a system is the velocity of the CM times the total mass:

PNET = MVCM.

We have also discussed reference frames that are related by a constant velocity vector (i.e. relative motion).

Now consider putting yourself in a reference frame in which the CM is at rest. We call this the CM reference frame.

In the CM reference frame, VCM = 0 (by definition) and therefore PNET = 0.


Lecture 15, Act 2 Force and Momentum

Two men, one heavier than the other, are standing at the center of two identical heavy planks. The planks are at rest on a frozen (frictionless) lake.

The men start running on their planks at the same speed.

Which man is moving faster with respect to the ice?

(a) heavy (b) light (c) same


Lecture 15, Act 2 Conceptual Solution

The external force in the x direction is zero (frictionless):

The CM of the systems can’t move!

x

X

X

X

X

CM CM


Lecture 15, Act 2 Conceptual Solution

The external force in the x direction is zero (frictionless):

The CM of the systems can’t move!

The men will reach the end of their planks at the same time, but lighter man will be further from the CM at this time.

The lighter man moves faster with respect to the ice!

X

X

X

X

CM CM


Lecture 15, Act 2 Algebraic Solution

Consider one of the runner-plank systems:

There is no external force acting in the x-direction:

Momentum is conserved in the x-direction!

The initial total momentum is zero, hence it must remain so.

We are observing the runner in the CM reference frame!

x

Let the mass of the runner be m and the plank be M.

m

M

Let the speed of the runner and the plank with respect to the ice be vR and vP respectively.

vR

vP


Lecture 15, Act 2 Algebraic Solution

The speed of the runner with respect to the plank is V = vR + vP (same for both runners).

x

m

M

vR

vP

MvP = mvR (momentum conservation)

Plugging vP = V - vR into this we find:

v VM

m MR

So vR is greater if m is smaller.


Example 1: Using CM Reference Frame

A glider of mass m1 = 0.2 kg slides on a frictionless track with initial velocity v1,i = 1.5 m/s. It hits a stationary glider of mass m2 = 0.8 kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What are the final velocities?

m1 m2 v1,i v2,i = 0

x m1

VCM

m2

m1 v1,f v2,f m2

+ = CM


Example 1...

Four step procedure

First figure out the velocity of the CM, VCM.


Example 1...

If the velocity of the CM in the “lab” reference frame is VCM, and the velocity of some particle in the “lab” reference frame is v, then the velocity of the particle in the CM reference frame is v*:

v* = v - VCM (where v*, v, VCM are vectors)

VCM

v

v*


Example 1...

Calculate the initial velocities in the CM reference frame (all velocities are in the x direction):


Example 1 continued...

Now consider the collision viewed from a frame moving with the CM velocity VCM. ( jargon: “in the CM frame”)

m1 m2 v*1,i v*2,i

x m2 m1

m1 v*1,f v*2,f m2


Energy in Elastic Collisions:

Use energy conservation to relate initial and final velocities.

The total kinetic energy in the CM frame before and after the collision is the same:

But the total momentum is zero:

So:

2f,2

22

2

2f,1

21

1

2i,2

22

2

2i,1

21

1

*vmm2

1*vm

m2

1*vm

m2

1*vm

m2

1

2f,1

2i,1 *v*v

2i,222

i,11 *vm*vm

2f,1

21

21

2i,1

21

21

*vmm2

1

m2

1*vm

m2

1

m2

1

(and the same for particle 2)

Therefore, in 1-D: v*1,f = -v* 1,i v*2,f = -v*2,i


Example 1...

v*1,i v*2,i

x

m1 m2

m1

m1

m2

m2

v*1,f = -v* 1,i v*2,f = -v*2,i


Example 1...

So now we can calculate the final velocities in the lab reference frame, using:

v = v* + VCM

v* = v - VCM

Four easy steps! No need to solve a quadratic equation!!


Lecture 15, Act 3 Moving Between Reference Frames

Two identical cars approach each other on a straight road. The red car has a velocity of 40 mi/hr to the left and the green car has a velocity of 80 mi/hr to the right.

What are the velocities of the cars in the CM reference frame?

(a) VRED = - 20 mi/hr (b) VRED = - 20 mi/hr (c) VRED = - 60 mi/hr

VGREEN = + 20 mi/hr VGREEN = +100 mi/hr VGREEN = + 60 mi/hr


Lecture 15, Act 3 Moving Between Reference Frames

The velocity of the CM is:

x

Vm m

mhrCM

80 40

2 mi /

= 20 mi / hr

20mi/hr

CM

80mi/hr 40mi/hr

So VGREEN,CM = 80 mi/hr - 20 mi/hr = 60 mi/hr

So VRED,CM = - 40 mi/hr - 20 mi/hr = - 60 mi/hr

The CM velocities are equal and opposite since PNET = 0 !!


As a safety innovation, Volvo designs a car with a spring attached to the front so that a head on collision will be elastic. If the two cars have this safety innovation, what will their final velocities in the lab reference frame be after they collide?

Lecture 15, Act 3 Aside

20mi/hr

CM

80mi/hr 40mi/hr

x


Lecture 15, Act 3 Aside Solution

v*GREEN,f = -v* GREEN,i v*RED,f = -v*RED,i

v*GREEN,f = -60 mi/hr v*RED,f = 60 mi/hr

v´ = v* + VCM

v´GREEN,f = -60 mi/hr + 20 mi/hr = - 40 mi/hr

v´RED,f = 60 mi/hr + 20 mi/hr = 80 mi/hr

v*GREEN,i = 60 mi/hr

v*RED,i = -60 mi/hr


Summary: Using CM Reference Frame

: Determine velocity of CM

: Calculate initial velocities in CM reference frame

: Determine final velocities in CM reference frame

: Calculate final velocities in lab reference frame

VCM =

21 mm

v* = v - VCM

v*f = -v*i

v = v* + VCM

(m1v1,i + m2v2,i)


Interesting Fact

We just showed that in the CM reference frame the speed of an object is the same before and after the collision, although the direction changes.

The relative speed of the blocks is therefore equal and opposite before and after the collision. (v*1,i - v*2,i) = - (v*1,f - v*2,f)

But since the measurement of a difference of speeds does not depend on the reference frame, we can say that the relative speed of the blocks is therefore equal and opposite before and after the collision, in any reference frame.

Rate of approach = rate of recession

v*1,i v*2,i

v*1,f = -v*1,i v*2,f = -v*2,i

This is really cool and useful too!


Basketball Demo.

Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M). Drop these from some height. The height reached by the small ball after they “bounce” is ~ 9 times the original height!! (Assumes M >> m and all bounces are elastic).

Understand this using the “speed of approach = speed of recession” property we just proved.

v

v

v

v

3v

v

(a) (b) (c)

m

M

Drop 2 balls


More comments on energy

Consider the total kinetic energy of the system in the lab reference frame:

E m v m vLAB 1

2

1

21 1

22 2

2but

so

(same for v2)

= KREL = KCM = PNET,CM = 0

CM2

CM1

Vv

Vv v*1

v*2

1CM21

2CM11

21 2Vv Vvv v* v*

2211CM2

CM21222

211LAB mmVmm

2

1m

2

1m

2

1E Vv* v* v* v*


More comments on energy...

Consider the total kinetic energy of the system in the LAB reference frame:

= KREL = KCM

So ELAB = KREL + KCM

KREL is the kinetic energy due to “relative” motion in the CM frame.

KCM is the kinetic energy of the center of mass.

This is true in general, not just in 1-D

2CM21222

211LAB Vmm

2

1m

2

1m

2

1E v* v*


More comments on energy...

ELAB = KREL + KCM

Does total energy depend on the reference frame??

YOU BET!

KREL is independent of the reference frame, but

KCM depends on the reference frame (and = 0 in

CM reference frame).


Recap of today’s lecture

Inelastic collisions in two dimensions (Text: 8-6)

Explosions

Elastic collisions in one dimension (Text: 8-6)

Center of mass reference frame (Text: 8-7)

Colliding carts problem

Some interesting properties of elastic collisions

Killer bouncing balls

Documents

Physics 211: Lecture 15 - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect15.pdfPhysics 211: Lecture 15, Pg 24 Lecture 15, Act 2 Force and Momentum Two men, one heavier than the other,