collision (tumbukan)

Physics 211: Lecture 15, Pg 1

Physics 211: Lecture 15Physics 211: Lecture 15

Today’s AgendaToday’s Agenda

Elastic collisions in one dimension Analytic solution (this is algebra, not rocket science!)

Center of mass reference frame Colliding carts problem Two dimensional collision problems (scattering) Solving elastic collision problems using COM and inertial reference

frame transformations

Some interesting properties of elastic collisions Center of mass energy and energy of relative motion


Momentum Conservation: ReviewMomentum 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

EXTddt

ddtP 0 FEXT 0


Comment on Energy ConservationComment on Energy Conservation

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

Mechanical Energy is lost: (remember what this is??)» Heat (bomb)» Bending of metal (crashing cars)

Kinetic energy is not conserved since dissipative work is done during an inelastic collision! (here, KE equals mechanical energy)

Total momentum, PT, along a certain direction is conserved when there are no external forces acting in this direction.

F = ma = dPT/dt says this has to be true!! (Newton’s Laws) In general, momentum conservation is easier to satisfy than

mechanical energy conservation. Remember: in the absence of external forces, total energy

(including heat…) of a system is always conserved even when mechanical energy is not conserved.

How much do two objects that inelastically collide heat up?


Lecture 15, Lecture 15, Act 1Act 1CollisionsCollisions

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)(a) 1

(b) (b)

(c)(c) 2

2


Lecture 15, Lecture 15, Act 1Act 1SolutionSolution

No external forces in the x direction, so PX is constant.

v m m

2

v2mPF m m v / 2

x

mvPI


Lecture 15, Lecture 15, Act 1Act 1SolutionSolution

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, Lecture 15, Act 1Act 1Another solutionAnother solution

We can write

m m

m m

K12

mv2 P2m

2

P is the same before and after the collision.

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

2K

K

F

I


Lecture 15, Lecture 15, Act 1Act 1Another Question: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, Lecture 15, Act 1Act 1 Another QuestionAnother 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 CollisionsElastic 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 motionsafter the collision as well as before.

all 3D collision problems can be solved in 2 dimensions by using center of mass inertial reference frame

Start with a simpler 1-D problem

Initial Final


Elastic Collision in 1-DElastic Collision in 1-Dwhat has to happenwhat has to happen

v1,i v2,i

initial

x

m1m2

v1,fv2,f

finalm1

m2

Why is this elastic?

Kinetic energy potential energy kinetic energyThe spring is conservative

Maybe, it depends…


Elastic Collision in 1-DElastic Collision in 1-Dthe spring is conservativethe spring is conservative

x

Conserve PX: (no external forces!)

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

Conserve Kinetic Energy: (it’s elastic!)

1/2 m1v21,i + 1/2 m2v2

2,i = 1/2 m1v21,f + 1/2 m2v2

2,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!

v1,i v2,i

before

m1 m2

v1,fv2,f

after


Elastic Collision in 1-DElastic Collision in 1-D

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

First, describe the solution to the problem using algebra. Useful analysis and useful formulae

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

1/2 m1v21,i + 1/2 m2v2

2,i = 1/2 m1v21,f + 1/2 m2v2

2,f

Airtrack

Collisionballs

momentum

energy


Elastic Collision in 1-DElastic Collision in 1-Dspecial casespecial case: equal masses: equal masses

If the masses of the two objects are equal the algebra is not too bad. Let’s see what we get…

Divide through by m = m1 = m2

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

1/2 m1v21,i + 1/2 m2v2

2,i = 1/2 m1v21,f + 1/2 m2v2

2,f

momentum

energy

v1,i + v2,i = v1,f + v2,f

v21,i + v2

2,i = v21,f + v2

2,f


Elastic Collision in 1-DElastic Collision in 1-Dspecial casespecial case: equal masses: equal masses

Now just rearrange equations to bring v1,i and v1,f to left hand side, and v2,i and v2,f to rhs

Divide through energy equation by momentum equation which gives

Particles just trade velocities in 1-D elastic collision of equal mass objects (let equation talk to you…)

momentum

energy

v1,i + v1,f = v2,f + v2,i

v1,i - v1,f = v2,f - v2,i

v21,i - v2

1,f = v22,f - v2

2,i

(v1,i - v1,f )(v1,i + v1,f ) = (v2,f - v2,i )(v2,f + v2,i )

v2,f = v1,i v1,f = v2,i v1,i - v1,f = v2,f - v2,i


Elastic Collision in 1-DElastic Collision in 1-Dgeneral case: general case: unequal massesunequal masses

Conserve linear momentum and mechanical energy, but now the masses are different:

Divide through energy equation by momentum equation which gives

Now solving these two linear equations is only a bit more complicated

momentum

energy

v1,i + v1,f = v2,f + v2,i

m1(v1,i - v1,f ) = m2(v2,f - v2,I )

m1(v21,i - v2

1,f ) = m2(v22,f - v2

2,I )

m1(v1,i - v1,f )(v1,i + v1,f ) =m2 (v2,f - v2,i )(v2,f + v2,i )

m1(v1,i - v1,f ) = m2(v2,f - v2,I )


Elastic Collision in 1-DElastic Collision in 1-Dgeneral case: general case: unequal massesunequal masses

Algebra just gave us the following equations based on conservation of momentum and mechanical energy:

Now just solve for final velocities, v1,f and v2,f in terms of v1,i and v2,i

2 1 21, 2, 1,

1 2 1 2

2f i i

m m mv v v

m m m m

v1,i + v1,f = v2,f + v2,i

m1(v1,i - v1,f ) = m2(v2,f - v2,I )

1 2 12, 1, 2,

1 2 1 2

2f i i

m m mv v v

m m m m

v2,f = v1,i v1,f = v2,i

When m1 = m2


Another way to solve elastic collision Another way to solve elastic collision problems:problems:

CM Reference FrameCM Reference Frame We have shown that the total momentum of a system of

particles is the velocity of the CM times the total mass:

PPNET = MVVCM.

We have also discussed reference frames that are related by a constant velocity vector (i.e.they’re in 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, VVCM = 0 (by definition) and

therefore PPNET = 0. This is a cool mathematical tool that makes the algebra

solving this much simpler (it doesn’t change the physical situation)


Lecture 15, Lecture 15, Act 2Act 2Force and MomentumForce 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 somewhere in Wisconsin.

The men start running on their planks at the same speed. Which man is moving faster with respect to the ice?

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


Lecture 15, Lecture 15, Act 2Act 2Conceptual SolutionConceptual Solution

The external force in the x direction is zero (frictionless): The CM of the systems can’t move! Aha! this is the key!!

x

XX

XX

XX

XX

CM CM


Lecture 15, Lecture 15, Act 2Act 2Conceptual SolutionConceptual 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.

His motion doesn’t count as much, since he is less massive The lighter man moves faster with respect to the ice!

XX

XX

XX

XX

CM CM


Lecture 15, Lecture 15, Act 2Act 2Algebraic SolutionAlgebraic 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, Lecture 15, Act 2Act 2Algebraic SolutionAlgebraic Solution

The velocity 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, it’s zero!)

Plugging vP = vR - V into thiswe find:

v VM

m MR

So vR is greater if m is smaller.


Example 1: Using CM Reference FrameExample 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?

x

m1m2 v1,i vv2,i = 0

VVCM

m1m2

m1vv1,f vv2,fm2

+ = CM

Video

of

CM frame


Example 1...Example 1...

Four step procedure

First figure out the velocity of the CM, VVCM.

» VVCM = (m1vv1,i + m2vv2,i), but vv2,i = 0 in this case so

VVCM = vv1,i

So VVCM = 1/5 (1.5 m/s) = 0.3 m/s

1

1 2m m

m

m m1

1 2

(for vv2,i = 0 only)



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

v*v* = vv - VVCM (where v*v*,, vv, VVCM are vectors)

VVCM

v v

v*v*

This is the “lab” frame velocity

This is the CM frame velocityIf you were traveling along with the CM, you

would see the velocity of the mass to be less than in the lab frame in this case


Example 1...Example 1... Calculate the initial velocities in the CM reference frame

(all velocities are in the x direction):

v*v*1,i = vv1,i - VVCM = 1.5 m/s - 0.3 m/s = 1.2 m/s

v*v*2,i = vv2,i - VVCM = 0 m/s - 0.3 m/s = -0.3 m/s

v*v*1,i = 1.2 m/s

v*v*2,i = -0.3 m/s


Example 1 continued...Example 1 continued...

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

m1m2 v*v*1,i v*v*2,i

xm2 m1

m1v*v*1,f v*v*2,f

m2

Movie


Energy in Elastic Collisions: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, it’s elastic!! (look how we write this…)

But the total momentum is zero, both initial and final:

So:

2f,2

22

2

2f,1

21

1

2i,2

22

2

2i,1

21

1

*vmm21

*vmm21

*vmm21

*vmm21

2f,1

21

21

2i,1

21

21

*vmm21

m21

*vmm21

m21

2f,1

2i,1 *v*v (and the same for particle 2)

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

2i,22

2

i,11 *vm*vm Likewise for final v’s



v*v*1,i v*v*2,i

x

m1m2

m1m2

v*v*1,f = - v*v*1,i = -1.2m/s v*v*2,f = - v*v*2,i =.3 m/s

m1m2

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

Calculate the final velocities in the CM frame:



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

vv1,f = v*v*1,f + VVCM = -1.2 m/s + 0.3 m/s = -0.9 m/s

vv2,f = v*v*2,f + VVCM = 0.3 m/s + 0.3 m/s = 0.6 m/s

vv1,f = -0.9 m/s

vv2,f = 0.6 m/s

v v = v*v* + VVCM

v*v* = vv - VVCM

Four easy steps! No need to solve a quadratic equation!!Especially important in 2D


Lecture 15, Lecture 15, Act 3 Act 3 Moving Between Reference FramesMoving 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, Lecture 15, Act 3 Act 3 Moving Between Reference FramesMoving 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, Lecture 15, Act 3 Act 3 AsideAside

20mi/hr

CM

80mi/hr 40mi/hr

x


Lecture 15, Lecture 15, Act 3 Act 3 Aside SolutionAside Solution

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

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

v´ v´ = v*v* + VVCM

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

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

v*v*GREEN,i = 60 mi/hr

v*v*RED,i = -60 mi/hr


Summary: Using CM Reference Frame Summary: Using CM Reference Frame

: Determine velocity of CM

: Calculate initial velocities in CM reference frame

: Determine final velocities in CMreference frame

: Calculate final velocities in lab reference frame

VVCM =

21 mm

v*v* = vv - VVCM

v*f = -v*i

v v = v*v* + VVCM

(m1vv1,i + m2vv2,i)


Interesting FactInteresting 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,iv*2,f = -v*2,i

This is really cool and useful too!


Basketball Demo.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 M

Drop 2 balls


Basketball Demo.Basketball Demo.

Initially, both balls drop with same velocity, since they started at the same time

The basketball bounces off the floor and has the same speed in the upward direction as it had going downward just before it hit the floor

Now the basketball and the golf ball collide. In CM frame, the golf ball has velocity almost -2v prior to collision.

In lab frame the golf ball recoil is reduced by v, and equals 3v

v

v

v

(a)

m

M M

v

(b)

3v

v

(c)

Drop 2 balls


important comments on energy illustrated here…important comments on energy illustrated here…

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

so

(same for v2)

E m v m vLAB 1

2

1

21 12

2 22 but

CM2

CM1

Vv

Vv v*1

v*2

1CM21

2CM11

21 2Vv Vvv v* v*

= KREL = KCM= PNET,CM = 0

2211CM2

CM21222

211LAB mmVmm

21

m21

m21

E Vv* v* v* v*


More comments on energy...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-DThis is true in general, not just in 1-D

2CM21

222

211LAB Vmm

21

m21

m21

E v* v*


More comments on energy...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 lectureRecap of today’s lecture

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

Look at textbook problems Look at textbook problems Chapter 8: # 71, 73, 75

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