Physics 111 lecture 15 today s agenda
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Physics 111: Lecture 15 Today’s Agenda. 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.

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Physics 111 lecture 15 today s agenda

Physics 111: Lecture 15Today’s Agenda

  • 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

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.


Comment on energy conservation

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.


Lecture 15 act 1 collisions

Lecture 15, Act 1Collisions

  • 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 finalkinetic energy of the system?

(a)1

(b)

(c)2


Lecture 15 act 1 solution

Lecture 15, Act 1Solution

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

v

m

m

m

m

v / 2

x


Lecture 15 act 1 solution1

Lecture 15, Act 1Solution

  • Compute kinetic energies:

v

m

m

m

m

v / 2


Lecture 15 act 1 another solution

Lecture 15, Act 1Another solution

  • We can write

  • P is the same before and after the collision.

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

m

m

m

m


Lecture 15 act 1 another question

Lecture 15, Act 1Another 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 question1

Lecture 15, Act 1Another 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 Collisions

  • Elastic means that kineticenergy is conserved as well as momentum.

  • This gives us more constraints

    • We can solve more complicated problems!!

    • Billiards (2-D collision)

    • The colliding objectshave separate motionsafter the collision as well as before.

  • Start with a simpler 1-D problem

Initial

Final


Elastic collision in 1 d

m2

m1

final

v2,f

v1,f

Elastic Collision in 1-D

m2

m1

initial

v1,i

v2,i

x


Elastic collision in 1 d1

ConservePX:

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

Conserve KineticEnergy:

1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f

Should be no problem 2 equations & 2 unknowns!

Elastic Collision in 1-D

m1

m2

before

v1,i

v2,i

x

after

v2,f

v1,f

Suppose we know v1,i andv2,i

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


Elastic collision in 1 d2

Elastic Collision in 1-D

Airtrack

Collision

balls

  • 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


Cm reference frame

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

Lecture 15, Act 2Force 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

Lecture 15, Act 2Conceptual Solution

  • The external force in the xdirection is zero (frictionless):

    • The CM of the systems can’t move!

X

X

X

X

x

CM

CM


Lecture 15 act 2 conceptual solution1

Lecture 15, Act 2Conceptual Solution

  • The external force in the xdirection 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

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

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

m

vR

vP

M

Lecture 15, Act 2Algebraic 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


Lecture 15 act 2 algebraic solution1

Plugging vP=V - vR into thiswe find:

Lecture 15, Act 2Algebraic Solution

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

MvP = mvR(momentum conservation)

m

vR

vP

So vR is greater if m is smaller.

M

x


Example 1 using cm reference frame

Example 1: Using CM Reference Frame

Video

of

CM 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?

m2

v1,i

v2,i = 0

m1

VCM

+ = CM

m2

m1

x

m2

v2,f

m1

v1,f


Example 1

Example 1...

  • Four step procedure

    • First figure out the velocity of the CM, VCM.

      • VCM = (m1v1,i + m2v2,i), butv2,i= 0 so

        VCM = v1,i

  • So VCM = 1/5 (1.5 m/s) = 0.3 m/s

Step 1

(for v2,i= 0only)


Example 11

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, VCMare vectors)

  • v

  • VCM

  • v*


Example 12

Example 1...

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

Step 2

v*1,i = v1,i - VCM = 1.5 m/s - 0.3 m/s = 1.2 m/s

v*2,i = v2,i - VCM = 0 m/s - 0.3 m/s = -0.3 m/s

v*1,i= 1.2 m/s

v*2,i = -0.3 m/s


Example 1 continued

Example 1 continued...

Movie

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

m2

v*1,i

v*2,i

m1

m2

m1

x

m2

v*2,f

m1

v*1,f


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:

  • But the total momentum is zero:

  • So:

(and the same for particle 2)

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


Example 13

Example 1...

Step 3

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

m2

v*1,i

v*2,i

m1

m2

m1

x

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

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

m2

m1


Example 14

v1,f= -0.9 m/s

v2,f = 0.6 m/s

Example 1...

v* = v - VCM

Step 4

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

v = v* + VCM

v1,f = v*1,f + VCM = -1.2 m/s + 0.3 m/s = -0.9 m/s

v2,f = v*2,f + VCM = 0.3 m/s + 0.3 m/s = 0.6 m/s

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


Lecture 15 act 3 moving between reference frames

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 frames1

= 20 mi / hr

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

20mi/hr

80mi/hr

- 40mi/hr

CM

Lecture 15, Act 3 Moving Between Reference Frames

  • The velocity of the CM is:

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

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

x


Lecture 15 act 3 aside

20mi/hr

CM

Lecture 15, Act 3 Aside

  • 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?

80mi/hr

- 40mi/hr

x


Lecture 15 act 3 aside solution

Lecture 15, Act 3 Aside Solution

v*GREEN,i = 60 mi/hr

v*RED,i = -60 mi/hr

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

v*GREEN,f = -60 mi/hrv*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


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

(m1v1,i + m2v2,i)

  • VCM =

Step 1

Step 2

v* = v - VCM

Step 3

v*f = -v*i

Step 4

v = v* + VCM


Interesting fact

Interesting Fact

v*1,i

v*2,i

  • 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*2,f = -v*2,i

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

This is really cool and useful too!


Basketball demo

Basketball Demo.

Drop 2 balls

  • 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.

3v

m

v

v

v

v

M

v

(a)

(b)

(c)


More comments on energy

v*1

v*2

v*

v*

v*

v*

v*

v*

= PNET,CM = 0

= KCM

= KREL

More comments on energy

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

but

so

(same for v2)


More comments on energy1

v*

v*

More comments on energy...

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

= KCM

= KREL

So ELAB = KREL + KCM

KCM is the kinetic energy of the center of mass.

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

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


More comments on energy2

More comments on energy...

ELAB = KREL + KCM

  • Does total energy depend on the reference frame??

  • YOU BET!

KRELis independent of the reference frame, but KCM depends on the reference frame (and = 0 in CM reference frame).


Recap of today s lecture

Recap 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 Chapter 11: # 63, 67, 71


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