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Physics 111: Lecture 15 Today’s AgendaPowerPoint Presentation

Physics 111: Lecture 15 Today’s Agenda

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Physics 111: Lecture 15 Today’s Agenda

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- Elastic collisions in one dimension
- Center of mass reference frame
- Colliding carts problem

- Some interesting properties of elastic collisions
- Killer bouncing balls

- 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.

- 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)

- Energy is lost:
- 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.

- 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

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

v

m

m

m

m

v / 2

x

- Compute kinetic energies:

v

m

m

m

m

v / 2

- 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

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

- 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 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

m2

m1

final

v2,f

v1,f

m2

m1

initial

v1,i

v2,i

x

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!

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

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

- 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.

- 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

- The external force in the xdirection is zero (frictionless):
- The CM of the systems can’t move!

X

X

X

X

x

CM

CM

- 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

- 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

- 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

Plugging vP=V - vR into thiswe find:

- 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

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

- Four step procedure
- First figure out the velocity of the CM, VCM.
- VCM = (m1v1,i + m2v2,i), butv2,i= 0 so
VCM = v1,i

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

- First figure out the velocity of the CM, VCM.
- So VCM = 1/5 (1.5 m/s) = 0.3 m/s

Step 1

(for v2,i= 0only)

- 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*

- 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

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

- 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

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

v1,f= -0.9 m/s

v2,f = 0.6 m/s

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!!

- 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

= 20 mi / hr

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

20mi/hr

80mi/hr

- 40mi/hr

CM

- 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

20mi/hr

CM

- 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

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

: 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

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!

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)

v*1

v*2

v*

v*

v*

v*

v*

v*

= PNET,CM = 0

= KCM

= KREL

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

but

so

(same for v2)

v*

v*

- 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

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).

- 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