Physics 2111 Unit 12 - cod.edu · Physics 2111 Unit 12 Today’s Concepts: a) Elastic Collisions b)...

Physics 2111

Unit 12

Today’s Concepts:

a) Elastic Collisions

b) Center-of-Mass Reference Frame

c) 2D collisions

Mechanics Lecture 12, Slide 1

Your Comments

Didn't understand the center of mass frame!

A bit confused on calculating the velocities after a 2D collision.

the slide on 2D elastic collision didn't do a good job explaining it - in my opinion

2D conservation of momentum

lets do some examples like the double mass half speed checkpoint one where we do not know the final velocities

The whole idea of the center of mass reference frame and how to begin with these problems is really unclear to me. A large array of examples of these types of problems and how to start them to at least get a pattern of recognition for how to approach these would be really helpful. Also a clarification on how to quickly and easily tell if a problem or situation is elastic or inelastic would be appreciated as that isn't super clear to me either.

more examples to be done in class

I keep thinking of elastic collisions as objects that stick together. Stop it me! Lol. Got anything, like an example that helps you remember the difference btw these two collisions?


Where are we?


Tool #1 (Newton’s Laws)

Tool #2 (Newton’s Laws Rephrased)

Tool #2.5 (Work-Energy Rephrased)

Tool #3 (Newton’s Law Rephrased)

Tool #3.5 (Impulse-Momentum Rephrasd)

Tool #3 (Just more examples)

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Prelecture

This collision is:

A) perfectly inelastic

B) perfectly elastic

C) neither perfectly elastic or in-elastic

D) depends on v

A block having mass M slides on a frictionless surface and collides with a block of mass 2M which is initially at rest. After the collision the small block is at rest and the larger block is moving to the right with speed V/2.


CheckPoint

A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. Compare the initial and final kinetic energies of the system.

A) Kinitial > Kfinal

B) Kinitial = Kfinal

C) Kinitial < Kfinal

initial

final

Perfectly Elastic Collisions


WNC = 0 = DME

If DPE = 0 (and it normally is) then

DKE = 0

KEo = KEf

What does this imply?

(v1o – v2o) = - (v1f – v2f)

Example 12.1 (Elastic Collision)

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 glider of mass m2 = 0.8 kgmoving to the left at v2i = 0.2m/sec 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?


m1m2

m2m1

x

m1m2

Velocities of Recession and Approach


Special Case


What if v2o = 0?

v2f = (v1o)*2m1/(m1+m2)

v1f = (v1o)*(m1 – m2)/(m1+m2)


Question

What is the final velocity of m1

if m1 = m2?

A. 2*v1o

B. v1o

C. zero

D. -v1o

E. -2*v1o

v2f = (v1o)*2m1/(m1+m2)

v1f = (v1o)*(m1 – m2)/(m1+m2)


Question

What is the final velocity of m1

if m1 <<< m2?

A. 2*v1o

B. v1o

C. zero

D. -v1o

E. -2*v1o

v2f = (v1o)*2m1/(m1+m2)

v1f = (v1o)*(m1 – m2)/(m1+m2)

Example 12.2 (Special Case Formulas)

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?


m1m2

m2m1

x

m1m2

Velocities of Recession and Approach


Center-of-Mass Frame

In the CM reference frame, vCM = 0

1 1 2 2

1 2

...

...CM

m v m vV

m m

=

tot

tot

P

M=

In the CM reference frame, PTOT = 0


CheckPoint

Before Collision

Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other.

The velocity of the center of mass of this system before the collision is

A) Toward the left B) Toward the right C) zero

m 2m

2v v


CheckPoint

after Collision

Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other.

The velocity of the center of mass of this system after the collision is

A) Toward the left B) Toward the right C) zero

m 2m

?


Center of Mass Frame & Elastic Collisions

When viewed in the CM frame, the speed of both objects is the same before and after an elastic collision. They just reverse directions.


m1m2

m1m2v*

1,iv*

2,i

m2v*1, f v*

2, fm1

Example 12.3 (Center of Mass Calculation)



m1m2

m2m1

x

m1m2

Example 12.3 (cont 1)


Four step procedure:

First figure out the velocity of the CM, vCM.

vCM = (m1v1,i + m2v2,i), so

So vCM = (0.2kg*1.5 m/s + 0.8kg*(-0.2m/sec))/(1kg)

= 0.14 m/s

21

1

mm

Step 1:


Now consider the collision viewed from a frame moving with the CM velocity VCM.


m1m2

m1m2

v*1,i

v*2,i

m2

v*1, f

v*2, f

m1

v*1,i = v1,i - vCM = 1.5 m/s - 0.14 m/s = 1.36 m/s

v*2,i = v2,i - vCM = -0.2 m/s - 0.14 m/s = -0.34 m/s

v* = v - vCM

Step 2:



vCM

v

v*

v = v* vCM

v*1,i = 1.36 m/s

v*2,i = -0.34 m/s

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


v*1, f = -v*

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

2,i

Use the fact that the speed of each block is the samebefore and after the collision in the CM frame.

Step 3:

x


m1m2

m1 m2

V*1,i

V*2,i

v*1, f = - v*

1,i = -1.36m/s v*2, f = - v*

2,i =.34 m/s

m2V*1, f V*

2, fm1



Calculate the final velocities back in the lab reference frame:

v1, f = v*1, f vCM = -1.36 m/s 0.14 m/s = -1.22 m/s

v2, f = v*2, f vCM = 0.34 m/s 0.14 m/s = 0.48 m/s

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

Step 4:

vCM

v

v*

v = v* VCM

v1, f = -1.22 m/s

v2, f = 0.48 m/s

Example 12.3 (Center of Mass Calculation)



m1m2

m2m1

x

m1m2

Example 12.4 (Mechanical Energy)

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 glider of mass m2 = 0.8 kgmoving to the left at v2i = 0.2m/sec What are the values of the mechanical energy in

- the lab frame of reference and

- the center of mass frame of reference?


m1m2

m2m1

x

m1m2

Example 12.5 (Spring compression)

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 glider of mass m2 = 0.8 kgmoving to the left at 0.2m/sec. A spring with k = 20N/m is attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What is the maximum compression of the spring?


m1m2

m2m1

x

m1m2

Example 12.6 (2D inelastic blocks)


m1

m2

2m/secTwo blocks, m1=3kg and m2 = 2kg,

are sliding across a frictionless

floor as shown. They collide and

stick together.

What is the direction and

magnitude of their final velocity?

5m/sec

30o

Example 12.7 (2D elastic pucks)


m1

Two pucks, m1=2kg and m2 = 1.2kg, collide elastically

while sitting on a frictionless table. They collide

slightly off center, so that m1 is set upwards while m2

is sent downwards.

What is the values of angles f and a?

v1o = 3m/sec

m2

f

a

Physics 2111 Unit 12 - cod.edu · Physics 2111 Unit 12 Today’s Concepts: a) Elastic Collisions b)...

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