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•  Spring break starts tomorrow! •  Finish up collisions today •  Start Chapter 10 (Rotational Motion).

Collisions and rotational kinematics

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Collision summary mA!vAi +mB

!vBi =mA!vAf +mB

!vBfIn all collisions, momentum is conserved:

mAvAxi +mBvBxi =mAvAxf +mBvBxfmAvAyi +mBvByi =mAvAyf +mBvByf

This is a vector equation. Components are conserved.

In perfectly inelastic collisions, final velocities are the same

!vAf =!vBf

12mAvAi

2 + 12mBvBi

2 = 12mAvAf

2 + 12mBvBf

2For elastic collisions, kinetic energy is conserved.

vAxf = vBxfvAyf = vByfComponents:

Remember momentum and velocities have signs

However kinetic energy is always positive

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Elastic collision example Two air carts have equal mass and springs attached. Cart A is moving toward cart B with speed vAi. Cart B is initially at rest. What is the speed of the carts after the collision?

The carts are on a track and have springs (conservative force). This is an elastic collision in 1D which gives us two equations:

BfBAfABiBAiA vmvmvmvm +=+2BfB2

12AfA2

12BiB2

12AiA2

1 vmvmvmvm +=+

Since masses are equal, can cancel them out. Also, vBi=0

BfAfAi vvv += 2Bf

2Af

2Ai vvv +=

Substituting gives 02or 2 BfAf2Bf

2Af

2BfBfAf

2Af =+=++ vvvvvvvv

Either vAf or vBf is 0 and the other must equal vAi. Since the carts can’t pass through each other, cart A must stop and cart B travels with velocity vAi.

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A big ball, mass M=10m, speed v, strikes a small ball, mass m, at rest. Could the following occur: The big ball comes to a complete stop and the small ball takes off with speed 10v?

Clicker question 1 Set frequency to BA

A.  Yes, this could occur B.  No, it would violate conservation of momentum C.  No, it would violate conservation of energy D.  Need more information.

m10mv

m10m v10

Momentum is conserved:

Pi =10mvPf =10mv

Kinetic energy is not conserved:

Ki =1210mv

2 = 5mv2

K f =12m(10v)

2 = 50mv2

We can lose kinetic energy during a collision (that is the definition of an inelastic collision) but we cannot gain kinetic energy.

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Other examples of elastic collisions Newton’s cradle is a good example of elastic collisions

Need to conserve both kinetic energy and momentum

The basketball and tennis ball experiment is a little complicated to work out so I will just set it up and then jump to the answer. The full details will be in the posted slides.

The text book works out the results of elastic collisions in more depth.

An elastic example A tennis ball is put on top of a basketball and both are dropped from a height of 1 m. How high will the tennis ball go? Assume all collisions are elastic.

Can use conservation of energy to get velocity after falling 1 m. becomes K1 +U1 = K2 +U2

mgh = 12mv

2 so v = 2gh = 2 ⋅10 m/s2 ⋅1 m = 4.5 m/s

Basketball hits first. What happens?

If no friction losses then by conservation of energy the upward speed after bounce equals the downward speed before bounce

Is momentum conserved? Only if you consider Earth as part of the system

Or, elastic collisions conserve kinetic energy. Same answer.

An elastic example So we have a 1D elastic collision between basketball moving at vB = 4.5 m/s and tennis ball moving at vT = –4.5 m/s

The collision conserves momentum: mBvB1 +mTvT1 =mBvB2 +mTvT2

Elastic collision conserves kinetic energy:

12mBvB1

2 + 12mTvT1

2 = 12mBvB2

2 + 12mTvT2

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Solve this problem in the reference frame where the basketball is at rest

vB1 = 0 m/svT1 = −9.0 m/s

mTvT1 =mBvB2 +mTvT2 mTvT12 =mBvB2

2 +mTvT22This gives and

( )( ) T1

BT

TT1

BT

BTT1T2T1B2

2 vmmmvmm

mmvvvv+

=+−

+=+=

Substitute back in to get

An elastic example mTvT1 =mBvB2 +mTvT2 mTvT1

2 =mBvB22 +mTvT2

2We have and

Divide the equations to get vB2 = vT1 + vT2mBvT1 +mBvT2 =mTvT1 −mTvT2

mBvB2 =mTvT1 −mTvT2 =mT vT1 − vT2( )mBvB2

2 =mTvT12 −mTvT2

2 =mT vT12 − vT2

2( ) =mT vT1 − vT2( ) vT1 + vT2( )Factoring:

Substitute in to get

mTvT2 +mBvT2 =mTvT1 −mBvT1 mT +mB( )vT2 = mT −mB( )vT1Rearrange:

becomes

( )( ) T1

BT

BTT2 vmm

mmv+−

=Solve:

An elastic example

vT2 =mT −mB( )mT +mB( )

vT1So in the basketball rest frame we have a post collision tennis ball velocity of

vT2 =.07 kg−.4 kg( )

.07 kg + .4 kg( )−9.0 m/s( ) = 6.3 m/s

This is with respect to the basketball which is moving up at 4.5 m/s. Therefore the tennis ball speed with respect to the ground is 6.3 m/s + 4.5 m/s = 10.8 m/s.

mB = .4 kg mT = .07 kgvB1 = 0 m/s vT1 = −9.0 m/sMasses and initial velocities:

Basketball and tennis ball Before collision, the tennis ball is moving down at 4.5 m/s

After collision the tennis ball is moving up at 10.8 m/s

What height will the tennis ball reach?

( ) m 8.5m/s 102m/s 8.10

2 so 21

2

222 =

⋅=== g

vhmghmv

19 foot height from just a 3 foot drop?!

Inelastic vs elastic collisions We have two balls with different material properties:

The sad ball just goes squash and lies there like a lump

We will use each ball connected to the end of a pendulum to try to knock over a block of wood.

The happy ball bounces right back like a champ

Both balls have the same mass and velocity just before hitting the block and therefore the same momentum: p = mv

Two balls of equal mass m are used as the bob of a pendulum and are released from the same point. Which ball is more likely to knock over the block of wood?

Clicker question 2 Set frequency to BA

A.  The ball that splats on the wood (sad ball) B.  The ball that bounces off the wood (happy ball) C.  It doesn’t make a difference which ball is used

Sad ball ends with p ≈ 0 so by conservation of momentum, the block gains momentum Δp ≈ mv (it gets an impulse of mv).

Happy ball ends with p ≈ −mv (because it bounces off). By conservation of momentum, the block gains momentum of Δp ≈ 2mv.

mballvball,i + 0 =mballvball,f +mblockvblock,fConservation of momentum:

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Angular kinematics Early in the semester, we dealt with kinematics which involved displacement, velocity, and acceleration.

Angular kinematics is the same thing but for objects which are rotating (rather than translating).

For something to rotate, it must have an axis about which it rotates like the axle for a wheel.

Only sensible place for the origin is along the axis.

r θ

Use polar coordinates (r,θ) instead of Cartesian coordinates (x,y). Axis is in the z-direction.

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Angular kinematics θ is a measure of angular position

r θ

1 revolution (1 rev) = 2π radians (2π rad) = 360 degrees (360°)

Δθ Δθ is a measure of angular displacement

In angular motion (and many other areas of physics) it is much more convenient to use radians (abbreviated rad) rather than degrees or revolutions to measure angles.

To convert you just need to remember

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Angular velocity Angular velocity tells us how fast (and in what direction) something is spinning.

ωz, avg =ΔθΔt

ωz =dθdt

Counterclockwise is positive

The z subscript indicates the axis is in the z-direction (and the rotation is therefore in the xy plane)

Also have angular acceleration which describes how the spinning rate changes

αz, avg =ΔωΔt

αz =dωdt

=d 2θdt2

r θ Δθ

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Angular kinematics The same equations which were derived for constant linear (also known as translational) acceleration apply for constant angular (also known as rotational) acceleration

ωz =ω0z +αztθ =θ0 +ω0zt +

12αzt

2

ωz2 =ω0z

2 + 2αz θ −θ0( )

Constant angular acceleration only!

Constant linear acceleration only!

vx = v0 x + atx = x0 + v0 xt +

12 axt

2

vx2 = v0 x

2 + 2ax x − x0( )

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Relationships to linear velocity If we want the linear displacement or velocity of a point on a rotating object, we need r and either θ or ω. What is the speed of the tire rim if the radius is 0.35 m and the tire rotates at 20 rad/s?

Radians measure distance around a unit circle

Therefore (only for θ in radians!) s = rθ

r θ

s

Time derivative of both sides (r is constant): dsdt

= r dθdt

v = rω . Linear speed is radius times angular speed Rim speed is v = rω = 0.35 m ⋅20 rad/s = 7 m/s