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05/15/22 Dr. Sasho MacKenzie - HK 376 1 Impacts Impacts An impact occurs when two objects are in contact for a very short period of time.

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Impacts. An impact occurs when two objects are in contact for a very short period of time. Contact vs. Impacts. In general, if two objects are in contact long enough to precisely measure the forces between the objects throughout the contact time, then it is NOT considered an impact. - PowerPoint PPT Presentation

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Page 1: Impacts

04/22/23Dr. Sasho MacKenzie - HK

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ImpactsImpacts

An impact occurs when two objects are in contact for a very short period of time.

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Contact vs. ImpactsContact vs. Impacts• In general, if two objects are in contact long

enough to precisely measure the forces between the objects throughout the contact time, then it is NOT considered an impact.– Jumping: foot-ground or Pitching: hand-ball

• If the contact time is very small and the forces cannot be directly measured, then it is considered an impact.– Golf: ball-clubhead or Baseball: bat-ball

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Impacts or CollisionsImpacts or Collisions• Impacts are commonly referred to

as collisions in physics.• Collisions can be elastic, meaning

they conserve energy and momentum, inelastic, meaning they conserve momentum but not energy, or totally inelastic (or plastic), meaning they conserve momentum and the two objects stick together.

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ImpactsImpacts• When analyzing a contact as an impact,

it is assumed that the collision is an instantaneous event.

• For example, when a ball bounces off the floor, we would consider the ball to have negative velocity just before impact and then positive velocity the next instant.

• Impact analysis requires combining three concepts– Relative Velocity, Coefficient of Restitution,

and Conservation of Momentum.

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Relative VelocityRelative Velocity• When dealing with impact, we need to

know what the velocities of the objects are relative to each other.

• To determine relative velocity, you subtract one velocity from the other.

• Essentially, you are saying if one object’s velocity was zero, what would the velocity of the other object be?

• Vbat – vball (example for baseball hitting)• The relative velocity of “impact” is simply

the absolute value of the above equation.

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Relative Velocity: Heading a Soccer Relative Velocity: Heading a Soccer BallBall

+4 m/s

Ball relative to Player

-10 - (+4) = -14 m/s

Player relative to Ball

+4 - (-10) = +14 m/s

-10 m/s

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Coefficient of Restitution Coefficient of Restitution (e)(e)

• It is a measure of the elasticity of two objects that impact. Values can range from 0 (no bounce) to almost 1 (highest bounce).

• It is calculated using the ratio of the relative velocity of separation to the relative velocity of approach.

• e = - ( V – v ) ..… 1 ( V – v )

It can also be measured as the ratio of the restitution impulse to the compression impulse.

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Conservation of Linear Conservation of Linear MomentumMomentum

• The forces of impact are internal to the system (M + m). The above law states that if no external forces act on the system, then the momentum of the system is the same before and after impact.

• MV + mv = MV + mv ….. 2

• v = M[ V(1+e) – ev] + mv ….. 3 M + m

• Combining equations 1 and 2, we can solve for the velocities of the objects post impact.

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Special CaseSpecial Case• In certain cases, when one of the

objects is at rest prior to impact (such as golf), the equation can be simplified. The velocity of the golf ball is:

• v = MV(1+e) M + m

Where: v is the velocity of the golf ball after impact, m is the mass of the golf ball, V is the velocity of the clubhead prior to impact, M is the mass of the clubhead, and e is the COR.

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Applying the FormulaApplying the Formula• Would a golf ball go farther if you

increased clubhead mass, or increased it’s velocity?

• We can graphically display how each of these variables would affect the velocity of the golf ball after impact.

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Increasing Clubhead MassIncreasing Clubhead MassB

all V

eloc

ity

Max Ball Velocity

Clubhead Mass

M is on both the top and bottom of the equation, so when M gets large, increases in M show very small changes in ball velocity

v = MV(1+e) M + m

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Increasing the Velocity of the Increasing the Velocity of the ClubClub

Bal

l Vel

ocity

Velocity of Golf Club

V is only on the top of the equation, so the bigger it gets, the bigger ball velocity gets.

Ball velocity increases as club velocity increases.

v' = MV(1+e) M + m

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Clubhead Mass Clubhead Velocity Ball Mass e Ball Velocity0.5 40 0.05 0.83 66.5

1 40 0.05 0.83 69.72 40 0.05 0.83 71.4

10 40 0.05 0.83 72.81000 40 0.05 0.83 73.2

0.5 40 0.05 0.83 66.50.5 45 0.05 0.83 74.90.5 50 0.05 0.83 83.20.5 70 0.05 0.83 116.50.5 1000 0.05 0.83 1663.6

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What is the Clubhead What is the Clubhead Velocity?Velocity?

With what velocity must a 0.25 kg clubhead contact a 0.05 kg golf ball in order to send the ball off the tee with a velocity of 80 m/s? e = .81

v = MV(1+e) M + m

Therefore, V = v (M+m) M (1+e)

V = 80 (.25+.05) .25 (1 + .81)

V = 53 m/s

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The Equations of ImpactThe Equations of Impact1)(

vVvVe

2vmVMmvMV

4)1(

mMeMVv

3])1([

mMmveveVMv

Equations 1 and 2 combine to give …

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• A golf ball is dropped from a height of 2 m onto the floor. If it rebounds with a velocity of 3.15 m/s, then what was the coefficient of restitution (e) associated with the impact?

• Before (e) can be determined, the velocity with which the ball hit the floor must be known, then we can apply the formula

• e = -v/v

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Finding impact velocity of Finding impact velocity of dropped balldropped ball

• If we can figure out how long it takes a ball to fall 2 m, then we can find velocity knowing that v=at, and a, in this case is -9.81 m/s2.

• Plotting a graph of the ball’s velocity allows us to determine the time that it spends accelerating to the floor.

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Since acceleration is a constant -9.81, that means that that the velocity curve will have a constant negative slope of -9.81.

V = at = -9.81t

The area under the velocity curve yields displacement. Since the area of a triangle is ½ height x base, we get … d = ½ (-9.81t)(t) = -4.905t2 . Since d=2, we can find t2 = -4.905t2, therefore, t = 0.64 s. Since Vf = Vi + at,Vf = 0 + (-9.81)(0.65) = -6.3 m/s

time of ball impact with floor

velocity

t

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Knowing that the golf ball hit the floor with a velocity of -6.3 m/s and rebounded with a velocity of 3.15 m/s, the coefficient of restitution can be calculated

e = -v/v = -3.15/-6.3 = 0.5

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Baseball ExampleBaseball Example• In which case will the baseball leave the

bat with a greater velocity and thus travel farther?

A. The pitcher throws the ball at -40 m/s and the batter swings at 20 m/s.

B. The pitcher throws the ball at -20 m/s and the batter swings at 40 m/s.

• We could use the impact equations to plug in numbers and find the answer or we could think our way through it.

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Relative Velocity of ImpactRelative Velocity of Impact• In both examples, the relative velocity of

impact is going to be the same.A. V – v = 20 – (-40) = 60 m/sB. V – v = 40 – (-20) = 60 m/s• The objects do not know what their

velocities are, so the characteristics of impact will be the same.

• The same forces, for the same period of time in both A and B.

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Impulse during ImpactImpulse during Impact• Impulse = Ft• If the forces are the same and the length

of time the forces act are the same, then the impulses must be the same for both cases.

• Impulse = change in momentum = mv• Therefore, the change in velocity of the

ball (v) equals the Impulse divided by the ball’s mass

Ft = mv v = Ft / m

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Ball’s Change in Velocity (Ball’s Change in Velocity (v)v)• If the impulse is the same for both A and

B, then the v is the same for both A and B.

• Since final velocity is equal to initial velocity plus change in velocity, we can come to a conclusion.

• vf = vi + v, where v = Ft / m• vf = vi + Ft / m

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ConclusionConclusion• vf = vi + Ft / m

Same for both A and B

The ball’s initial velocity is different in A and B.

A. vf = -40 + Impulse/massB. vf = -20 + Impulse/mass

In case B, the ball will leave the bat with a greater positive velocity, and thus travel further.