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5/11/2015 Dr. Sasho MacKenzie - HK 3761 Impacts An impact occurs when two objects are in contact for a very short period of time.

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Presentation on theme: "5/11/2015 Dr. Sasho MacKenzie - HK 3761 Impacts An impact occurs when two objects are in contact for a very short period of time."— Presentation transcript:

1 5/11/2015 Dr. Sasho MacKenzie - HK 3761 Impacts An impact occurs when two objects are in contact for a very short period of time.

2 5/11/2015 Dr. Sasho MacKenzie - HK 3762 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. –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

3 5/11/2015 Dr. Sasho MacKenzie - HK 3763 Impacts 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.elasticenergy momentuminelastic

4 5/11/2015 Dr. Sasho MacKenzie - HK 3764 Impacts 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.

5 5/11/2015 Dr. Sasho MacKenzie - HK 3765 Relative 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? V bat – v ball (example for baseball hitting) The relative velocity of “impact” is simply the absolute value of the above equation.

6 5/11/2015 Dr. Sasho MacKenzie - HK 3766 Relative Velocity: Heading a Soccer Ball +4 m/s Ball relative to Player (+4) = -14 m/s Player relative to Ball +4 - (-10) = +14 m/s -10 m/s

7 5/11/2015 Dr. Sasho MacKenzie - HK 3767 Coefficient of Restitution (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.

8 5/11/2015 Dr. Sasho MacKenzie - HK 3768 Conservation of Linear Momentum 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.

9 5/11/2015 Dr. Sasho MacKenzie - HK 3769 Special 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.

10 5/11/2015 Dr. Sasho MacKenzie - HK Applying 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.

11 5/11/2015 Dr. Sasho MacKenzie - HK Increasing Clubhead Mass Ball Velocity 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

12 5/11/2015 Dr. Sasho MacKenzie - HK Increasing the Velocity of the Club Ball Velocity 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

13 5/11/2015 Dr. Sasho MacKenzie - HK 37613

14 5/11/2015 Dr. Sasho MacKenzie - HK What is the Clubhead 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 (1 +.81) V = 53 m/s

15 5/11/2015 Dr. Sasho MacKenzie - HK The Equations of Impact Equations 1 and 2 combine to give …

16 5/11/2015 Dr. Sasho MacKenzie - HK 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

17 5/11/2015 Dr. Sasho MacKenzie - HK Finding impact velocity of dropped ball If we can figure out how long it takes a ball to fall 2 m, then we can find velocity knowing that  v=a  t, and a, in this case is m/s 2. Plotting a graph of the ball’s velocity allows us to determine the time that it spends accelerating to the floor.

18 5/11/2015 Dr. Sasho MacKenzie - HK Since acceleration is a constant -9.81, that means that that the velocity curve will have a constant negative slope of V = a  t =  t The area under the velocity curve yields displacement. Since the area of a triangle is ½ height x base, we get … d = ½ (-9.81  t)(  t) =  t 2. Since d=2, we can find  t 2 =  t 2, therefore,  t = 0.64 s. Since V f = V i + a  t, V f = 0 + (-9.81)(0.65) = -6.3 m/s time of ball impact with floor velocity  t

19 5/11/2015 Dr. Sasho MacKenzie - HK 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

20 5/11/2015 Dr. Sasho MacKenzie - HK Baseball 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.

21 5/11/2015 Dr. Sasho MacKenzie - HK Relative Velocity of Impact In both examples, the relative velocity of impact is going to be the same. A.V – v = 20 – (-40) = 60 m/s B.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.

22 5/11/2015 Dr. Sasho MacKenzie - HK Impulse during Impact Impulse =  F  t 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 = m  v Therefore, the change in velocity of the ball (  v) equals the Impulse divided by the ball’s mass  F  t = m  v  v =  F  t / m

23 5/11/2015 Dr. Sasho MacKenzie - HK Ball’s Change in Velocity (  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. v f = v i +  v, where  v =  F  t / m v f = v i +  F  t / m

24 5/11/2015 Dr. Sasho MacKenzie - HK Conclusion v f = v i +  F  t / m Same for both A and B The ball’s initial velocity is different in A and B. A. v f = Impulse/mass B. v f = Impulse/mass In case B, the ball will leave the bat with a greater positive velocity, and thus travel further.


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