ics Thursday, ober 14, 2004 Ch 9: Collisions Elastic Perfectly Inelastic Inelastic We ll finish up energy conservation after the break
Announcements Sunday, 6:30-8 pm in CCLIR 468 Midnight Madness: Monday 8 - midnight in NSC 118/119
Announcements Don t forget to read over the lab and prepare for the short quiz.
Announcements Tuesday,. 19, 2004 11:50 am 1:05 pm Ch. 5, 6, 9.1-9.4 Hint: Be able to do the homework (graded AND recommended) and you ll do fine on the exam! You may bring one 3 X5 index card (handwritten on both sides), a pencil or pen, and a scientific calculator with you. I will put any constants and mathematical formulas that you might need on a single page attached to the back of the exam.
Announcements Tuesday,. 19, 2004 11:50 am 1:05 pm Ch. 5, 6, 9.1-9.4 Format: Three sections (like homework). You do any two of the three. One essay question. Required for all. One section of multiple choice. 5 questions roughly evenly divided over Required for all.
Announcements Mastering ics Practice Exam #1 Score 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 Exam % = 0.91N + 65.7 0 10 20 30 Extra Points
We ve looked at a simple collision in which two balls strike each other head-on. The momentum is ALWAYS conserved for an isolated system. The result of such a collision usually fits into one of three categories: ELASTIC INELASTIC PERFECTLY INELASTIC
ELASTIC Both momentum & kinetic energy are conserved. K 1i = 1 m v 2 2 1 1i p 1i =m 1 v 1i v m 1i v 2i 1 m 2 K 2i = 1 m v 2 2 2 2i p 2i =m 2 v 2i K 1 f = 1 m v 2 2 1 1 f v 1f m 1 m 2 v 2f K 2 f = 1 m v 2 2 2 2 f p 1f =m 1 v 1f p 2f =m 2 v 2f
K 1i = 1 m v 2 2 1 1i p 1i =m 1 v 1i v m 1i v 2i 1 m 2 ELASTIC K 2i = 1 m v 2 2 2 2i p 2i =m 2 v 2i K i = 1 m v 2 + 1 m v 2 2 1 1i 2 2 2i p i = m 1 v 1i + m 2 v 2i K f = 1 m v 2 2 1 1 f + 1 m v 2 2 2 2 f p f = m 1 v 1f + m 2 v 2f K 1 f = 1 m v 2 2 1 1 f v 1f m 1 m 2 v 2f K 2 f = 1 m v 2 2 2 2 f p 1f =m 1 v 1f p 2f =m 2 v 2f
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? Motion Diagrams: Before: a 2 = 0 v 2 = 0 Problem #1 v 1 a 1 = 0 After: v 2 a2 = 0 v 1 a 1 = 0
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? Pictorial Representation: Before: v 1,i Unknowns: v 1, f & v 2, f After: v 2,i = 0 v 1, f v 2, f Knowns: m 1 = m 2 = 2.0 kg v 1,i = +10 m s ˆx v 2,i = 0 p f = p i K f = K i
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? Mathematical Representation: By conservation of momentum: mv 1i = mv 1 f + mv 2 f v 2 f = v 1i v 1 f
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? Mathematical Representation: By conservation of kinetic energy: 1 2 mv 1i 2 v 2 f 2 = 1 2 mv 1 f = v 2 v 1i 2 1 f 2 + 1 mv 2 2 2 f (v 1i v 1 f ) 2 = v 2 2 v 1i 1 f
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? (v v ) 2 = v 2 v 2 1i 1 f 1i 1 f (v 1i v 1 f ) 2 = (v 1i v 1 f )(v 1i + v 1 f ) (v 1i v 1 f ) = (v 1i + v 1 f ) 2v 1 f = 0 v 1f = 0
A 2-kg object (1) moves with a speed of 10 m/s straight towards another 2-kg object (2) that is initially at rest. What are the final speeds of these two objects after the collision if they collide elastically? We also know that momentum must be conserved. The initial momentum is (2 kg) (10 m/s) to the right, So that must be the final momentum as well. Which means v 1f = 0 v 2f = 10 m/s to the right.
When two equal mass objects collide elastically, they swap velocities! Let s now see what happens in the general case for two unequal masses. (Skip proof)
K i = K f ELASTIC 1 2 m 1 v 2 + 1 1i 2 m 2 v 2 = 1 2i 2 m 1 v 2 1 f + 1 2 m 2 v 2 2 f Rearranging terms, we get m (v 2 1 1 f v 2 ) = m (v 2 1i 2 2 f v 2 ) 2i m 1 (v 1 f v 1i )(v 1 f + v 1i ) = m 2 (v 2 f v 2i )(v 2 f + v 2i )
ELASTIC m 1 (v 1 f v 1i )(v 1 f + v 1i ) = m 2 (v 2 f v 2i )(v 2 f + v 2i ) But we know from conservation of momentum p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1 f + m 2 v 2 f m 1 (v 1 f v 1i ) = m 2 (v 2 f v 2i ) Elastic - derive rel. vel. Eqn.
ELASTIC m 1 (v 1 f v 1i )(v 1 f + v 1i ) = m 2 (v 2 f v 2i )(v 2 f + v 2i ) m 1 (v 1 f v 1i ) = m 2 (v 2 f v 2i ) Plugging the 2nd equation into the first, we get v 1 f + v 1i = v 2 f + v 2i And finally, rearranging this expression... Elastic - derive rel. vel. Eqn.
ELASTIC v v = ( v v ) 2 f 1 f 2i 1i The relative velocity of recession after the collision of two objects that collide elastically is equal in magnitude and opposite in direction to the relative velocity of approach before the collision. Elastic - Relative Velocity
Worksheet Problem #1
Thurs Worksheet Problem #1 A golf ball is fired at a bowling ball initially at rest and bounces back elastically. Compared to the bowling ball, the golf ball after the collision has 1. more momentum but less kinetic energy. 2. more momentum and more kinetic energy. 3. less momentum and less kinetic energy. 4. less momentum but more kinetic energy. 5. none of the above PI, Mazur (1997)
PERFECTLY INELASTIC The two objects stick together after the collision, so their final velocities are identical. K 1i = 1 m v 2 2 1 1i p 1i =m 1 v 1i v m 1i v 2i 1 m 2 K 2i = 1 m v 2 2 2 2i p 2i =m 2 v 2i K i = 1 m v 2 + 1 m v 2 p 2 1 1i 2 2 2i i = m 1 v 1i + m 2 v 2i K 1 f = 1 m v 2 2 1 f p 1f =m 1 v f m 1 v 1f = v 2f = v f v K f 2 f = 1 m v 2 2 2 f m 2 p 2f =m 2 v f
PERFECTLY INELASTIC The momentum of the system is conserved. p i = p f p = m v + m v = ( m + m ) v = p i 1 1i 2 2i 1 2 f f The kinetic energy of the system is NOT conserved NOT conserved. K i K f
Worksheet #2 CQ2: perfectly inelastic
Thurs Worksheet #2 If all three collisions in the figure shown are totally inelastic, which cause(s) the most damage? 1. I 2. II 3. III 4. I & II 5. I & III 6. II & III 7. all three cause equal amounts of damage PI, Mazur (1997)
INELASTIC Most real world collisions fall into this category which falls somewhere between the elastic collisions and the perfectly inelastic collisions. The momentum of the system is conserved. p = m v + m v = m v + m v = p i 1 1i 2 2i 1 2 f 2 2 f f The kinetic energy of the system is NOT conserved p i = p NOT conserved. K i K f f
A 5 kg object moves with a velocity of +10 m/s straight towards a 3 kg object that is initially at rest. After the collision, the 3 kg object is observed to have a velocity of +12 m/s. What is the velocity of the 5 kg object after the collision? Was the collision elastic? If not, how much energy was lost? Motion Diagrams: Before: v 1 a 1 = 0 After: a 2 = 0 v 2 = 0 v 1 a 1 = 0? 5 kg 3 kg collision Problem #2 v 2 a2 = 0
A 5 kg object moves with a velocity of +10 m/s straight towards a 3 kg object that is initially at rest. After the collision, the 3 kg object is observed to have a velocity of +12 m/s. What is the velocity of the 5 kg object after the collision? Was the collision elastic? If not, how much energy was lost? Pictorial Representation: Before: v 1,i Unknowns: v 2, f After: x v 2,i = 0 v 1, f v 2, f Knowns: m 1 = 5.0 kg m 2 = 3.0 kg v 1,i = +10 m s ˆx v 2,i = 0 v 2, f = p f = p +12 m s ˆx i? 5 kg 3 kg collision (P2)
A 5 kg object moves with a velocity of +10 m/s straight towards a 3 kg object that is initially at rest. After the collision, the 3 kg object is observed to have a velocity of +12 m/s. What is the velocity of the 5 kg object after the collision? Was the collision elastic? If not, how much energy was lost? Mathematical Representation: p i = m 1 v1i + m 2 v2i = (5 kg)(10 m/s ˆx) + 0 = 50 N/s ˆx p f = p i = 50 N/s ˆx = m 1 v1 f + m 2 v2 f
A 5 kg object moves with a velocity of +10 m/s straight towards a 3 kg object that is initially at rest. After the collision, the 3 kg object is observed to have a velocity of +12 m/s. What is the velocity of the 5 kg object after the collision? Was the collision elastic? If not, how much energy was lost? 50 N/s ˆx = (5 kg) v 1 f + (3 kg)(12 m/s ˆx) (5 kg) v 1 f = (50 36) N/s ˆx v 1 f = 2.8 m/s ˆx
Was the collision elastic? If not, how much energy was lost? Now, let s look at the kinetic energy... K = 1 m v 2 + 1 m v 2 = 1 (5 kg)(10 i 2 1 1i 2 2 2i 2 m/s)2 + 0 = 250 J K f = 1 m v 2 2 1 1 f + 1 m v 2 2 2 2 f K f K f = 1 (5 kg)(2.8 2 m/s)2 + 1 (3 kg)(12 m/s)2 2 = (19.6 + 216) J = 235.6 J ΔK = K f K i = (235.6 250) J = 14.4 J inelastic
Data gathered from the scene of an accident is put into a computer model to produce a simulation of the accident. The following accident did not happen.
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