1. MOMENTUM AND COLLISION CHAPTER 6 Momentum video

2. LINEAR MOMENTUM, P, IS THE PRODUCT OF AN OBJECT’S MASS, M, AND VELOCITY, V. P = MV, WHERE P IS THE MOMENTUM, M IS THE MASS IN KG, AND V IS THE VELOCITY OF THE OBJECT IN M/S. THE UNIT FOR MOMENTUM, P, IS KGM/S. MOMENTUM HAS THE SAME DIRECTION AS THE VELOCITY. THUS, MOMENTUM IS A VECTOR QUANTITY. MOMENTUM DESCRIBES AN OBJECT'S MOTION. THE GREATER THE MASS OF AN OBJECT, THE GREATER THE MOMENTUM. Can a bullet and an elephant have a large momentum? How?

3. MOMENTUM IS OFTEN DISCUSSED IN SPORTS. FOOTBALL PLAYERS QUITE FREQUENTLY REFER TO HOW MUCH MOMENTUM THEY HAVE. IT IS ADVANTAGEOUS IN FOOTBALL NOT ONLY TO HAVE A LOT OF MASS, BUT HAVE A LOT OF SPEED AS WELL. THE MORE MOMENTUM A PLAYER HAS, THE MORE DIFFICULT IT IS TO STOP THE PLAYER. The same applies to Rugby players

4. SOME RUGBY VIDEOS

5. YOU ARE SLIDING DOWN A HILL ON A SLED, OR COASTING DOWN A HILL ON YOUR BICYCLE. YOU ARE ACCELERATING BECAUSE OF GRAVITY. THE FASTER YOU MOVE, THE MORE MOMENTUM YOU HAVE AND THE MORE DIFFICULT IT IS TO COME TO A STOP. IN GOING DOWN THE HILL, YOU ARE ‘PICKING UP SPEED’ OR ‘GAINING MOMENTUM’.

6. THE MOMENTUM OF A BOWLING BALL, AND THUS ITS ABILITY TO KNOCK OVER PINS, DEPENDS ON THE BALL’S MASS AND SPEED. MOMENTUM IS TRANSFERRED FROM THE BALL TO THE PIN AS THE VELOCITY OF THE BALL DECREASES WHILE THE VELOCITY OF THE PIN INCREASES. IS THE PIN MORE LIKELY TO MOVE RAPIDLY WHEN THE BALL TRAVELS AT A HIGH SPEED OR A LOW SPEED? WHY? THE HIGH SPEED BALL SINCE THERE IS MORE MOMENTUM AVAILABLE TO TRANSFER TO THE PIN TO CAUSE A VELOCITY CHANGE.

7. BOWLING BALL 1 HAS A MASS OF 3 KG. BOWLING BALL 2 HAS A MASS OF 7 KG. THE TWO BOWLING BALLS MOVE AT THE SAME SPEED, V = 4 M/S. WHICH BALL IS MORE LIKELY TO MOVE THE PIN? SUPPORT YOUR ANSWER MATHEMATICALLY AND EXPLAIN. M 1 = 3 KGM 2 = 7 KGBALL 2 IS MORE LIKELY TO MOVE THE PIN SINCE IT V 1 = 4 M/SV 2 = 4 M/SHAS MORE MASS AT THE SAME VELOCITY AND P 1 = MVP 2 = MVTHEREFORE HAS MORE MOMENTUM TO TRANSFER. P 1 = (3)(4)P 2 = (7)(4) P 1 = 12 KGM/SP 2 = 28 KGM/S

8. A SMALL OBJECT MOVING WITH A VERY HIGH VELOCITY HAS A LARGE MOMENTUM. HAIL FALLING FROM VERY HIGH CLOUDS CAN BE DANGEROUS, AS THEIR HIGH VELOCITY GIVES THEM ENOUGH MOMENTUM TO HURT YOU.

9. A PENNY HAS A MASS OF 2.5 G. IF YOU DROP THE PENNY OFF THE EMPIRE STATE BUILDING, IT REACHES A VELOCITY OF 48.6 M/S WHEN IT HITS THE GROUND. WHAT IS THE PENNY’S MOMENTUM WHEN IT REACHES THE GROUND? M = 0.0025 KG V = 48.6 M/S P = MV. P = (0.0025)(48.6) P = 0.1215 KGM/S

10. WHICH HAS THE GREATEST MOMENTUM: BALL 1 OF MASS 3.1 KG TRAVELING AT 15 M/S OR BALL 2 OF MASS 8.2 KG TRAVELING AT 5.4 M/S? SUPPORT YOUR ANSWER MATHEMATICALLY. M 1 = 3.1 KGM 2 = 8.2 KGBALL 1 HAS MORE MOMENTUM. V 1 = 15 M/SV 2 = 5.4 M/S P 1 = MVP 2 = MV P 1 = (3.1)(15)P 2 = (8.2)(5.4) P 1 = 46.5 KGM/SP 2 = 44.28 KGM/S

11. A CHANGE IN MOMENTUM,  P, IS AN OBJECT’S MASS TIMES A CHANGE IN ITS VELOCITY,  V.  P = P F – P I AND P F = MV F, P I = MV I THEN  P = MV F – MV I A BALL WITH A MASS OF 0.11 KG AND A VELOCITY OF 3.5 M/S STRIKES A WALL AND BOUNCES STRAIGHT BACK WITH A VELOCITY OF 1.6 M/S. WHAT IS THE CHANGE IN MOMENTUM OF THE BALL? M = 0.11 KG V I =+ 3.5 M/S V F = - 1.6 M/S Δ P = MV F - MV I Δ P = 0.11(-1.6) – 0.11(3.5) Δ P = -0.561 KGM/S

12. FROM NEWTON’S 2 ND LAW, F= MA, AND SUBSTITUTING A AV =  V/  T: F = M  V /  TAND SINCE  V = V F – V I F = M (V F – V I ) /  TF = MV F – MV I /  T AND SINCE  P = MV F – MV I F =  P /  T F =  P /  T SHOWS THAT FORCE, F, IS THE TIME RATE OF CHANGE OF MOMENTUM AND THE OBJECT’S RATE OF CHANGE OF MOMENTUM IS IN THE DIRECTION OF THE FORCE. MULTIPLYING BOTH SIDES OF F = M  V /  T BY  T GIVES: F  T = M  V. F  T IS CALLED IMPULSE. IMPULSE, F  T, IS THE PRODUCT OF A FORCE AND THE TIME INTERVAL DURING WHICH IT ACTS. Impulse = Change in momentum Impulse video

13. SINCE  P = M  V AND F  T = M  V, THEN F  T =  P, WHICH MEANS THAT THE IMPULSE EQUALS THE CHANGE IN MOMENTUM,  P. THUS, A CHANGE IN MOMENTUM,  P, TAKES FORCE AND TIME. F  T =  P (F  T = MV F – MV I ) IS CALLED THE IMPULSE-MOMENTUM THEOREM AND CAN ONLY BE APPLIED TO CASES THAT INVOLVE A CONSTANT FORCE. F  T =  P SHOWS THAT A NET EXTERNAL FORCE APPLIED CONSTANTLY TO AN OBJECT FOR A CERTAIN TIME INTERVAL WILL CAUSE A CHANGE IN THE OBJECT’S MOMENTUM EQUAL TO THE PRODUCT OF THE FORCE AND TIME INTERVAL.

14. IMPULSE-MOMENTUM THEOREM A CHANGE IN MOMENTUM OVER A LONGER TIME REQUIRES LESS FORCE. THE IMPULSE-MOMENTUM THEOREM EXPLAINS WHY FOLLOW-THROUGH IS IMPORTANT IN SPORTS AND EVERYDAY ACTIVITIES, SUCH AS BILLIARDS, BASEBALL, SOFTBALL, SOCCER, TENNIS, GOLF, AND PUSHING A SHOPPING CART. THE FOLLOW-THROUGH ALLOWS MORE CONTACT INCREASING THE TIME INTERVAL, AND THUS DECREASING THE FORCE NECESSARY TO OBTAIN A CHANGE IN MOMENTUM EQUIVALENT TO A CHANGE IN MOMENTUM FORMED FROM A STRONGER FORCE OVER A SHORTER TIME INTERVAL. EXTENDING THE TIME INTERVAL OVER WHICH A CONSTANT FORCE IS APPLIED ALLOWS A SMALLER FORCE TO CAUSE A LARGER CHANGE IN MOMENTUM THAN WOULD RESULT IF THE FORCE WERE APPLIED FOR A VERY SHORT TIME.

15. NETS AND GIANT AIR MATTRESSES USED BY FIREFIGHTERS TO CATCH PEOPLE WHO MUST JUMP OUT OF TALL BUILDING USE THIS PRINCIPLE FOR AN INCREASED TIME INTERVAL FOR DECREASING THE FORCE. HIGHWAY SAFETY ENGINEERS ALSO USE THE IMPULSE- MOMENTUM THEOREM FOR VEHICLE STOPPING TIMES AND DISTANCES.

16. IT TAKES LONGER TO STOP A CAR ON A WET OR ICY ROAD BECAUSE THE TOTAL FORCE OPPOSING THE CAR’S MOTION IS LESS WHEN THE FORCE OF FRICTION BETWEEN THE ROAD AND TIRES IS REDUCED. THIS LOWER FORCE MUST BE APPLIED FOR A LONGER TIME IN ORDER TO CAUSE THE SAME CHANGE IN MOMENTUM, F  T =  P F  T = MV F – MV I.

17. A 1400 KG CAR MOVING WEST WITH A VELOCITY OF 15 M/S COLLIDES WITH A UTILITY POLE AND IS BROUGHT TO REST IN 0.30 SECONDS. A.FIND THE MAGNITUDE AND DIRECTION OF THE FORCE EXERTED ON THE CAR DURING THE COLLISION. B.FIND THE MAGNITUDE AND DIRECTION OF THE FORCE EXERTED ON THE POLE DURING THE COLLISION. C.HOW FAR HAS THE CAR TRAVELED DURING THE 0.30 SECONDS? a. m = 1400 kgb. By Newton’s 3 rd Law, F = 70000N, west v i =+ 15 m/s v f = 0 m/s c. Δx = ½ (v f + v i ) Δt Δt = 0.30 secΔx = ½ (0 + 15) 0.30 FΔt = mv f - mv i Δx = 2.25 m F(0.30) = 1400(0) – 1400(15) F = -70000 N = 70000 N, east

18. A 2250 KG CAR TRAVELING TO THE EAST IS SLOWED UNIFORMLY FROM AN INITIAL VELOCITY OF 20 M/S BY A 8450 N BRAKING FORCE ACTING OPPOSITE THE CAR’S MOTION. USE THE IMPULSE-MOMENTUM THEOREM TO ANSWER THE FOLLOWING QUESTION: A.WHAT IS THE CAR’S VELOCITY AFTER 4 SECONDS? B.HOW FAR DOES THE CAR MOVE DURING 4 SECONDS? C.HOW LONG DOES IT TAKE THE CAR TO COME TO A COMPLETE STOP? A. M = 2250 KG V I =+ 20 M/S V F = ? M/S B. ΔX = ½ (V F + V I ) ΔT ΔT = 4 SECΔX = ½ (20 + 4.97) 4 F = -8450 NΔX = 50 M FΔT = MV F - MV I -8450(4) = 2250 V F – 2250(20) V F = 4.98 M/S, EAST C. ΔT = ? V F = 0 M/S FΔT = MV F - MV I -8450(ΔT) = 2250 (0) – 2250(20) ΔT =5.33 SEC

19. NOTES - SECTIONS 6.2

20. THE LAW OF CONSERVATION OF MOMENTUM STATES THAT WHEN NO NET EXTERNAL FORCES ARE ACTING ON A SYSTEM OF OBJECTS, THE TOTAL VECTOR MOMENTUM OF THE SYSTEM REMAINS CONSTANT. IN A COLLISION BETWEEN TWO OBJECTS, THE MOMENTUM OF EACH OBJECT CHANGES, BUT THE TOTAL MOMENTUM OF THE TWO OBJECTS TOGETHER REMAINS THE SAME. MOMENTUM CAN BE TRANSFERRED FROM ONE OBJECT TO ANOTHER DURING A COLLISION. THE LAW OF CONSERVATION OF MOMENTUM ALSO APPLIES TO SITUATIONS WHERE THE INITIAL MOMENTUM IS ZERO. SEE SKATER EXAMPLE ON PAGE 217.

21. THE INITIAL TOTAL MOMENTUM BEFORE A COLLISION/INTERACTION WILL ALWAYS EQUAL THE FINAL TOTAL MOMENTUM AFTER A COLLISION/INTERACTION.  P I =  P F FOR OBJECT 1 COLLIDING OR INTERACTING WITH OBJECT 2: P 1,I + P 2,I = P 1,F + P 2,F INITIAL MOMENTUM OF THE TWO OBJECTS = FINAL MOMENTUM OF THE 2 OBJECTS (BEFORE THE COLLISION)(AFTER THE COLLISION) M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F

22. A BALL IS MOVING EAST WHILE ANOTHER BALL IS MOVING WEST. THESE TWO BALLS COLLIDE. HOW DO YOU INDICATE THE OPPOSITE DIRECTIONS IN YOUR EQUATION? YOU USE POSITIVE AND NEGATIVE SIGNS.

23. A 76 KG BOATER, INITIALLY AT REST IN A STATIONARY 45 KG BOAT, STEPS OUT OF THE BOAT AND ONTO THE DOCK. IF THE BOATER MOVES OUT OF THE BOAT WITH A VELOCITY OF 2.5 M/S TO THE RIGHT, WHAT IS THE FINAL VELOCITY (MAGNITUDE AND DIRECTION) OF THE BOAT? BOATERBOAT RIGHT + AND LEFT - M1 = 76 KGM2 = 45 KG V1,I = 0 M/SV2,I = 0 M/S V1,F =+ 2.5 M/S, RV2,F = ? M1V1,I + M2V2,I = M1V1,F + M2V2,F 0 = 76(2.5) + 45 V2,F V2,F = -4.22 M/S = 4.22 M/S, LEFT

24. A 1.5 KG BALL WHOSE VELOCITY IS 8.00 M/S SOUTHWARD COLLIDES WITH A 2.00 KG BALL TRAVELLING NORTHWARD WITH A VELOCITY OF 3.00 M/S. IF THE VELOCITY OF THE 2.00 KG BALL IS 6.50 M/S SOUTHWARD AFTER IMPACT, WHAT IS THE VELOCITY OF THE 1.5 KG BALL? M1 = 1.5 KGM2 = 2.00 KGSOUTH + AND NORTH - V1,I = +8.00 M/S, SV2,I = -3.00 M/S, N V1,F = ?V2,F = +6.50 M/S, S M1V1,I + M2V2,I = M1V1,F + M2V2,F (1.5)(8) + (2)(-3) = (1.5) V1,F + (2)(6.5) V1,F = - 4.67 M/S = 4.67 M/S, N

25. IN SOME COLLISIONS, THE PARTICLES STICK TOGETHER AFTER THE COLLISION SO THAT THEIR FINAL VELOCITIES ARE THE SAME. THUS V 1,F = V 2,F AND THE RIGHT HAND SIDE OF M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F BECOMES M 1 V 1,I + M 2 V 2,I = (M 1 + M 2 ) V F.

26. A TRUCK WITH A MASS OF 3000 KG TRAVELS AT A VELOCITY OF 2.5 M/S, EAST. IT HITS A STATIONARY CAR HEAD ON. THE CAR HAS A MASS OF 1500 KG. THE VEHICLES CONNECT AND ROLL TOGETHER AFTER IMPACT. WHAT IS THE VELOCITY OF THE CONNECTED VEHICLES? M1 = 3000 KGM2 = 1500 KGEAST + AND WEST - V1,I = +2.5 M/S, EASTV2,I = 0 M/SVF = ? M1V1,I + M2V2,I = (M1 + M2) VF. (3000)(2.5) = (3000 + 1500) VF VF = 1.67 M/S, EAST

27. VIDEO CAR CRASH TECH EGG CRASH! DESIGNING A COLLISION SAFETY DEVICE

28. TYPES OF COLLISIONS THERE ARE MANY DIFFERENT TYPES OF COLLISIONS/INTERACTIONS. THREE OF WHICH ARE ELASTIC, INELASTIC, AND PERFECTLY INELASTIC. IN BOTH OF THESE COLLISION/INTERACTION TYPES, MOMENTUM IS CONSERVED,  P I =  P F. INELASTIC AND ELASTIC COLLISIONS VIDEO

29. ELASTIC COLLISION 1. IN AN ELASTIC COLLISION, KINETIC ENERGY IS ALSO CONSERVED,  KE I =  KE F, SO THAT NO KINETIC ENERGY IS LOST. THE TOTAL KINETIC ENERGY OF THE PARTICLES BEFORE THE COLLISION IS EQUAL TO THE TOTAL KINETIC ENERGY AFTER THE COLLISION. THE OBJECTS BOUNCE AFTER THE COLLISION SO THAT THEY MOVE SEPARATELY. ALSO, THE OBJECTS RETURN TO THEIR ORIGINAL SHAPE AFTER THE COLLISION. FOR TWO OBJECTS, OBJECT 1 AND OBJECT 2:  P I =  P F  KE I =  KE F P 1,I + P 2,I = P 1,F + P 2,F KE 1,I + KE 2,I = KE 1,F + KE 2,F M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F ½ M 1 V 1,I 2 + ½ M 2 V 2,I 2 = ½ M 1 V 1,F 2 + ½ M 2 V 2,F 2

30. IN INELASTIC AND PERFECTLY INELASTIC COLLISIONS, SOME KINETIC ENERGY IS ‘LOST’ BY TRANSFORMATION INTO OTHER FORMS OF ENERGY SUCH AS HEAT, WORK OR SOUND. A COLLISION IS PERFECTLY INELASTIC IF THE PARTICLES STICK TOGETHER AFTER THE COLLISION SO THAT THEIR FINAL VELOCITIES ARE THE SAME. A COLLISION IS INELASTIC IF THE PARTICLES DEFORM DURING THE COLLISION, AND THE OBJECTS MOVE SEPARATELY AFTER THE COLLISION. FOR TWO OBJECTS, OBJECT 1 AND OBJECT 2:  P I =  P F P 1,I + P 2,I = P 1,F + P 2,F M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F THE DECREASE IN KINETIC ENERGY DURING THE COLLISIONS CAN BE FOUND: KE I = KE 1,I + KE 2,I KE F = KE 1,F + KE 2,F KE I = ½ M 1 V 1,I 2 + ½ M 2 V 2,I 2 KE F = ½ M 1 V 1,F 2 + ½ M 2 V 2,F 2 DECREASE IN KE = KE I - KE F FOR PERFECTLY INELASTIC, THE FINAL VELOCITIES OF THE TWO OBJECTS ARE THE SAME: V 1,F = V 2,F SO THE FORMULA M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F BECOMES M 1 V 1,I + M 2 V 2,I = (M 1 + M 2 )V F

31. IN ANALYZING THE EFFECTS OF COLLISIONS, YOU CAN ALWAYS APPLY THE PRINCIPLE OF CONSERVATION OF MOMENTUM. IF THE COLLISION IS ELASTIC, YOU CAN ALSO APPLY THE PRINCIPLE OF CONSERVATION OF KINETIC ENERGY. EXAMPLE OF A PERFECTLY INELASTIC COLLISION IN ONE DIMENSION: ONE CAR RUNS INTO THE BACK OF ANOTHER CAR, AND AFTER THE COLLISION, THEY STICK TOGETHER. EXAMPLE OF AN ELASTIC COLLISION IN TWO DIMENSION: A MOVING STEEL BALL OR BILLIARD BALL STRIKES A STATIONARY ONE. AS A RESULT OF THE COLLISION, BOTH BALLS HAVE MOTION BUT THE PATH OF THE FIRST BALL IS DEFLECTED AT THE POINT OF COLLISION. MOST COLLISIONS ARE NEITHER ELASTIC NOR PERFECTLY INELASTIC. MOST COLLISIONS ARE INELASTIC COLLISIONS, WHERE THE COLLIDING OBJECTS BOUNCE AND MOVE SEPARATELY AFTER THE COLLISION, BUT THE TOTAL KINETIC ENERGY DECREASES IN THE COLLISION. IN MOST COLLISIONS INVOLVING CARS, THE CARS BOUNCE AWAY FROM EACH OTHER (REBOUND) AND ARE DAMAGED OR DEFORMED IN THE COLLISION. TWO FOOTBALL PLAYERS OR TWO DESTRUCTION DERBY DRIVERS COLLIDE INELASTICALLY SUCH THAT THE TOTAL ENERGY IS CONSERVED, BUT TOTAL KINETIC ENERGY IS NOT. THE MOTION OF THE FOOTBALL PLAYERS OR CARS IS HALTED (KINETIC ENERGY IS REDUCED TO ZERO) AS KINETIC ENERGY IS CONVERTED TO HEAT OR WORK (BENDING THE PLAYERS’ PADDING OR FRAMES OF THE CARS).

32. A BALL HAVING A MASS OF 0.25 KG AND A VELOCITY OF 0.20 M/S EASTWARD COLLIDES WITH ANOTHER BALL HAVING A MASS OF 0.10 KG AND A VELOCITY OF 0.10 M/S, ALSO EASTWARD ALONG THE SAME STRAIGHT LINE. AFTER THE COLLISION, THE MORE MASSIVE BALL HAS A VELOCITY OF 0.15 M/S EASTWARD. A. WHAT KIND OF COLLISION IS THIS – ELASTIC OR PERFECTLY INELASTIC? B. WHAT IS THE VELOCITY OF THE LESS MASSIVE BALL?

33. m 1 = 0.25 kgm 2 = 0.10 kg v 1,i = 0.20 m/s eastwardv 2,i = 0.10 m/s eastward v 1,f = 0.15 m/s eastwardv 2,f = ____?____ m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f (0.25)(+0.20) + (0.10)(+0.10) = (0.25)(+0.15) + (0.10) v 2,f v 2,f\ = 0.225 m/s eastward

34. A 0.015 KG MARBLE MOVING TO THE RIGHT AT 0.225 M/S MAKES A HEAD-ON COLLISION WITH A 0.030 KG SHOOTER MARBLE MOVING TO THE LEFT AT 0.180 M/S. AFTER THE COLLISION, THE MARBLES MOVE SEPARATELY. THE SMALLER MARBLE MOVES TO THE LEFT AT 0.315 M/S. A. WHAT KIND OF COLLISION IS THIS – ELASTIC OR PERFECTLY INELASTIC? B. WHAT IS THE VELOCITY OF THE 0.030 KG MARBLE AFTER THE COLLISION?

35. M 1 = 0.015 KGM 2 = 0.030 KGRIGHT (-) LEFT (+) V 1,I = 0.225 M/S RIGHTV 2,I = 0.180 M/S LEFT V 1,F = 0.315 M/S LEFTV 2,F = ___?_____ M 1 V 1,I + M 2 V 2,I = M 1 V 1,F + M 2 V 2,F (0.015)(-0.225) + (0.030)(+ 0.180) = (0.015)(+ 0.315) + (0.030) V 2,F V 2,F = -0.09 M/S = 0.09 M/S RIGHT

36. A FREIGHT CAR WITH A MASS OF 300,000 KG TRAVELS AT A VELOCITY OF 2.5 M/S, EAST. IT COLLIDES WITH A STATIONARY CAR HAVING MASS OF 150,000 KG ON A HORIZONTAL TRACK. THE CARS CONNECT AND ROLL TOGETHER AFTER IMPACT. A. WHAT KIND OF COLLISION IS THIS? PERFECTLY INELASTIC B. WHAT IS THE VELOCITY OF THE CONNECTED CARS? C. WHAT IS THE DECREASE IN KINETIC ENERGY DURING THE COLLISION?

37. A FREIGHT CAR WITH A MASS OF 300,000 KG TRAVEL AT A VELOCITY OF 2.5 M/S, EAST. IT COLLIDES WITH A STATIONARY CAR HAVING MASS OF 150,000 KG ON A HORIZONTAL TRACK. THE CARS CONNECT AND ROLL TOGETHER AFTER IMPACT. A. WHAT KIND OF COLLISION IS THIS? PERFECTLY INELASTIC B. WHAT IS THE VELOCITY OF THE CONNECTED CARS? C. WHAT IS THE DECREASE IN KINETIC ENERGY DURING THE COLLISION? M 1 = 300,000 KGM 2 = 150,000 KG V 1,I = 2.5 M/S, EASTV 2,I = 0 M/S V 1,F = ___?_____V 2,F = __?______V 1,F = V 2,F M 1 V 1,I + M 2 V 2,I = (M 1 + M 2 )V F (300,000 KG)(+ 2.5) + (150,000)(0) = (300,000 + 150,000) V F V F = 1.67 M/S EAST KE I = ½ M 1 V 1,F 2 + ½ M 2 V 2,I 2 KE F = ½ M 1 V 1,I 2 + ½ M 2 V 2,F 2 KE I = ½(300,000)( 2.5) 2 KE F = ½ (300,000)(1.67) 2 + ½ (150,000)(1.67) 2 KE F = 627502.5 J KE I = 937,500 J DECREASE IN KE = KE I - KE F DECREASE IN KE = 937500 - 627502.5 DECREASE IN KE = 309997.5 J