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Mo’ mentum. Whiteboard warmup! A 145-g softball is pitched horizontally toward home plate with a speed of 165 m/s. The batter hits a direct line drive,

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Presentation on theme: "Mo’ mentum. Whiteboard warmup! A 145-g softball is pitched horizontally toward home plate with a speed of 165 m/s. The batter hits a direct line drive,"— Presentation transcript:

1 Mo’ mentum

2 Whiteboard warmup! A 145-g softball is pitched horizontally toward home plate with a speed of 165 m/s. The batter hits a direct line drive, back toward the pitcher. The bat exerts an average force of 650 N on the ball, and is in contact with the ball for 0.08 seconds. a)Determine the impulse delivered to the ball by the bat. b)Determine the final velocity of the ball (magnitude and direction) (you may ignore the impulse delivered by the Earth during this time)

3 final vfvf initial vivi Don’t forget positive and negative directions!

4 Closed Systems A closed system is a choice of a system in which the net impulse is zero. In a closed system, the external forces exerted on the system are balanced. The Law of Conservation of Momentum… when applied to a closed system… (only for a closed system) gives a very useful result!

5 For a closed system, This means that if we choose a system that has no net impulse delivered to it, The total momentum of a closed system does not change! Often very useful for analyzing collisions and explosions!

6 Try It! A 650-kg cannon fires a 50-kg cannonball horizontally, as shown. If the cannonball fires at a speed of 28 m/s, what is the recoil speed of the cannon?

7 Two identical train cars of mass m are on a smooth, level track. One car is at rest initially while the other moves toward it, as shown below. When the cars collide, the coupling mechanism latches, causing the cars to stick together and move as a unit. By using momentum conservation, represent the final velocity of the locked cars in terms of the initial velocity of the first car.

8 Conservation on Momentum If you choose a system in which the net external impulse is zero… The vector sum of the momenta before equals the vector sum of the momenta after! Momentum is a vector – be mindful of positive and negative directions. Sketching before and after scenarios with velocity vectors will help you avoid careless mistakes!

9 Whiteboard Collision Time 1.5 kg 4 kg 2.4 m/s0.8 m/s 1.5 kg 4 kg 2.3 m/s ?? ? What is the final velocity of the red puck?

10 Start with the LoCoM 1.5 kg 4 kg 2.4 m/s 0.8 m/s 1.5 kg 4 kg 2.3 m/s ? p i red + p i blue = p f red + p f blue m red v red i + m blue v blue i = m red v red i + m blue v blue i v red f = -1.6 m/s 1.5 kg 4 kg 2.4 m/s 0.8 m/s 1.5 kg 4 kg 2.3 m/s1.6 m/s

11 If you drop your keys, their momentum increases as they fall. Is momentum still conserved in this scenario? Explain your answer, and use a picture to support your answer. If we choose just the keys as the system to analyze, then the Earth delivers an impulse that increases their momentum! The LoCoM is true! If we choose the keys and Earth as the system to analyze, then the system maintains a net momentum of zero! The keys and Earth fall toward one another! The LoCoM still works! FΔt Δp v v p p

12 Two pucks, of masses m and 2m, are initially at rest on a frictionless sheet of ice. A constant force F is exerted horizontally on the smaller puck by a stream of air from a hose (that follows it), over a horizontal distance D. This is then repeated for the larger puck, until it has traveled the same amount of horizontal distance D. Compare the final momenta of the two pucks after they have traveled this distance, and explain how you have reached your conclusion. D

13 The same force is exerted on each puck, for the same amount of distance. But, the more massive puck will take longer to reach the end! (The more massive puck will have a smaller acceleration) Same for both The more massive one takes longer! The more massive one will have a greater final momentum The more massive puck will have a larger impulse delivered to it

14 2-D Collision Analysis! A car with a mass of 950 kg and a speed of 16 m/s approaches an intersection, as shown below. A 1,300-kg minivan traveling at 21 m/s is heading for the same intersection. The car and the minivan collide and stick together. Find the speed and direction of the wrecked vehicles just after the collision, assuming that frictional forces can be ignored.

15 Most Important Tip: Stay Organized!!! pxpx Car pypy Minivan System 15,200 kg m/s 0 0 27,300 kg m/s 15,200 kg m/s 27,300 kg m/s 15,200 kg m/s 27,300 kg m/s 31,246 kg m/s 60.9° The total momentum of the system after the collision must also have these components. Now we can make a momentum vector triangle! Finally, divide by the combined mass of the system to find its final velocity 13.9 m/s

16 24 kg 12 kg 4 kg 8 kg ? θ = ? 30° 16 m/s 6 m/s Whiteboard Watermelon Encore! Calculate the exit angle of the smallest piece, as well as the velocity of the biggest piece after the explosion!

17 12 kg 4 kg 8 kg ? θ 30° 16 m/s 6 m/s The total momentum after the explosion must equal the total momentum before the explosion (ZERO!) Since the big piece has zero x-momentum, this means that the two smaller pieces must have exactly opposite x-momentum. 4 * 16 * cosθ = 8 * 6 * cos(30)θ = 49.5° Since the total y-momentum must equal zero, the y-momentum of the big piece must be opposite to the sum of the y-momentum of the smaller pieces. 4 * 16 * sin(49.5) + 8 * 6 * sin(30) = 12 * v v = 6 m/s


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