Impulse AP Physics

End Slide What is Impulse? What do you think of when you hear the term Impulse? (you don’t have to relate it to physics) Sudden change in behavior? To do something “on a whim”?

End Slide What is Impulse? Impulse will cause a change in motion for a given mass. Impulse depends on how much net force is being applied and how long that force is applied to a mass. Impulse = Fnet*t The units are Newton*seconds (N•s)

End Slide Impulse Examples A baseball is thrown with a force of 800 N for 0.4 sec. What is the impulse on the baseball? A hockey player makes a slap shot, exerting a constant force of 30.0N on a hockey puck for 0.16 s. What is the impulse given to the puck? 320 N•s 4.8 N•s

End Slide Impulse by Graphing Looking at the graph to the right, what is the impulse over the given time period? Graphically, how do you find impulse?

What is the Impulse here? End Slide What is the Impulse here? From 0s to 3s? From 0s to 5s? From 2s to 5s? From 2s to 4s? From 1s to 3s? From 4s to 5s? From 1s to 5s? From 3s to 5s? 4.5 N•s 12 N•s 10 N•s 6.0 N•s 4.0 N•s 4.0 N•s 11.5 N•s 7.5 N•s

What is the Impulse here? End Slide What is the Impulse here? Example: From 2s to 5s? 1/2*(4s-2s)*(4N-2N) = 2 N•s (4s-2s)*(2N-0N) = 4 N•s (5s-4s)*(4N-0N) = 4 N•s 2N•s + 4N•s + 4N•s = 10 N•s

Check your understanding… End Slide Check your understanding… I start pushing a table and I add more and more force on in the first second. I get to pushing it as hard as I can for one second and somebody helps me put more force for another second. We both get to our max force for one second and begin to decrease our force on the table until we stop pushing all together. What is the total impulse?

100N•s+200N•s+200N•s+50N•s+300N•s+150N•s = 1000 N•s End Slide 300 N•s 50 N•s 200 N•s 200 N•s 150 N•s 100 N•s 100N•s+200N•s+200N•s+50N•s+300N•s+150N•s = 1000 N•s

End Slide If friction is -200 N, how long will it take to equal out the impulse put into the table? 1000 N•s -1000 N•s

End Slide Work it out J = -1000 N•s -1000 N•s = (-200N)*t t = 5 s

Momentum AP Physics

End Slide Momentum Momentum (p) is the product of an object’s mass (m) and velocity (v). Since velocity has a direction, momentum has a direction; thus momentum is a vector. For a 2500 kg car moving at 30.0 m/s North, what is the momentum? p = 75,000 kg*m/s North

End Slide Momentum Examples What is the momentum of your book with a mass of 1.5 kg sitting on your table? What is the momentum of a 3.5 kg rock when it gets to the bottom of a 30 m cliff after being dropped? Find the momentum of a 0.25 kg ball rolling with a velocity of -3.0 m/s. 0 kg*m/s -84.9 kg*m/s -0.75 kg*m/s

Change in Momentum (Dp) End Slide Change in Momentum (Dp) Final Momentum (pf) minus Initial Momentum (po). Dp = pf - po For a 2500 kg car moving at 30.0 m/s North and then slows to 10.0 m/s, what is the change in momentum (Dp)? Dp = pf - po = 25,000 kg*m/s – 75,000 kg*m/s Dp = -50,000 kg*m/s

Change in Momentum Examples End Slide Change in Momentum Examples What is the change in momentum of a 3.5 kg rock when it dropped off of a 30 m cliff? Dp = -84.9 kg*m/s – 0 kg*m/s = -84.9 kg*m/s Find the change in momentum of a 0.25 kg ball initially rolling with a velocity of –3.0 m/s but accelerates and is now rolling with a velocity of +3.0 m/s. Dp = (0.75 kg*m/s) – (-0.75 kg*m/s) = 1.50 kg*m/s

Relationship between Impulse and Change in Momentum End Slide Relationship between Impulse and Change in Momentum Newton’s Second Law Fnet = ma Fnet = m(vf – vo) Fnet*t= m(vf – vo) Fnet*t= mvf – mvo J = Dp a = Dv = vf – vo J = Fnet*t Dp = mvf – mvo t t t t t

Real Life Applications for J = Dp AP Physics

Examples of wanting to stop End Slide Examples of wanting to stop If you want to stop, you will have some mass, some initial velocity, and you want your final velocity to be 0 m/s. How is “damage” minimized on your body in the following examples?

Wanting to Stop Car wreck Rock Climbing with a Nylon rope End Slide Wanting to Stop Car wreck Use of air bags Crumple zones Rock Climbing with a Nylon rope Nylon allows stretch Getting hit in boxing “Riding the punch”

What do they have in common? End Slide What do they have in common? J = Dp and “Dp” will be constant To minimize “damage”, you want to minimize force If you increase the time that the force is being applied, the average force will decrease

Happy/Sad Ball Demo Which has more change in momentum? End Slide Happy/Sad Ball Demo Which has more change in momentum? How do you know? Which type of material would you rather make a wrecking ball out of? Why? What are “elastic” and “inelastic” collisions?

Why is cork in bats against the rules in baseball? End Slide Why is cork in bats against the rules in baseball? Cork is easily compressed and allows the rest of the bat to compress. If the bat is able to compress more, the bat acts more like the happy ball than the sad ball. This allows more change in momentum to be given to the baseball and thus, an unfair advantage.

Using Impulse Graphs AP Physics

Example End Slide A 7.0-kg bowling ball is rolling down the alley with a velocity of 2.0 m/s. For each impulse, a and b, as show in Figure 9-3, find the resulting speed and direction of the motion of the bowling ball. 5.0 N•s -5.0 N•s

“+” is “toward pins” and “-” is “away from pins” m=7.0kg; vo=2.0m/s End Slide “+” is “toward pins” and “-” is “away from pins” m=7.0kg; vo=2.0m/s Part A; J = 5.0 N•s J = Dp = mvf-mvo 5.0N•s = 7kg*vf – 7kg*2m/s 5 = 7vf – 14 vf = 2.71m/s 2.71m/s towards pins Part B; J = -5.0 N•s J = Dp = mvf-mvo -5.0N•s = 7kg*vf – 7kg*2m/s -5 = 7vf – 14 vf = 1.29m/s 1.29m/s towards pins 14 + + 14 14+ + 14 7 7 7 7

A 30 kg mass undergoes the following Impulse A 30 kg mass undergoes the following Impulse. Assuming the mass started at 0 m/s, what is its final velocity? End Slide 1000 N•s

1000 N•s = (30 kg)vf – (30 kg)*(0 m/s) End Slide J = 1000 N•s J = Dp mvf – mvo 1000 N•s = (30 kg)vf – (30 kg)*(0 m/s) 1000 = 30vf - 0 vf = 33.3 m/s = mvf – mvo 30 30

End Slide Given a 1.5 kg mass has an initial velocity of 3.0 m/s, what is the mass’s final velocity for your time interval? From 0s to 3s? From 0s to 5s? From 2s to 5s? From 2s to 4s? From 1s to 3s? From 4s to 5s? From 1s to 5s? From 3s to 5s? 6.0 m/s 11 m/s 9.7 m/s 7.0 m/s 5.7 m/s 5.7 m/s 10.7 m/s 8.0 m/s

m = 1.5 kg and vo = 3 m/s Example: From 2s to 5s? J = mvf – mvo End Slide Example: From 2s to 5s? J = mvf – mvo 10 N•s = (1.5kg)vf – (1.5kg)*(3m/s) 10 = 1.5vf – 4.5 14.5 = 1.5vf vf = 9.7 m/s J = 10 N•s 4 N•s 2 N•s 4 N•s

End Slide A rocket with mass 57 grams is fired from rest. The engine’s Force vs. time graph is given to the right. What is the final velocity of the rocket after it’s fired off? 46.3 m/s

End Slide Richard strikes a 0.058-kg golf ball with an average force of 272 N and gives it a velocity of 62.0 m/s. How long was Richard’s club in contact with the ball? J = Dp Fnet*t = m(vf – vo) 272 t = 0.058 * 62.0 t = 0.013 s 272 272

Example End Slide A 0.145-kg baseball is moving at 35 m/s when it is caught by a player. Find the change in momentum. ∆𝒑=𝒎( 𝒗 𝒇 − 𝒗 𝒐 ) =𝟎.𝟏𝟒𝟓(𝟎−𝟑𝟓) → ∆𝒑=−𝟓.𝟎𝟖 𝒌𝒈⋅ 𝒎 𝒔 If the ball is caught with a mitt held in a stationary position so that the ball stops in 0.050 s, what is the average force exerted on the ball? → 𝑭 𝒏𝒆𝒕 = ∆𝒑 𝒕 = −𝟓.𝟎𝟖 𝟎.𝟎𝟓𝟎 𝑱=∆𝒑 → 𝑭 𝒏𝒆𝒕 𝒕=∆𝒑 → 𝑭 𝒏𝒆𝒕 =𝟏𝟎𝟏.𝟓𝑵 If, instead, the mitt is moving backward so that the ball takes 0.500 s to stop, what is the average force exerted by the mitt on the ball? 𝑭 𝒏𝒆𝒕 = ∆𝒑 𝒕 = −𝟓.𝟎𝟖 𝟎. 𝟓𝟎𝟎 → 𝑭 𝒏𝒆𝒕 =𝟏𝟎.𝟏𝟓𝑵

Questions: Which has more momentum, a supertanker tied to a dock or a falling raindrop? When you jump from a height to the ground, you let your legs bend at the knees as your feet hit the floor. Explain why you do this in terms of the concepts learned in this unit. An archer shoots arrows at a target. Some of the arrows stick in the target, while others bounce off. Assuming that the masses of the arrows are the same, which arrows produce the bigger impulse on the target?

End Slide A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. What is the change in momentum of the ball? ∆𝒑=𝒎( 𝒗 𝒇 − 𝒗 𝒐 ) =𝟎.𝟏𝟕𝟒(𝟑𝟖−(−𝟐𝟔)) ∆𝒑=𝟏𝟏.𝟏𝟒 𝒌𝒈⋅ 𝒎 𝒔 What is the impulse delivered by the bat? 𝑱=∆𝒑 → 𝑱=𝟏𝟏.𝟏𝟒 𝑵⋅𝒔 If the bat and softball are in contact for 0.80 ms, what is the average force that the bat exerts on the ball. → 𝑭 𝒏𝒆𝒕 = ∆𝒑 𝒕 = 𝟏𝟏.𝟏𝟒 𝟎.𝟎𝟎𝟎𝟖𝟎 𝑱=∆𝒑 → 𝑭 𝒏𝒆𝒕 𝒕=∆𝒑 𝑭 𝒏𝒆𝒕 =𝟏𝟑,𝟗𝟎𝟎𝑵

Conservation of Momentum End Slide Conservation of Momentum 𝑭 𝑨𝑩 A B 𝑭 𝑩𝑨 𝑵𝒆𝒘𝒕𝒐 𝒏 ′ 𝒔 𝟑𝒓𝒅 𝑳𝒂𝒘 𝑭 𝑨𝑩 =− 𝑭 𝑩𝑨 𝑰𝒔 𝑨 𝒊𝒏 𝒄𝒐𝒏𝒕𝒂𝒄𝒕 𝒘𝒊𝒕𝒉 𝑩 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒕𝒉𝒂𝒕 𝑩 𝒊𝒔 𝒊𝒏 𝒄𝒐𝒏𝒕𝒂𝒄𝒕 𝒘𝒊𝒕𝒉 𝑨? 𝑭 𝑨𝑩 ∗𝒕=− 𝑭 𝑩𝑨 ∗𝒕 𝑭 ∗𝒕=𝑱 ∆𝒑 𝑨 + ∆𝒑 𝑩 =𝟎 𝑱 𝑨 =− 𝑱 𝑩 𝑪𝒐𝒏𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝑱=∆𝒑 ∆𝒑 𝑨 =− ∆𝒑 𝑩 𝒎 𝟏 𝒗 𝒇𝟏 − 𝒗 𝒐𝟏 + 𝒎 𝟐 𝒗 𝒇𝟐 − 𝒗 𝒐𝟐 =𝟎

End Slide A B Cart A of mass 6.0 kg is moving at 3.5 m/s and collides with Cart B with mass 4.0 kg that is at rest. Cart A moves off at 0.7 m/s. What is the final velocity of Cart B? 𝒎 𝑨 𝒗 𝒇𝑨 − 𝒗 𝒐𝑨 + 𝒎 𝑩 𝒗 𝒇𝑩 − 𝒗 𝒐𝑩 =𝟎 𝟔.𝟎 𝟎.𝟕−𝟑.𝟓 +𝟒.𝟎 𝒗 𝒇𝑩 −𝟎 =𝟎 𝟔.𝟎 −𝟐.𝟖 +𝟒.𝟎 𝒗 𝒇𝑩 =𝟎 −𝟏𝟔.𝟖+𝟒.𝟎 𝒗 𝒇𝑩 =𝟎 𝟒.𝟎 𝒗 𝒇𝑩 =𝟏𝟔.𝟖 → 𝒗 𝒇𝑩 =𝟒.𝟐 𝒎 𝒔

End Slide A 35.0-g bullet strikes a 5.0-kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and the bullet fly off together at 8.6 m/s. What is the original speed of the bullet? 𝒎 𝟏 𝒗 𝒇𝟏 − 𝒗 𝒐𝟏 + 𝒎 𝟐 𝒗 𝒇𝟐 − 𝒗 𝒐𝟐 =𝟎 𝟎.𝟎𝟑𝟓 𝟖.𝟔− 𝒗 𝒐𝟏 +𝟓.𝟎 𝟖.𝟔−𝟎 =𝟎 𝟎.𝟑𝟎𝟏−𝟎.𝟎𝟑𝟓 𝒗 𝒐𝟏 +𝟒𝟑=𝟎 𝟒𝟑.𝟑𝟎𝟏−𝟎.𝟎𝟑𝟓 𝒗 𝒐𝟏 =𝟎 𝟎.𝟎𝟑𝟓 𝒗 𝒐𝟏 =𝟒𝟑.𝟑𝟎𝟏 → 𝒗 𝒐𝟏 =𝟏𝟐𝟒𝟎 𝒎 𝒔

End Slide 1.5 kg 4.5 kg A thread holds a 1.5-kg cart and a 4.5-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5-kg cart a speed of 27 cm/s to the left. What is the velocity of the 4.5-kg cart? 𝒎 𝟏 𝒗 𝒇𝟏 − 𝒗 𝒐𝟏 + 𝒎 𝟐 𝒗 𝒇𝟐 − 𝒗 𝒐𝟐 =𝟎 𝟏.𝟓 −𝟎.𝟐𝟕−𝟎 +𝟒.𝟓 𝒗 𝒇𝟐 −𝟎 =𝟎 −𝟎.𝟒𝟎𝟓+𝟒.𝟓 𝒗 𝒇𝟐 =𝟎 𝟒.𝟓 𝒗 𝒇𝟐 =𝟎.𝟒𝟎𝟓 → 𝒗 𝒇𝟐 =𝟎.𝟎𝟗 𝒎 𝒔 𝒗 𝒇𝟐 =𝟗.𝟎 𝒄𝒎 𝒔