We last left off by talking about how the area under a force vs. time curve is impulse.

Transcription

1 Lecture 11 Ipulse and Moentu We last left off by talking about how the area under a force vs. tie curve is ipulse. Recall that for our golf ball we had a strongly peaked force curve: F F avg t You have to have a changing force because as you initiate contact, you have a rapid increase and as you take the club away, you lose that contact. If we find the ipulse (Δp), what we ve found is the area under the F avg curve since ipulse Δp F avg Δt. So now that we know all about ipulse, we can talk about conservation of oentu. Reeber that conservation eans initial final. Recall that we did a specific exaple where ass was conserved. Let s look at a situation where we have a collision. Collisions are the archetypal situation for studying oentu. We have two balls of equal ass, one oving toward the other at rest. The initial oentu of this syste is easy. It s the oentu of ball 1. Now consider the final oentu. Ball 1 strikes ball 2 and it shoots off with a lot of oentu while ball 1 oves with just a little oentu. So the final oentu is the addition of these two. In this way we can see the initial and final oentu atch. It is just a atter of how the coponents of the syste share the oentu. So how do we justify this conservation. It s not exactly an obvious conclusion. We can start fro Newton s Second Law: a "v v f # v i p f # p i "p "p

2 Fro here we can ask the question, what happens when 0? "p 0 # "p 0,so p f p i The significance of this is that when there is no net external force in a syste, oentu is conserved. Now let s do an exaple to clarify soe of the details. We take a syste of two blocks separated by a spring. 1 2 If we allow the spring to expand and push the blocks away fro one another we can use the conservation of oentu to analyze the syste. The spring force is an internal force to the syste (no external forces, like friction), so we can say that 0, and hence, Δp 0. We then define our initial and final conditions. Initially, the syste is at rest. In the final state you have two separate contributions to the oentu, p 1 and p 2. Now, let s assign soe quantities to the proble. Let 1 1.0kg and 2 2.0kg. If 1 has a velocity of 1.8 /s to the left, what is the velocity of 2? Because the only force acting in this proble is the spring which is internal to the syste we can eploy conservation of oentu. Recall that we can use the conservation of oentu since there are no external forces (i.e. friction). "P 0 p 1 p 2 P i P f 0 p 2 # p 1 p 2 p 1 2 v 2 1 v 1 v 2 1 v 1 2 (1.0kg)(1.8 /s) (2.0kg) 0.9 /s, to the right

3 Another good exaple to try: A 2000 kg car is traveling north at 15 /s when it overtakes and crashes into a 5000 kg truck also traveling north and oving with a speed of 10 /s. Find the velocity of the cobined wreckage the instant after the collision. Initially we have the oentu of the car and truck separately. In the final state we have the collective velocity of the car and truck stuck to each other. P i p car + p truck P f p car+truck Since there are no external forces (the forces of the car crashing into the truck are internal) ΔP 0, so P i P f Pi car v car + truck v truck Pf ( car + truck )V car v car + truck v truck ( car + truck )V V ( car v car + truck v truck )/( car + truck ) ( )/(7000) 11.4 /s This has the right units, and it also falls between 10/s and 15/s. A 10g bullet is fired into a 1kg wood block, where it lodges. Subsequently, the block slides 4.00 across a wood floor (µ k 0.2 for wood on wood). What was the bullet s speed? First we need to define our initial and final conditions: Initially we have the bullet oving and the block at rest: B v B M 1kg V P i p bullet P f P block+bullet ΔP 0 since there are no external forces.

4 P i P f p bullet p block +bullet v (M + )V v (M + )V We have all the asses, but we don t know the velocity of the ass+block syste. So we have to go back to our kineatics equations. We can list our knowns and unknowns:? v f 0 Δx 4.0 a? But we can find a using Newton s Second Law! a We should go ahead and ake a free body diagra: ( ) x " f a x f µn f N W # "µn a x ( ) y N " W a y 0 N W g a x " f " µg "µg "0.2(9.8) 1.96 "2.0/ s2 Now we can go back to find the velocity using kineatic equations:

5 v a"x v 2 # 2a"x 0 # (2)(#2 /s 2 )(4.0) 4 /s Now recall the original solution we cae up with for the bullet: v v (M + )V (1kg +.01kg)(4 /s) (.01kg) 404 /s