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Linear Momentum Momentum and Force Conservation of Momentum Collisions and Impulse. Before we get into chapter 7, let’s step back and think about where we have come from. What’s in your Physics toolbox?. English!. algebra and trig. kinematics (motion without worrying about forces).
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Linear Momentum Momentum and Force Conservation of Momentum Collisions and Impulse
Before we get into chapter 7, let’s step back and think about where we have come from. What’s in your Physics toolbox? English! algebra and trig kinematics (motion without worrying about forces) • dynamics (Newton’s laws, forces) (including rotational dynamics) conservation of energy (work and energy) We are about to add another tool—conservation of momentum.
Chapter 7 Linear Momentum 7.1 Momentum and its Relation to Force The momentum is of an object of mass m is defined as v v p p Momentum is a vector quantity! You must pay attention to all our rules on the handling of vectors.
This gives another OSE: A force is required to change an object’s momentum. The correct statement of Newton’s second law is actually F=dp/dt (or F=p/t if you prefer to avoid calculus), from which F=ma can be derived.
before after Example 7-1. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it without splashing it back (kind of a fake problem, but that’s OK). What is the force exerted by the water on the car. After we finish section 7.3 I’ll give you a litany for momentum problems. For now, I’ll modify our “usual” steps. Step 1: draw before and after sketch. You can draw a fancy sketch, but I suggest you save time and draw point masses. Make sure you have SEPARATE before and after parts.
vi vf=0 m m x x before after Step 2: label point masses and draw velocity or momentum vectors (your choice). Hint: draw unknown velocity (or momentum) vectors with components that appear to be positive, to avoid putting extra – signs into your work. Step 3: choose axes, lightly draw in components of any vector not parallel to an axis. Step 4: OSE. OSE: Fx=px/t
x x vi vf=0 FWC m m before during Steps 5 and 6 are not applicable to this problem. Step 7: solve. In a time of t=1s, m=1.5 kg of water hits the car. FWC means force on water by car This drawing is not part of the “official” procedure; it is to help you visualize the force.
FWC,x = -30 N FCW,x= 30 N FWC,x = -30 N FCW,x= 30 N Steps 5 and 6 are not applicable to this problem. Step 7: solve. In a time of t=1s, m=1.5 kg of water hits the car. FWC means force on water by car FWC,x=pWx/t FWC,x = (PWfx –PWix) / t FWC,x = (mvWfx – mvWix) / t FWC,x = m (vWfx – vWix) / t FWC,x = 1.5 kg (0 m/s– 20 m/s) / 1 s
By way of example, I have evidently told you it is now OK to not include in your diagram “things” that appear in your equations. • You can show either p or v. The one you don’t show will appear in your equations but not your diagram. That’s now “legal.” • You are not explicitly required to show the forces that appear in F=p/t. • If the problem asks for a force, it is a good idea to show it.
7.2 Conservation of Momentum The momentum of a single particle is denoted by lowercase p. For a system of particles, we define the total momentum P by where M is the total mass of all particles and Vcm is the velocity of the center of mass of the system (we will define the center of mass later in this chapter). A note on notation: I will use arrows for vectors whenever possible, but it is usually not convenient in a block of text. Lowercase p is the momentum ofaa single particle. Script P is the total momentum of a collection ofparticles on the Phys. 23 OSE sheet. I will usePin my notes.
It has been observed experimentally and verified over and over that in the absence of a net external force, the total momentum of a system remains constant. The above is a verbal expression of the Law of Conservation of Momentum. It sounds like an experimental observation, which it is… …which implies maybe we just haven’t done careful enough experiments, and that maybe some day we will find the “law” is not true after all. But the Law of Conservation of Momentum is much more fundamental than just an experimental observation.
If you assume that the laws of physics are invariant under coordinate transformations, then the Law of Conservation of Momentum follows mathematically and inevitably. Every time I let you choose your coordinate system, I have used that assumption of invariance. If the assumption is false, then the laws of physics will be different for everybody, and there is no point in doing physics. Any violation of the Law of Conservation of Momentum would be as revolutionary (if not more so) as Einstein’s relativity. Most likely, any “new” laws of physics would contain all our “old” ones, which would still work under “normal” circumstances.
Our definition of system momentum, combined with the relationship between force and momentum, gives and our OSE expressing conservation of momentum is Example 7-3.A moving railroad car, mass=M, speed=Vi1, collides with an identical car at rest. The cars lock together as a result of the collision. What is their common speed afterward?
Vi1 Vi2=0 Vf? M M M M before after Step 1: draw before and after sketch. Step 2: label point masses and draw velocity or momentum vectors (your choice). If I had made a pictorial sketch (i.e., drawn railroad cars), at this point I would probably re-draw the sketch using two point masses. For this example, I will stick with the above sketches.
x x Vi1 Vi2=0 Vf? M M M M before after Step 3: choose axes, lightly draw in components of any vector not parallel to an axis. Step 4: OSE. Caution: do not automatically assume the net external force is zero. Verify before using! I will assume the friction in the wheels is negligible, so the net force can be “zeroed out” here.
x x Vi1 Vi2=0 Vf? M M M M before after 0 Step 5 will be applicable after we study section 7.3. Step 6: write out initial and final sums of momenta (not velocities). Zero out where appropriate.
x x Vi1 Vi2=0 Vf? M M M M before after Vf= Vi1 / 2 Step 7: substitute values based on diagram and solve. p2mfx= p1ix (2m)(+Vf)= m(+Vi1) Vf= mVi1 / 2m
7.3 Collisions and Impulse The net force on an object is Fnet = F=p/t. From this, we see that a force applied for a time t produces a change of momentum p: Fnet t= p. Impulse is defined as Fnet t. (Note that Fnet and Fexternal and Fnet, external mean the same thing in this class.) I recently revised the OSE sheet to include the subscript “net” on F. The above equation is valid if the force changes little during the time t.
If the force depends on time, we have to integrate (but we won’t do that in Physics 31). Demonstrate impulse with ball and bat. You’ll see a more thorough discussion of impulse in the next section. Important: here is your litany for momentum problems. Here is a link to a site that has lots of introductory-level problems on momentum and impulse. (Here is a sample problem. Click the letter of your choice. Find out if you are correct, or what your mistake was.)