Retrospective Before we continue with momentum and impulse, let’s step back and think about where we have come from.  English! What tools do you have in your toolbox?  algebra and trigonometry  kinematics (motion without worrying about forces)  dynamics (Newton’s laws, forces) rotational dynamics is a subset of this  English!  algebra and trigonometry  kinematics (motion without worrying about forces)  dynamics (Newton’s laws, forces) rotational dynamics is a subset of this  conservation of energy (work and energy) We have just added another tool—conservation of momentum.

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.

In 1905, mathematician Emmy Noether proved the following theoremIn 1905, mathematician Emmy Noether proved the following theorem: For every continuous symmetry of the laws of physics, there must exist a conservation law. For every conservation law, there must exist a continuous symmetry. The conservation law corresponding to time translational symmetry is the Law of Conservation of Energy (we have seen a “special case” of this law—conservation of mechanical energy). The conservation law corresponding to space translational symmetry is the Law of Conservation of Momentum. These conservation laws “emerge from symmetry concepts far deeper than Newton’s laws.”

Impulse Example: 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. Important: here is your litany for momentum problems.here 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. beforeafter

Step 2: label point masses and draw velocity or momentum vectors (your choice). beforeafter vivi v f =0 m m 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. xx Step 4: OSE. OSE: F x =p x /t

Steps 5 and 6 are not applicable to this problem. Step 7: solve. F WC,x =p Wx /t F WC means force on water by car F WC,x = (P WFx – P Wix ) / t In a time of t=1s, m=1.5 kg of water hits the car. F WC,x = (mv WFx – mv Wix ) / t F WC,x = m (v WFx – v Wix ) / t F WC,x = 1.5 kg (0 m/s – 20 m/s) / 1 s F WC,x = -30 N  F CW,x = 30 N

Conservation of Momentum Example: A moving railroad car, mass=M, speed=V i1, collides with an identical car at rest. The cars lock together as a result of the collision. What is their common speed afterward?

Step 1: draw before and after sketch. V i1 V i2 =0 MM before VfVf MM after 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.

V i1 V i2 =0 MM before VfVf MM after Step 3: choose axes, lightly draw in components of any vector not parallel to an axis. xx Step 4: OSE. OSE: P f = P i because F ext =0 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.

Step 5 will be applicable later. V i1 V i2 =0 MM before VfVf MM after xx Step 6: write out initial and final sums of momenta (not velocities). Zero out where appropriate. P fx = P ix p 2mfx = p 1ix + p 2ix 0

Step 7: substitute values based on diagram and solve. V i1 V i2 =0 MM before VfVf MM after xx (2m)(+V f ) = m(+V i1 ) V f = mV i1 / 2m V f = V i1 / 2