Collision between Football Players A and B: Calculation of Total Change in Mechanical Energy

1. Football player A (100 kg) is moving at 4.5 m/s when he collides and hangs onto player B (150 kg), who is moving at –2.0 m/s. Assuming the collision happens in the air (that is, the ground doesn’t influence it), calculate their total change in mechanical energy (from the instant before the collision to the instant afterward). A.DEmech = 0 J B.DEmech = – 300 J C.DEmech = –1.27 x 103 J D. Not enough information. E. None of the above. Oregon State University PH 211, Class #25

2. Tarzan is caught in the path of stampeding elephants, so Jane swings down on the end of a vine, to try to haul him up to safety. The vine’s length is 25 m, and Jane starts her swing with the vine horizontal. Jane’s mass is 54 kg; Tarzan’s is 82 kg.To what height above the ground will the pair swing after she grabs him? (Treat the vine as mass-less and each person as a point mass. Ignore air drag.)A. 3.94 mB. 6.95 mC. 9.93 mD. 12.9 mE. None of the above.A James Bond movie has a similar scene with skiers—and they apparently violate at least one physics principle that applies here. (Mr. Bond does indeed seem to be above all laws.) Oregon State University PH 211, Class #25

3. Another Way to Store Potential Energy The oscillation of a mass on an ideal spring continues undiminished if there is no friction—or other external force—doing work on the mass. That is, mechanical energy is conserved in an ideal spring. We get that energy back—by releasing the spring (just as we get the work back from gravity, by releasing an object to fall). So, how much energy can we store in such a spring? We know the work done when pushing directly upward, against the constant force of local gravity (W = –FG·s = –|mg||Dy|cosf = mgDy). But unlike local gravity, the force exerted by a spring to oppose displacement of a mass is a function of that displacement: Fspring.mass = –kxmass So, to find the work invested in the displacement, we must integrate: W = – ∫Fspring.mass.xdx = – ∫–kxmassdx = (1/2)kx2 Oregon State University PH 211, Class #25

4. Spring (or “elastic”) Potential Energy • This is the elastic potential energy associated with the displacement of a stretched or compressed ideal spring: • Uelas = US = ½k(x)2 • Note the SI units will be (N/m)·m2, or N·m, or J. • Note where x must be measured from—always from the “rest position” of the spring (i.e. where the spring is neither stretched nor compressed). That position is always defined as x = 0. • In the absence of external forces that do work on the system, Conservation of Mechanical Energy can still hold: • ΔEmech = ΔK + ΔUg+ ΔUs = 0 Oregon State University PH 211, Class #25

5. So the fully-developed Work-Energy Theorem now looks like this—now with three “sub-accounts” in the Mechanical Energy “bank:” • Emech = KT + UG + US • = (1/2)mv2 + mgh + (1/2)kx2 • Any change in the total bank balance is due to “deposits” and/or “withdrawals”—work done by external forces (forces other than gravity or ideal springs): Wext = Emech • Or, arranged differently: Emech.f = Emech.i + Wext • Fully detailed: (1/2)mvf2 + mghf + (1/2)kxf2 • = (1/2)mvi2 + mghi + (1/2)kxi2 + Wext Oregon State University PH 211, Class #25