A hot-air balloon rises vertically at velocity vBE, as measured from the earth. The balloonist hurls a rock horizontally outward at velocity vRB as measured from the balloon. Which path best represents the velocity vRE, of the rock as measured from the earth? 10/9/17 OSU PH 211, Before Class 7

This is not news—we have a sense that this must be true: The velocity of the rock (from the earth’s point of view) is the combination (the vector sum, in fact) of the velocity of the rock from the balloon’s point of view and the velocity of the balloon from the earth’s point of view. 10/9/17 OSU PH 211, Before Class 7

Try another such “thought experiment…” You and a friend are traveling in a convertible (open-top) car. It is moving horizontally along a straight road at some steady vCE, “the velocity of the car as measured with respect to the earth.” As your friend drives, you sit in the passenger seat and toss a ball vertically upward (from your point of view). That is, vBC, “the velocity of the ball as measured with respect to the car,” is straight upward. What does a standing observer see as you release the ball? That is, what is vBE, “the velocity of the ball as measured with respect to the earth?” The standing observer would see you launch the ball up and forward in an arc that brings it back to your hand as you travel beneath it. 10/9/17 OSU PH 211, Before Class 7

Again, this combination (the vector sum): The velocity of the ball from earth’s point of view is the velocity of the ball from the car’s point of view plus the velocity of the car from the earth’s point of view: vBE = vBC + vCE And notice: The subscripts in the vector sum on the right form a sort of chain—where the “inner” ones (C in this case) match: BC…CE And the outer ones are the subscripts of the desired sum on the left. – “From the earth’s point of view” applies to anyone not moving at all compared to the earth—that’s the standing observer here. –”From the car’s point of view” applies to anyone not moving at all compared to the car—that’s you and your friend, riding in the car. From your point of view, every part of the car (seats, steering wheel, dashboard, mirrors, etc.) stays at the same position compared to you. 10/9/17 OSU PH 211, Before Class 7

Relative Motion (Galilean Relativity): vAC = vAB + vBC Definitions vBC = “the velocity of B relative to C” or “the velocity of B with respect to C” or “the velocity of B from C’s point of view” (The observer is always at rest with respect to him/herself.) and: vBC = – vCB (If B is moving east from C’s point of view, C is moving west from B’s point of view.) Common applications: • Motion from a moving vehicle (see previous examples). • Impact speed when moving objects collide (see next examples). • Motion in a moving medium (water, air, etc.)—in Monday’s class. 10/9/17 OSU PH 211, Before Class 7

Impact velocities of collisions (vBA = –vAB, so ||vAB|| = ||vBA||) Head-on: Car A moving east at 60 mph: vAE = 60Ð0° Car B moving west at 45 mph: vBE = 45Ð180° What is vAB (A’s motion as observed by B)? vAB = vAE + vEB (use the subscript “chain rule”) = vAE + (–vBE) (vEB = –vBE) vAE = 60 –vBE = 45 vAB = 105 B sees A moving east (approaching B) at 105 mph. 10/9/17 OSU PH 211, Before Class 7

Same collision, observed from A rather than B: Head-on: Car A moving east at 60 mph: vAE = 60Ð0° Car B moving west at 45 mph: vBE = 45Ð180° What is vBA (B’s motion as observed by A)? vBA = vBE + vEA (use the subscript “chain rule”) = vBE + (–vAE) (vEA = –vAE) –vAE = 60 vBE = 45 vBA = 105 A sees B moving west (approaching A) at 105 mph. 10/9/17 OSU PH 211, Before Class 7

Impact velocities of collisions (vBA = –vAB, so ||vAB|| = ||vBA||) Merging: Car A on eastbound fwy. at 60 mph: vAE = 60Ð0° Car B merging onto fwy. at 45 mph: vBE = 45Ð10° What is vAB (A’s motion as observed by B)? vAB = vAE + vEB (use the subscript “chain rule”) = vAE + (–vBE) (vEB = –vBE) vAE = 60 10° vAB –vBE = 45 B sees A approaching B from Ж26.5° (from back-left) at about 17.5 mph. 10/9/17 OSU PH 211, Before Class 7

Same collision, observed from A rather than B: Merging: Car A on eastbound fwy. at 60 mph: vAE = 60Ð0° Car B merging onto fwy. at 45 mph: vBE = 45Ð10° What is vBA (B’s motion as observed by A)? vBA = vBE + vEA (use the subscript “chain rule”) = vBE + (–vAE) (vEA = –vAE) –vAE = 60 10° vBA vBE = 45 A sees B approaching B from Ð153.5° (from front-right) at about 17.5 mph. 10/9/17 OSU PH 211, Before Class 7