1. Relative Motion: Suppose you are on a train platform as the train rushes through the station without stopping. Someone on board the train is pitching a ball, throwing it has hard as they can towards the back of the train. If the train’s speed is 60 mph and the pitcher is capable of throwing at 60 mph, what is the speed of the ball as you see it from the platform? 60 mph Train Pitcher a)60 mph to the right b)120 mph to the right c)0 mph (not moving) d)60 mph to the left

2. Correct Answer: C Before it was thrown the ball was moving to the left at 60 mph, because it was moving with the train’s velocity. If the pitcher had simply dropped the ball then its horizontal velocity would have continued to be 60 mph to the left, just as the pitcher’s was. Otherwise our experience would be that everything in the train would get left behind as soon as it started moving. So when the pitcher throws the ball to the right at 60 mph, it must be understood, from the point of view of someone standing on the platform, that the ball is being accelerated not from a state of rest (no motion), but from a state of motion 60 mph to the right. It seems clear that two velocities one of 60 mph to the right and one of 60 mph to the left (or –60 mph) must cancel each other out.

3. Now suppose the pitcher on the train threw the ball so that its motion was not entirely horizontal. Perhaps the motion is angled upwards at an angle of 20 degrees to the horizontal. Now we have two vectors, v(train) which describes the velocity of the train as seen by the person on the platform, and v(pitch) which describes the velocity of the ball as seen by the pitcher on the train. What is the velocity V of the ball as seen by the person on the platform? v(train) Train Pitcher v(pitch) a)V = v(pitch) – v(train) b) V = v(pitch) b)V = v(train) – v(pitch) d) V = v(pitch) + v(train)

4. Correct Answer: D In the first problem what we essentially did was to add the velocities of the train and the ball 60 mph + (-60mph) = 0 mph It makes sense that we can still do the same thing with two dimensional vectors. Before the ball is thrown the man on the platform sees it moving with the same speed as the train. Its velocity is v(train). When the pitcher throws it he is adding to its motion, in some sense. The new velocity, as seen by the man on the platform, is the old velocity + the new velocity, or v(train) + v(pitch). v(pitch) v(train) v(pitch) V V = v(train) + v(pitch)

5. Now, what velocity does the pitcher on the train see the person on the platform moving with? Recall that the pitcher is on a train which is moving with a velocity of 60 mph to the left. a)60 mph to the right b)60 mph to the left c)0 mph (not moving) d)120 mph

6. Answer: A The pitcher on the train is not actually aware of his or her own motion, because if they drop the ball or throw it, it behaves just as it normally would, falling straight down or moving with its usual speed (as that person sees it). But if they look out the window they will notice that everything they can see appears to be “falling behind.” It is as if everything else in the world that is not on the train is moving towards the rear of the train. Therefore if the person on the train is moving to the left at 60 mph, they will see the person on the platform, the platform itself, and everything else not on the train moving to the right at 60 mph. In short when you are in a vehicle moving with velocity V you think you are at rest, but everything outside the vehicle is moving with velocity –V.

7. So now, if the person on the platform pitches a ball with velocity v(pitch), what velocity, V will the person on the train see the ball moving with? v(pitch) v(train) Train Pitcher on platform a)V = v(train) + v(pitch) b) V = v(pitch) – v(train) c) V = v(train) – v(pitch) d) V = v(pitch)

8. Correct Answer: B We have already established that if you see someone pitch a ball then the velocity of the ball, as you see it, will be v(pitch) + v(pitcher) where v(pitcher) is the velocity of the person throwing the ball, as you see it, and v(pitch) is the velocity with which that person throws the ball. So what is the velocity of the pitcher in this case, as seen by the person on the train? It is simply v(pitcher) = –v(train). Therefore V = v(pitch) + v(pitcher) = v(pitch) – v(train)

9. So if you are on a place, moving northeast at 50 mph, and you look out the window, and see a car which is driving on a road which points due south at 30 mph, what of the following is correct about the relative velocity of the car, as seen by you from the plane? a)You see the car moving south and east b)You see the car moving north and east c)You see the car moving north and west d)You see the car moving south and west

10. Correct answer: D (south and west) v(plane) v(car) -v(plane) v(car) V = v(car) – v(plane)