Experiments with scanner: relative motion MUSE Workshop

WCPE Istanbul, July 1-6, 2012

Published by the MUSE group (More Understanding with Simple Experiments) , Physics Education Division (PED) of the European Physical Society (EPS)

Experiments with scanner: relative linear motion[1],[2]

SOLUTIONS

PROBLEM A

I. Mark on the graph or explain how to determine from the graph the following quantities:

  1. xF– initial position (at time t = 0) of the front of the car
  2. xB– initial position (at time t = 0) of the back of the car
  3. a– length of the car
  4. b – lengthof thescanned image of the car
  5. vs – speed of the scanner head slope of the tilted line
  6. va– speed of the car zero (slope of the car lines)
  7. ta– time needed for the scanner head to scan the car only (see graph)

PROBLEM B

Car will be placed on the scanner window with front part pointing in the direction of scanning, as in previous case and moved with constant speed during the scanning. The magnitude of the speed of the car will be smaller than the magnitude of the speed of the scanner head (vavs).The car will be moving in the same direction as the scanner head (to the right).

I. Make a prediction how will the scanned image of the car in this case compare with the image shown in problem A. Describe your prediction by drawing a sketch of the anticipated image. Sketch your prediction image under the sketch of the car from problem A, next to letter B. Make sure you draw all details including the letters that are painted on the bottom of the car.

Describe your reasoning in few sentences:

Car is moving in the same direction as scanner head => it will take more time for the car to pass the scanner head then before => the image will be longer than the original image (see the explanation how scanner works and equation on the first page). Letter on the bottom will appear stretched in horizontal direction but the order will be the same as before.

IV. Draw the x(t)graph for the motion of the car and the scanner head in this problem. You may assume that at t =0the car and the scanner head were already moving with constant speeds.

I. Mark on the graph or explain how to determine from the graph the following quantities:

  1. xF– initial position (at time t = 0) of the front of the car
  2. xB – initial position (at time t = 0) of the back of the car
  3. a– length of the car
  4. b – lengthof thescanned image of the car
  5. vs – speed of the scanner head slope of the scanner head line
  6. va– speed of the car slope of the car line (va < vs)
  7. ta – time needed for the scanner head to scan the car only

V. Check if characteristics of the image are consistent with the quantities that can be determined from the graph. Longer car on a scanned image is consistent with b > a on the graph.

PROBLEM C

Based on the scanned image obtained in the last experiment (Problem B) calculate the speed of the car as measured in the laboratory reference frame va and the speed of the car relative to the scanner head vrel (i.e. speed of the car as measured in the reference frame that is moving together with the scanner head). The speed of the scanner head was equal to vs = 7.5 cm/s. Show your calculation and use the same symbols as with the graphs.

Scanned image

The length of the stationary car is approximately equal to a = 9 cm and the length of the scanned image of the moving car is approximately equal to b =20 cm (see image on the first page and image above) => time needed for the scanner head to pass the moving car (i.e. distance b)is ta = b/vs = 2.67 s.

During the time that scanner head passed the car, the car moved by the distance equal to b – a (see graph in the previous problem). Therefore the speed of the car in the lab frame is equal to

va = (b-a)/ta = 4,1 cm/s

Knowing the speed of the scanner head and applying the Galilean transformation one can find the speed of the car relative to the scanner head:

vrel = va - vs = -3.4 cm/s.

Note that negative sign of vrel is consistent with what an observer in the scanner head reference frame would see.

Alternatively one can start from the scanned head reference frame. Viewed from the scanner head reference frame car moved bya (car’s length) during ta. So, the speed of the car relative to the scanner head is equal to vrel = -a/ta = -9cm/2.67s = -3.4 cm/s (negative sign was added to be consistent with the choice of the direction of axis).

Using the equations above one can express the length of the scanned image of the car in the following form:

b = a vs/(vs-va)

PROBLEM D

This time the car will be placed on the left edge of the scanner, outside the glass, as shown in the photo on the right. During the scanning the car will be moving with constant speed larger than the speed of the scanner head (vavs) and in the same direction as the head. So in this case the car starts behind the scanner head and pass it.

I. Make a prediction how will the scanned image of the car in this case compare with the image shown in problem A. Describe your prediction by drawing a sketch of the anticipated image. Sketch your prediction image under the sketch of the car from problem A, next to letter B. Make sure you draw all details including the letters that are painted on the bottom of the car.

or

Describe your reasoning in few sentences:

Since the car starts behind the scanner head, the first part of the car “seen” by the scanner is the front of the car. Sometime later the scanner “sees” the middle of the car and finally the back of the car. Therefore the scanned image is the mirror image of the car. Mirroring is clearly visible if you observe the letters that are written on the bottom of the car. Note that the length of the scanned image can be either longer, shorter or equal to the original length of the car, depending on the speed of the car relative to the scanner head.

Result of the experiment:

IV. Draw the x(t)graph for the motion of the car and the scanner head in this problem. You may assume that at t =0the car and the scanner head were already moving with constant speeds.

V. Check if characteristics of the image are consistent with the quantities that can be determined from the graph.

The equations that we derived in Problem C are valid also in this case (we leave readers to verify this). Note that in this case va > vs and therefore the equation

b = a vs/(vs-va)

gives negative b.This is consistent with observation if we interpret the negative sign as mirroring the image.

1

[1]The MUSE group (G. Planinsic, E. Sassi, L. Viennot) takes responsibility for the content of this paper (July 2012). The intellectual property remains with the author Gorazd Planinšič.

[2]Author wishes to acknowledge Eugenia Etkina and Bor Gregorčič for valuable discussions and suggestions.