Some silly questions for you, but you need to know a bit about physics to work these out.

- Mad Professor Zweistein has managed to build a car that has an unlimited to speed and incredible breaks that allow it to stop instantly. He only has one problem, the car is seven metres long and his garage can only accommodate a six-metre car. How fast does Professor Zweistein need to drive the car so that it will fit into the garage and when does he need to apply the breaks?
- During his attempt to get his car into the garage, Professor Zweistein is caught by a red light camera for driving through an intersection while the light was red. In court, Zweistein defended himself by saying that at the speed he was travelling the red light from the traffic light actually looked green (due to the doppler effect). Being a physics buff himself, the judge decided to take Professor Zweistein at his word. If the standard rate is $1 per km above the speed limit of 60 km/h, how much was the speeding fine issued by the judge to Professor Zweistein?

a) This question is about relativistic length contraction. As the velocity of the vehicle increases, its apparent length, as viewed from a stationary position contracts. The amount of contraction is calculated using L' = sqrt[1-((v*v)/(c*c)]*L where L' is the contracted length, v is the velocity of the vehicle , c is the speed of light and L is the original, uncontracted, length of the vehicle. As the vehicles stationary length is 7m and it has to contract to 6m to fit inside the garage, the velocity required can be calculated as

6 = sqrt[1-(v*v)/(c*c)]*7 which when rearranged gives sqrt{[1-(36/49)]*c*c} = v = 154523626 m/s. Now

1 m/s = 2.2369 miles per hour so the car would have to be doing 345653900 miles per hour to give a length to contraction of 1m, as viewed from the stationary garage.

b) The second part of the question asks when the brakes need to be applied. Once the brakes are applied, the vehicle is no longer travelling 345653900 miles per hour, relative to the stationary garage, its apparant length will return to the original 7m and so it will no longer fit in the garage. Therefore he should apply the brakes before he hits the back wall of the garage, but he will still not be able to fit the car inside!

Sent in by D Judson, University of Brighton

b) none- the judge took him at his word and didn't fine him.

Submittedby Dawn (m y s t i c a l . d a w n)

On question 44 about the mad Professor Zweistein, there are two parts. The first part involves two questions and a slightly complicated answer. However, assuming the first part was answered correctly, one can then answer the second part. Part B states that the Professor is caught on video tape running a red light. He is then taken to court for this offense. However, upon his hearing he defends himself by stating that at that speed the red light actually looked green. Well since the Judge is himself a physics buff, he realizes that the Professor is right. However, he then decides to give the professor a speeding ticket. The last Sentence of part B is "If the standard rate is $1 per km above the speed limit of 60 km/h, how much was the speeding fine issued by the judge to Professor Zweistein?" If you will notice, the professor was brought in for running a red light but was issued a fine for speeding. Since we have the answer as to how fast he was traveling from part A, we can then determine how much his fine was. The answer in part A states that the Professor was traveling at 154523626 m/s. Using a simple conversion of 1m/s = 3.6km/h we can deduce that the professor was traveling at 556285053.6Km/h. However, since the fine is $1 above 60km/h then you merely have to subtract 60km/h leaving you with a fine of $556284993.6.

Sent in by Michael Hutchins

The answer to Mindbenders 44 is incomplete at best. The statement that the car can instantaneously stop is not well defined. The reason is that instantaneous, or the phrase “at the same time” is dependent on the frame of reference you are in.

From the frame of reference of the garage, the car back end of the car enters the garage at the same moment that the front end touches the back wall. If the car stops instantaneously in this frame of reference, that is every point in the car stops at this point in time in the garage frame of reference, then an extremely compressed car stops and is inside the garage.

From the driver’s point of view the garage is compressed, but the front of the car touches the back of the garage long before the back of the car clears the door. If the car stops instantaneously in this frame of reference then the car is its normal size, but is never completely inside the garage.

Now, if the car stops instantaneously in the garage frame of reference at the moment the front of the car touches the back of the wall, what is it like from the driver’s point of view. He sees the front of the car stop just as it touches the wall, but all points behind the front keep moving. Points further back stop at progressively later times. Since the front has stopped and the back is still moving, the car is being compressed. The back stops just as it sneaks inside the door, and a highly compact car is left.

If the car stops instantaneously at the time the front touches the back wall from the driver’s point of view, what happens in the garage frame? Here the back stops first, long before the front reaches the back wall of the garage. Points further forward stop at progressively later times until the front stops just as it reaches the back of the garage. Because the front is now moving with the back stopped, the car is stretched from its apparent compressed form until it is its full length again.

The key point is that two events that take place at the same time in one frame of reference may take place at different times in another frame. This is almost always forgotten in problems like this.

Lee Loveridge, PhD