Lucky K.

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Physics (Newtonian Mechanics)

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Question:

You are in a car driving straight towards a wall 10kms away. You are traveling at a constant speed of 10km/h. At time t = 0, a bee starts flying towards your car at a constant speed of 20km/h. After some time, the bee reaches your car's windshield and immediately turns back towards the wall. When it reaches the wall, it again turns back towards the car. This continues till you reach the wall and the bee gets squashed between the windshield and the wall. Calculate the total distance traveled by the bee.

Lucky K.

Answer:

There are two ways to solve this question. The first one involves a bit advanced calculations while the second one is easier and smarter way of solving the question. Let us first understand the one with more calculations. Here we will try to calculate the distance traveled by the bee between each U-turn taken by the bee. For the first turn, the distance traveled can be calculated using the relative speed of the car and the bee and finding out at which point both will meet. $$ Distance\space traveled \space (car \space + \space bee) = 10kms $$ $$Time\space taken = \frac{10 kms}{(10+20)km/h} = \frac{1}{3}hours$$ $$\therefore Distance \space traveled \space by \space the \space bee, \space d_1 = \frac{20}{3}kms$$ The bee will travel the same distance back to the wall $$\therefore d_2 = \frac{20}{3}kms$$ At the instance the bee again reaches the wall, the distance between the bee and the wall will be $D'_1 = 10 - \left( \frac{2}{3}*10 kms \right) = \frac{10}{3}kms$ We can continue with this analysis till the end. The total distance traveled by the bee would be $$ D = d_1 + d_2 + d_3 + ....$$ $$ D = \frac{20}{3} + \frac{20}{3} + \frac{40}{9} + \frac{40}{9} + .... $$ This will form an infinite series which when solved will give the correct answer. We can calculate it to get the correct answer. Now we will come to the second method. For this method you need to read the question carefully. You will realize that the bee will keep flying at a constant speed till the car reaches the wall. The total time spent traveling for the car and the bee will be equal. As the car too is traveling at constant speed the time taken can be easily calculated $$Time \space taken = \frac{Distance}{Speed}$$ $$t = \frac{10kms}{10kms/h}= 1 \space hour$$ Hence the total distance traveled by the bee $$D = Speed*Time = 20*1 = 20kms$$ This question shows the importance of reading the question carefully and understanding it to solve it using the fastest way possible.

Physics

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Question:

You are standing on a conical pillar exactly at the equator of the earth. You have a gun from the future which can fire shots at any speed you enter into its settings. You need to shoot an object at the bottom of the pillar which is not visible from the top. Find the speed needed for the bullet to travel around the earth and hit the bottom of the pillar if you fire parallel to the ground below you. Height of the pillar = 10m Radius of Earth = 6400km

Lucky K.

Answer:

In this question you should realize that the horizontal and vertical axes can be solved separately with the total time taken being common for them. For the vertical axis, $$Initial \space Speed, \space u_y = 0$$ $$Distance \space traveled, \space d_y = 10m$$ $$Acceleration \space due \space to \space gravity, \space g = 10m/s^2$$ $$\implies 10m = \frac{1}{2}*10*t^2$$ $$\therefore t = \sqrt{2} For the horizontal axis, $$Initial \space Speed, \space u_x = x$$ $$Distance \space traveled, \space d_x = 2 * \pi * 6400000 m \space (Circumference \space of \space the \space earth)$$ $$Acceleration, a_x = 0$$ $$\implies 40192000 = x * t$$ $$\therefore x = \frac{4019200}{\sqrt{2}} \approx 28420035m/s^2$$ Hence you will need a lot of firepower to make a bullet go around the earth.

Physics

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Question:

It is the year 2200 and intergalactic travel is fairly common. You are going for a mission at a space station 10 light years away from Earth. You have the fastest space-jet available to mankind with a top speed of 0.99c (where c is the speed of light) relative to a stationary observer on Earth. You start on the day of your 21st birthday. How old will you be by the time you reach the space station? Hint: Einstein's Theory of Special Relativity Assumption: Special acceleration technique on earth helps you reach the speed of light instantaneously. Ignore the effects on the human body due to such acceleration. Similarly you decelerate instantaneously on the space station. Neglect the effects of the earth's motion on the relative speed.

Lucky K.

Answer:

It may seem that it is a basic question about distance, time and speed. However, from Einstein's Theory of Special Relativity, we know that time acts differently at speeds close to the speed of light. Let us first look at it from the perspective of an observer on earth. $$Time = \frac{Distance}{Speed}$$ $$\therefore Time = \frac{10 * c}{c} \space years = 10 \space years$$ For an observer sitting on earth, you would have taken 10 years to reach the space station. But for you, it would not be the same due to a phenomenon known as time dilation. So if you are moving at a speed v relative to the stationary observer, the relation between time taken as seen by you and the observer on earth would be given by $$\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where $\Delta t'$ is the time taken as seen by the stationary observer and $\Delta t$ is the time taken as seen by you. Hence for the given case, $$\Delta t = 10 * \sqrt{1-\frac{(0.99c)^2}{c^2}} = 10 *\sqrt{1 - 0.9801} \approx 10*0.1411\approx 1.411\space years$$ You will be 22 years and 150 days old when you reach the space station. This is a perfect example of how fascinating physics could be. While a lot of the readers might feel that such a space vehicle would never be created hence there is not much use of the theory. This small fact might be useful for you. The GPS in your phone would not be so accurate if the relativistic effects were not known.

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