# Changing Direction in a Vaccum [closed]

If a mass of one kilogram is traveling at one meter per second at 90 degrees, how much energy is required to get it to travel going 180 Degrees?

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## closed as too localized by user1504, Chris White, Waffle's Crazy Peanut, David Z♦Apr 2 '13 at 5:45

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That really depends on how you are going to deflect the motion.

This is the best case: you apply a radial force. For instance, you could attach a fixed string to your mass, effectively making it a pendulum (if that helps your imagination). What's important is that the mass will be moved in a circular path, because the taut string provides a fixed radius. The radial force will never do any work (since force and movement are always perpendicular to each other). If you release the string after the 90° turn has been completed, you will not have put in any work into the system.

It could be much worse though (and will in reality be somewhat worse). For instance, you could also slow the mass to rest and then accelerate it to the terminal velocity in the new direction. This would require you to put in an amount of work equal two twice $E_{kin}=mv^2/2$. Given your quantities, that would be $W=1\,\text{J}$.

But in theory, your initial energy is the same as the final energy, because kinetic energy is a scalar quantity and does not depend on the direction of travel. Therefore, if you do it cleverly, you don't need any energy (or work) at all.

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I presume the object is intended to move at same velocity in the new direction. Since the new direction is at 90 degree separation, You must apply a force for 1 sec at 225 degrees (optimum, since sin and cosine would become of same angle and later you'll see it cancels easily). So one component of the force (F1) would act along -90 degrees or 270 degrees that will cause the object to come to a halt along 90 degrees, and another component (F2) would make it accelerate to 1 m/s along 180 degrees. As given, m=1kg, v=-1m/s, F1 = 1kg*1m/s^2 = 1N. And F2 = 1kg*1m/s^2 = 1N. Hence, F=Sq.Root[(1N)^2 + (1N)^2] = 1N along 225 degrees (sin^2 (45) + cos^2 (45) = 1, that's why 45 degree is chosen). But again to calculate energy, you must calculate the work done. for F1, W1=F1.s1 where s1=slow down distance which is .5m (1m/s*1s - .5*1m/s^2*1s^2). so W1=1N*.5m=.5J. Similarly W2=.5J. Hence, W=W1+W2=1J. So energy required is 1J.

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Hi Khan. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. This markup is very useful for understanding math equations and parameters. Please have a look here for an intro or our FAQ for more info ;-) – Waffle's Crazy Peanut Apr 12 '13 at 18:05