Projectile motion on Ideas to Implementation question. (1 Viewer)

[nirukk]

This is from 2009. I don't get it why they use horizontal component instead of vertical. Can someone explain part b ?

 

[alexobern]

Part B is essentially your usual projectile, but with the force that the field is applying divided by the mass of the object instead of gravity, so use your deltay=uyt+1/2ayt^2, solve for t, it shouldn't matter that you use the vertical componenet, or as they have done, v=u+at where v=-u, solve for t, should get the same answer

 

[Parvee]

For part a) you use F=qE where E=V/d and you should get a force of 1.602x10^-16N down towards the 0V plate.

b)To work out the acceleration you equate F=ma and F=qE therefore a=qE/m and you should get 1.759x10^14 ms^2
Using v=u+at, v is 0 at max height therefore t=-u/a and you should get t=2.95x10^-8 seconds (this is time to max height)
To get time of flight you times this by two and you should get t=5.91x10^-8 seconds

 

[nirukk]

For part a) you use F=qE where E=V/d and you should get a force of 1.602x10^-16N down towards the 0V plate.

b)To work out the acceleration you equate F=ma and F=qE therefore a=qE/m and you should get 1.759x10^14 ms^2
Using v=u+at, v is 0 at max height therefore t=-u/a and you should get t=2.95x10^-8 seconds (this is time to max height)
To get time of flight you times this by two and you should get t=5.91x10^-8 seconds

That's how I done it, but the sample answers uses somehow horizontal component of the velocity and their answer is 2.9x10^-8 seconds.

 

[GoldyOrNugget]

5.91e-8 is the correct answer.

 

[kiinto]

That's how I done it, but the sample answers uses somehow horizontal component of the velocity and their answer is 2.9x10^-8 seconds.

I just checked the 2009 Sample Answers. They do use vertical velocity ('the perpendicular component of the velocity'), however for some inexplicable reason they do not follow the final step and multiply by two. Parvee's right though, and, if you did it that way, you are too.

 

[nirukk]

I just checked the 2009 Sample Answers. They do use vertical velocity ('the perpendicular component of the velocity'), however for some inexplicable reason they do not follow the final step and multiply by two. Parvee's right though, and, if you did it that way, you are too.

Thank you. I checked the marking guidelines now and they seem to have a reference to horizontal component of the velocity. Can you have a look at that too? Thanks again.

 

[kiinto]

Thank you. I checked the marking guidelines now and they seem to have a reference to horizontal component of the velocity. Can you have a look at that too? Thanks again.

Dunno bro. Seems like a typo to me. I don't see why the horizontal component would come into it, plus they contradicted themselves by doing a different thing in the sample answer.