#Projectile motion on an incline

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boreal fjord
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A body is projected up with speed v along the line of greatest slope of an inclined plane with inclination "A". If body collides elastically perpendicular to the plane, find the time after which the body passes through its point of projection.

hexed tendonBOT
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boreal fjord
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When the body collides with the plane, it will only have the vertical component of velocity.

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As the collision is elastic, this vertical velocity just flips it direction

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That means, the object shouldn't reach its original point of projection?

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This is what I mean.

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Do, we have to consider the fact that the object will roll down the incline to original point (and add this time)

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T1= time to go up the incline
T2= for vertically up and down motion
T3= Rolling down

T=T1+T2+T3

boreal fjord
shrewd sand
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shouldn’t it bounce back to like exactly where it started or something

boreal fjord
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So, at collision, you only have Vy

shrewd sand
shrewd sand
boreal fjord
boreal fjord
shrewd sand
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yeah ok

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the body should literally just retrace its path back to the start though

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like the velocity gets negated

boreal fjord
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That's what the solutions says

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But, it doesn't make sense to me

shrewd sand
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why not panda_think

boreal fjord
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The fact that Vy reverses direction at contact but Vx=0 so net velocity is in Y direction

shrewd sand
boreal fjord
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As Vx=0 so there is nothing to flip!

shrewd sand
shrewd sand
boreal fjord
shrewd sand
boreal fjord
shrewd sand
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ok wait can I tell you how I’m imagining the problem

boreal fjord
shrewd sand
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if I flip 0 then it is still 0

boreal fjord
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Can you show a diagram?

shrewd sand
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so if the velocity was (0, v) before then it’s (0, -v) now

shrewd sand
boreal fjord
boreal fjord
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0i-vj flips to 0i+vj

shrewd sand
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purple line is the velocity before collision, blue is after

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they are parallel have the same magnitude and are in opposite directions

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ok suppose you imagine there’s a video of the ball’s path from where it starts to where it hits the incline

shrewd sand
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what does that do? it reverses the velocity and doesn’t change the acceleration

boreal fjord
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Alright, I can imagine that.

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Okay, so it retraces its path

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so total time=2T where T is the time to go up

shrewd sand
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your original q was kinda vague there’s like a gazillion variables

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not sure if it’s just me who’s reading it wrong

boreal fjord
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$T=\frac{2v}{g \sqrt{1+3sin^2A}}$

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I found out T

slender talonBOT
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Agent Krasnov

boreal fjord
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So $2T=\frac{4v}{g \sqrt{1+3sin^2A}}$

slender talonBOT
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Agent Krasnov

boreal fjord
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Is the final answer

shrewd sand
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yeah that’d do it lol

boreal fjord
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Anyways thanks!

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I had a massive conceptual error.

shrewd sand
boreal fjord
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+close

tribal relicBOT
# boreal fjord +close
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shrewd sand
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:00000

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yooo