- #1

babaliaris

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- 15

Known:

1) The mass of the ball is ##m## (constant ##\frac{dm}{dt} = 0##)

2) ##v(0) = v_{0}##

3) Air drag force magnitude ##| \vec F_{D} | = B \cdot | \vec v(t) |## (##B \in R##)

4) The ramp is frictionless.

5) The magnitude of Earth's acceleration = ##g##

I'm not sure if

**θ**is known or not, and I can't show you the exercise since it is written in Greek.

Looking for the distance that the ball will travel until it stops:

Question: ##s(t)_{max} = ?## @ ##v(t)=0##

**My Solution (What is wrong with it?):**

##θ + γ + 90 = 180 <=> γ = 90 - θ##

##arg( \vec w ) = 180 + γ = 180 + 90 - θ = 270 - θ##

##\vec F_{D} = -B | \vec v(t)| \vec i##

##\vec w = mg cos(270-θ) \vec i + mg sin(270-θ) \vec j##

##\vec N = mg cos(90) \vec i + mg sin(90) \vec j = mg\vec j####Σ \vec F = m \vec a <=> \vec F_{D} + \vec N + \vec W = m \frac{d \vec v(t)}{dt}##

$$<=> -B | \vec v(t) | \vec i + mg \vec j + mg cos(270-θ) \vec i + mg sin(270-θ) \vec j =

m \frac{d | \vec v(t) | }{dt} \vec i$$

$$<=> [-B | \vec v(t) | + mg cos(270-θ)] \vec i + [mg + mg sin(270-θ)] \vec j =

m \frac{d | \vec v(t) | }{dt} \vec i$$##-B | \vec v(t) | + mg cos(270-θ) = m \frac{d | \vec v(t) | }{dt}##

##mg + mg sin(270-θ) = 0####mg + mg sin(270-θ) = 0 => sin(270-θ) = -1##

##<=> 270-θ = sin^{-1}(-1) = -90 <=> θ = 270+90 = 360##I don't think ##θ = 360## makes sense so I stopped here... This means that the ramp is not

a ramp at all ##θ = 360 = 0##.

Where is my mistake? Is it in my Physics understanding or a math mistake?

Thank you!

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