On-a-frictionless-horizontal-surface-assumed-to-be-the-x-y-plane-a-small-trolley-A-is-moving-along-a-straight-line-parallel-to-the-y-axis-see-figure-with-a-constant-velocity-of-3-1-m-s-At-

Question Number 20425 by Tinkutara last updated on 27/Aug/17

On a frictionless horizontal surface,

assumed to be the x - y plane, a small

trolley A is moving along a straight

line parallel to the y - axis (see figure)

with a constant velocity of (\sqrt{3} - 1) m / s .

At a particular instant, when the line

O A makes an angle of 45 ° with the

x - axis, a ball is thrown along the

surface from the origin O . Its velocity

makes an angle ϕ with x - axis and it

hits the trolley .

1 . The motion of the ball is observed from

the frame of the trolley . Calculate the

angle θ made by the velocity vector of

the ball with the x - axis in this frame .

2 . Find the speed of the ball with

respect to the surface, if ϕ = \frac{4 θ}{3}

Commented by Tinkutara last updated on 26/Aug/17

Answered by ajfour last updated on 28/Aug/17

(1 .)

l e t t h e t r o l l e y m o v e i n t h e l i n e

x = a a n d i s h i t b y t h e b a l l i n t i m e t .

l e t s p e e d o f t r o l l e y b e u, a n d t h a t

o f b a l l v .

a \tan ϕ = a + u t

a \sec ϕ = v t

\Rightarrow \frac{a \tan ϕ - a}{u} = \frac{a \sec ϕ}{v}

o r v \tan ϕ - v = u \sec ϕ

v \sin ϕ - v \cos ϕ = u

\Rightarrow \frac{v \sin ϕ - u}{v \cos ϕ} = 1 = \tan θ

S o θ = 45 °

(2 .) ϕ = \frac{4 θ}{3} = \frac{4}{3} (45 °) = 60 °

\frac{v \sin ϕ - u}{v \cos ϕ} = 1 = \tan θ

\Rightarrow \frac{v \sqrt{3}}{2} - (\sqrt{3} - 1) = \frac{v}{2}

\Rightarrow v (\sqrt{3} - 1) = 2 (\sqrt{3} - 1)

v = 2 m / s .

Commented by Tinkutara last updated on 26/Aug/17

Thank you very much Sir!

Thank you very much Sir!

Commented by Tinkutara last updated on 28/Aug/17

Thank you!

Thank you!