A wedge of mass
and angle
is placed on a smooth ground. As shown in the diagram
is a point on ground. A particle also having mass
is dropped from a height
at a horizontal distance
from
. It finally touches the ground at point
. Find
in metres.
Details and Assumptions
Take , metre , metre.
The collision of the ball with the wedge is elastic and the wedge is free to move
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Initial state
Image 1
Final state
Image 2
Since the collision was elastic we write v s e p e r a t i o n = v a p p r o a c h
hence we have v 1 + v 2 s i n ( θ ) = v c o s ( θ ) ( i )
Also applying conservation of momentum in x-direction we get 0 = m ( v 1 s i n ( θ ) + v s i n ( θ ) c o s ( θ ) ) − m ( v 2 )
⇒ v 2 = v 1 s i n ( θ ) + v s i n ( θ ) c o s ( θ ) ( i i )
From (i) and (ii) we get v 1 = 1 + s i n 2 ( θ ) v c o s 3 ( θ ) ( i i i )
Now we will find v x a n d v y . v x = v 1 s i n ( θ ) + v s i n ( θ ) c o s ( θ ) = 1 + s i n 2 ( θ ) 2 v s i n ( θ ) c o s ( θ ) v y = v 1 c o s ( θ ) − v s i n 2 ( θ ) = 1 + s i n 2 ( θ ) v ( c o s 2 ( θ ) − 2 s i n 2 ( θ ) )
Since t a n ( θ ) = 2 1 hence v x = 2 v , v y = 0
Also note that v = 2 g ( h − x t a n θ ) ⇒ v x = g ( h − x t a n θ )
Range = v x g 2 x t a n θ = 2 x t a n θ ( h − x t a n θ )
Now putting x = 1 , h = 2 2 m we get Range = 3
Now A B = R a n g e − x ⇒ A B = 3 − 1 ⇒ 3 A B = 3 − 3