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A ball starts falling freely from height h from a point on the inclined plane forming alpha with horizontal. after collision with the incline it rebounds elastically off the plane. Find time when it again strikes the incline & distance from point of collision along incline?

A ball starts falling freely from height h from a point on the inclined plane forming alpha with horizontal. after collision with the incline it rebounds elastically off the plane. Find time when it again strikes the incline & distance from point of collision along incline?

Grade:12th pass

2 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
7 years ago
Dear Student,

Just before collision at point A on the wedge,(see the figure below)

256-667_shear modulus.jpg

Just after the collision at point A (seet he figure below)
256-2048_shear modulus.jpg

The ball collides again on the wedge at point B.
So, applying second equation of moition in y direction:
0 = (√(2gh) * cos\alpha) * t + 0.5 * (-g cos\alpha) * t2
\rightarrowt = √(h/2g) (Time taken by ball to collide again)
\rightarrowLet distance from point of collision along the wedge = s
So, applying second equation of moition in x direction:
s = (√(2gh) sin\alpha) * t + 0.5 * (g sin\alpha) * t2
s = 1.25 h sin\alpha
Sarthak Mehta
37 Points
5 years ago
The velocity of the ball when it first strikes the wedge at point A is v = √(2gh) vertically downwards. 

The ball rebounds with velocity v at angle 2θ from the vertical. 

Taking point A as origin, the co-ordinates of the ball at some subsequent time t are 

x = (vsin2θ)t, y = (vcos2θ)t - ½gt^2. 

The equation of the slope of the wedge is y = -x.tanθ. 

The time T at which the ball next makes contact with the wedge (at point B) is found by inserting the co-ordinates of the ball into the equation of the slope : 

(vcos2θ)T - ½gT^2 = - (vsin2θ)T .tanθ 
v[1-2(sinθ)^2] - ½gT = - v.2(sinθ)^2 
T = (2v/g) = 2√(2h/g). 

The co-ordinates of B are found by inserting the value of T into the co-ordinates of the ball : 

x = (2v^2/g)sin2θ = 4h.sin2θ, y = 4hcos2θ - 4h = -4h(1 - cos2θ). 

The distance AB is given by 

AB^2 = x^2 + y^2 = (4h.sin2θ)^2 + (4h.(1 - cos2θ))^2 
= (4h)^2.[ (sin2θ)^2 + 1 - 2cos2θ + (cos2θ)^2 ] 
= (4h)^2.2( 1 - cos2θ ) 
= (4h)^2.2.2(sinθ)^2 
= (8h)^2.(sinθ)^2 

Therefore distance AB = 8h.sinθ.

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