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
9 years ago
Dear Student,

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

Just after the collision at point A (seet he figure below)

The ball collides again on the wedge at point B.
So, applying second equation of moition in y direction:
0 = (√(2gh) * cos) * t + 0.5 * (-g cos) * t2
t = √(h/2g) (Time taken by ball to collide again)
Let distance from point of collision along the wedge = s
So, applying second equation of moition in x direction:
s = (√(2gh) sin) * t + 0.5 * (g sin) * t2
s = 1.25 h sin
Sarthak Mehta
37 Points
8 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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