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A ball falls on an incline plane of inclination '@' from a height h above the point of impact and makes a perfectly elastic collision where will it hit the plane again



04-Sep-2016 2:05 AM
~2

Answers (2)


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Let the angle be θ

Assume that h be the height above the point of impact that the ball is dropped.
From Newton's third equation of motion

v² = u² + 2as

v² = 0 + 2(-9.8)(-h)

v = √(19.6h)

Also since we are dealing in a perfectly elastic collision 

angle of incidence = angle of reflection

angle of incidence = (90-θ) degrees above slope

therefore angle of reflection = (90-θ) degrees above slope.

The speed of the ball will be the same as before, as it is perfectly elastic.

x-component of velocity = √(19.6h) cos(90-θ)
y-component of velocity = √(19.6h) sin(90-θ)

taking the point of impact to be the origin and the slope now to be horizontal.

x = (√(19.6h) cos(90-θ))t
= (√(19.6h) sin(θ))t

y = (√(19.6h) sin(90-θ))t + 0.5(-9.8)t²
= (√(19.6h) cos(θ))t + 0.5(-9.8)t²

his slope when y = 0

0 = (√(19.6h) cos(θ))t + 0.5(-9.8)t²

t = 0 (initially when ball hits the slope)
t = (√(19.6h) cos(θ)) / 4.9

subbing into x:

x = ((√(19.6h) cos(θ)) / 4.9)((√(19.6h) sin(θ)))

x = (19.6h /4.9) cos(θ) sin(θ)

putting this into matrix form and doing a matrix rotation of angle θ clockwise.

x' = x cos(-θ) - y sin(-θ) = x cos(θ) + y sin(θ)
y' = x sin(-θ) + cos(-θ) = -x sin(θ) + y sin(θ)

x' = (19.6h /4.9) cos²(θ) sin(θ)
y' = -(19.6h /4.9) cos²(θ) sin²(θ)

these are the coordinates for where on the slope it will hit again, taking the origin at the point of impact.



04-09-2016 11:42
~6

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Let the velocity  with which the ball hits the 

Plane be v .

Before the collision v perpendicular to plane is -v cos(@).

After the collision v perpendicular to plane is +v cos(@).

Before the collision velocity parallel to plane is -v sin(@) which remains unchanged even after the collision.

For the ball to hit the plane displacement n the direction perpendicular to the plane should be zero while the acceleration perpendicular to plane is -g cos(@).

By writing the kinematics equation n direction perpendicular to plane... After time t;

0= v cos(@)*t- 1/2 (g cos@)*t²

=> t= 2*(v/g) which is the time after which the ball hits the plane.

Now displacement n direction parallel to plane is -l(say) velocity parallel to plane is 

-v sin(@) and acceleration in this direction is 

 - g sin(@)

Writing the  equation of displacement  parallel to plane...

l= v sin(@)*t + 1/2(g sin@)*t²

Where t= 2*(v/g)

=> l=  4*(v²sin(@)/g) this is the displacement of ball parallel to the plane which makes angle @ with horizontal.

=> Component of displacement along horizontal is l*cos(@)=x(say)

Component of displacement perpendicular to horizontal i.e  along vertical direction is l*sin(@)=y(say)

X-coordinate= -4*h*sin(2@)

Y-coordinate=-8*h*sin(@)*sin(@).

Thank you :)



14-09-2016 02:21
~0

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