IIT JEE , AIEEE Forum - Physics , Mechanics

vandya

A ball falls on an inclined plane of of inclination 'x' from a height 'h' above the point of impact and makes a perfectly elastic collission.Where will it hit the plane again?
(pls xplain)

raulrag009

It's an irodov question!!!!

THE ball will hit the plane with some velocity and rebound with the same velocity , follow a projectile path and hit the plane again

since the ball has travelled for height 'h'

it's velocity while hittin is

v²=0+2gh

v=

2gh

since the collision is elastic

the ball will rebound with the same velocity

It's initial velocities are

In X axis In Yaxis

u_x = -vsinx u_y= vcosx

a_x= -gsinx a_y= -gcosx

now on the inclined plane,after falling for the second time

It's displacement in y-axis will be zero {w.r.t inclined plane}

In Y axis

0=vcosxt - (1/2)gcosxt^2

t=2v/g

NOw

In X axis

Let R be the distance travelled before hitting on the inclined plane

R= -vsinxt - (1/2)dsinxt^2

put value of t

R = -2v^2sinx/g -4v^2sinx/2g

R = -8v^2sinx/2g

R = -8 (2gh)sinx/2g

R = -8hsinx

thus

mag(R)=8hsinx

elessar_iitkgp

Let the particle collide with the inclined plane with an initial velocity u₀. Then, u₀

2gh.
After the collision only the component of velocity perpendicular to the inclined plane reverses its direction. Let the velocity after the collision be v₀. As the collision is elastic, v₀ = u₀ =

2gh.
At any time t after the collision the position vector of the particle is
r = v₀t + (1/2)gt²
= v₀(sin

i +cos

j)t + (1/2)g(sin

i - cos

j)t²
= (v₀t + (1/2)gt²)sin

i+ (v₀t - (1/2)gt²)cos

j
The instantaneous coordinates of the particle are then,
x = (v₀t + (1/2)gt²)sin

y = (v₀t - (1/2)gt²)cos

Let at t = T, the particle strike the inclined plane again. At that time y(T) = 0 and x(T) = R
y(T) = 0

(v₀T - (1/2)gT²)cos

= 0
T = 2v₀/g
x(T) = R

R = (v₀T + (1/2)gT²)sin

= (4v₀²/g)sin

Substituting v₀ =

2gh,
R = 8hsin

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