IIT JEE , AIEEE Forum - Physics , Mechanics - gud question-rates for satisfactory ans-

annihilator

A particle bounces between two walls.If the collision with one fixed wall is elastic and the other moving wall has a velocity in such a way that the speed of the particle is constant ,then time between collisions,given velocity of particle is v and distance of walls at the first collision with the moving wall is d and coefficient of restitution is e,find the time till the second collision with the moving wall.

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is gravity acting??

annihilator

you can neglect gravity

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is the answer
d(1+e)/v
??

annihilator

yup u r rite .Could u share ur soln??

hash_include

yeah.. it'll be much easier with a diagram tho..
the main thing is to find the velocity with which the moving wall moves..
let it be v2
coeff of restitn = e = (v - v2)/(v+v2)
solvin, we get v2 = v(1-e)/(1+e)

let time after the ball collides wid the stationary ball be t1
obviously t1 = d/v

in that time t1, the moving wall would have moved v2 * t1 = d(1-e)/(1+e)
now, let t2 be time it takes the ball (after hitting the stationary wall) hits the moving wall again

we get :
v*t2 + v2 *t2 = d - [d(1-e)/(1+e)]

solving for t2, we get t2 = de/v

total time = t1 + t2 = d(1+e)/v

hope u understood it :)

annihilator

yup got it....thnx mate

hash_include

yes.. i shouldve explained that..
see after colliding with the stationary wall, the distance between the ball and the moving wall is
d - d(1-e)/(1+e)

wait... ill draw a diag and post it....

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Re:gud question-rates for satisfactory ans-

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