Collisions after effects!

we will be making use of newton's experimental law for co-efficient of restitution,concepts of impulse and simple equations of two-dimensional kinematics.

we will first attempt to find the velocity of spheres A and B after collision

the angle of impact with vertical is $\theta$ for sphere A

newton's experimental law gives us

$v_{1}cos\theta$ is the velocity of sphere A along line of impact,where $v_{1}=\sqrt{2gh}$

however,we note that the sphere is on a rigid floor which rules out vertical motion till free fall starts. the floor imparts an impulse that cancels the $vertical-component$ of impulse imparted by sphere C. THUS horizontal component of velocity imparted can be computed to be

$\sqrt{2gh}sin\theta cos\theta$ which is $\sqrt{\frac{gh}{2}}sin2\theta$

similarly for sphere B ,velocity immediately after collision is $\sqrt{\frac{gh}{2}}sin2\alpha$ however in the left direction towards the wall.

now sphere B collides with the wall elastically i.e., his velocity gets reversed in direction.

now using the fact that $v=\frac{s}{t}$ for uniform translational motion,we get to understand that time before sphere B starts falling is $2\sqrt{\frac{2h}{g}}cosec2\alpha$

now it is time to equations of motion

for sphere A velocity vector $v$ can be written as $\sqrt{\frac{gh}{2}}sin2\theta "i"$ and ( $gt" j"$ )

that for (B) is $\sqrt{\frac{gh}{2}}sin2\alpha "i"$ and ( $g(t-2\sqrt{\frac{2h}{g}}cosec2\alpha)"j"$ )

now the time at which these two are perpendicular is when their dot product is zero

taking dot product,

$\frac{gh}{2}sin2\theta sin2\alpha + g^{2}(t^{2} - 2t\sqrt{\frac{2h}{g}cosec2\alpha}) = 0$

solving this quadratic equation, $t = \sqrt{2gh}(\frac{2cosec2\alpha + \sqrt{4cosec^{2}2\alpha-sin2\theta in2\alpha}}{hsin2\theta sin2\alpha})$

we ignore the other root as we are concerned only about the bigger root

from here,we get ** $P+H+Y+S+I+C+S = 15$

hope u liked it :)