Question

A particle of mass M moving with a speed u strikes a smooth horizontal surface at an angle 45° the particle rebounds at an angle Alpha with speed V if coefficient of restitution is one by root 3 then angle Alpha is

Answer 1

velocity of the ball in vertical and horizontal directions just before it collide with the ground is given as

$v_x = ucos45$

$v_y = u sin45$

now as it collide with the floor the velocity in horizontal direction will remain the same while in vertical direction its velocity will change because its inelastic collision

so after collision the speed is given as

$v_x' = u cos45$

and in vertical direction

$v_y' = e* usin45 = \frac{1}{\sqrt3} * usin45$

so the final speed after collision is given as

$v^2 = v_x'^2 + v_y'^2$

$v^2 = \frac{u^2}{2} + \frac{u^2}{6}$

$v = \sqrt{\frac{2}{3}}u$

also the direction of rebound with vertical is given as

$\theta = tan^{-1}\frac{v_x'}{v_y'}$

$\theta = tan^{-1}\sqrt3 = 60 degree$

so it will rebound at an angle of 60 degree

Answer 2

Answer:

60°

Explanation:

Just keep it simple.....

We know the formula :  tan(α) = (1/e) * tan(Θ)

Given : e = (1/∛3)

           Θ = 45°

Hence , on putting values we get , tan(α) = ∛3 * tan(45°) = ∛3

∴ α = 60°

Thanks...