Question

A particle strikes a horizontal frictionless floor with a speed 'u' at an angle ' $\theta$ ' with the vertical and rebounds with a speed 'v' at an angle '$\alpha$' with the vertical. Find the value of 'v' if 'e' is the coefficient of restitution.

• A. $v=u\sqrt{e^{2}si{n}^2\theta +co{s}^2\theta }$
• B. $v=u\sqrt{e^{2}co{s}^2\theta +si{n}^2\theta }$
• C. $v=u\sqrt{e^{2}co{s}^2\theta +ta{n}^2\theta }$
• D. $v=u\sqrt{e^{2}co{t}^2\theta +co{s}^2\theta }$

Answer 1

The correct option is B $v=u\sqrt{e^{2}co{s}^2\theta +si{n}^2\theta }$

$u_x$ will remain constant $=usin\theta =v_x$

& $v_y=eucos\theta$

so $v=\sqrt{v_x^2+v_y^2}$

$=u\sqrt{si{n}^2\theta +e^{2}co{s}^2\theta }$