Question

A ball collides with a smooth and fixed inclined plane of inclination $\theta$ after falling vertically through a distance h. If it moves horizontally just after impact, the coefficient of restitution iis

• A. $ta{n}^2\theta$
• B. $co{t}^2\theta$
• C. $tan\theta$
• D. $cot\theta$

Answer 1

The correct option is C $tan\theta$

R.E.F image

The velocity of particle changes direction

along perpendicular axis

take a reference axis along plane

So, Initial velocity

$\overrightarrow{u_1}=-u(sin\theta \hat{i}+cos\theta \hat{j})$ $[u=\sqrt{2gh}]$

and final velocity :-

$\overrightarrow{V_1}=u^{\prime }(cos\theta \hat{i}+sin\theta \hat{j})$

momentum along x axis would be constant, as impulse

acts along the normal

$\Rightarrow -usin\theta =u^{\prime }(-cos\theta )$ [From conservation of momentum]

$\Rightarrow u^{\prime }=utan\theta$

That given $\overrightarrow{v_1}=utan\theta (-cos\theta \hat{i}+sin\theta \hat{j})$

Thus. coeff. of restitution. In this case :-

$e=\frac{|{\overrightarrow{v}}_1|}{|\overrightarrow{u_1}|}=\frac{utan\theta }{u}=tan\theta$ (Ans)