Question

A ball moving with speed v collides with a horizontal smooth surface at an angle theta with normal to surface as shown in the figure. If coefficient of restitution is ‘e’ , then find velocity after it re-bounces making same angle theta with the normal.

Answer 1

Problem:

A ball moving with speed $v$ collides with a horizontal smooth surface at an angle theta with normal to surface as shown in the figure. If coefficient of restitution is 'e', then find velocity after it re-bounces making same angle theta with the normal.

Solution:

Coefficient of restitution can be defined as (for example, for ball, bouncing from the floor):

$$e = \frac {v _ {Y f i n a l}}{v _ {y}}$$

and(as the surface is smooth with no friction):

$$v _ {x} = v _ {X f i n a l} = v \cos \theta$$

Where $v_{Yfinal}, v_y = v \sin \theta$ - projections of final and initial velocities on axis OY;

$v_{Xfinal}, v_x$ - projections of final and initial velocities on axis OX;

Thus,

$$v _ {f i n a l} = \sqrt {v _ {X f i n a l} ^ {2} + v _ {Y f i n a l} ^ {2}} = \sqrt {v _ {x} ^ {2} + (e v _ {y}) ^ {2}} = v \sqrt {s i n ^ {2} \theta + e ^ {2} c o s ^ {2} \theta}$$

Answer: $v_{final} = v\sqrt{\sin^2\theta + e^2\cos^2\theta}$ .