Question

A ball, of mass m, moving with a velocity, v, of 10 m/s, ‘collides’, at an angle of ‘incidence’ of 45° , with a smooth (frictionless) surface having a coefficient of restitution (e) of 1/√3 . The difference between the angle of ‘reflection’ and ‘incidence’ of the ball, and the change in the velocity of the ball, are equal, respectively, to

(1) (-15°) and (7 m/s) \(\hat{j}\)

(2)  (+15°) and (7 m/s) \(\hat{j}\)

(3)  (-15°) and (11.3 m/s) \(\hat{j}\)

(4)  (+15°) and  (11.3 m/s) \(\hat{j}\)

Answer 1

 (4)  (+15°) and  (11.3 m/s) \(\hat{j}\)

The initial horizontal (or x) and vertical (or y) component, of the velocity of the ball, are v cos 45° and v sin 450 respectively.

The surface being smooth, the final (i.e. after ‘reflection’) horizontal component, of the velocity of the ball, would still be v cos 45° . The vertical component, however, would be (e v sin 45° ). The vertical component, however, would be (e v sin 45°). If therefore, the final velocity (v) of the ball makes an angle θ with the horizontal, we have

The angle of ‘reflection’ is, therefore, (90°–30° ) = 60° . Hence the diference, between the angle of ‘reflection’ and the angle of incidence, is (60°–45° ) = +15° .

The final velocity, V is given by

The initial velocity, V, is given by

∴ Change in velocity