Rotational Motion Question 114
Question: A ball is projected in a direction inclined to the vertical and bounces on a smooth horizontal plane. The range of one rebound is$R$ . If the coefficient of restitution is $e$ , then range of the next rebound is
Options:
A) $R’=eR$
B) $R’=e^{2}R$
C) $R’=\frac{R}{e}$
D) $R’=R$
Show Answer
Answer:
Correct Answer: A
Solution:
[a]
$R=\frac{2u\cos \theta .u\sin \theta }{g}$
After impact horizontal component remains the same = u$\cos \theta $
The vertical component becomes $V=e(\sin \theta )$
New range ${R}’=\frac{2u\cos \theta .eu\sin \theta }{g}{R}’=eR$
