MCQ
$A$ ball is thrown vertically downwards with velocity $\sqrt {2gh} $ from $a$ height $h$. After colliding with the ground it just reaches the starting point. Coefficient of restitution is
- ✓$1 / \sqrt 2$
- B$1/2$
- C$1$
- D$\sqrt 2$
✓
Answer
Correct option: A.
$1 / \sqrt 2$
a
The ball has an initial velocity as $\sqrt{2 g h}$
The ball has an initial velocity as $\sqrt{2 g h}$
Thus the velocity just before it collides on the ground is given using $v^{2}=u^{2}+2 a s$
or
$v^{2}=2 g h+2 g h=4 g h$
or
$v_{i}=\sqrt{4 g h}$
Now the velocity changes to $v_{f}$ (say) after collision.
The ball just reaches the initial height $h.$
Thus we get
$0=v_{f}^{2}-2 g h$
or $v_{f}=\sqrt{2 g h}$
Thus the coefficient of restitution is given as
$e=\frac{v_{f}}{v_{i}}=\frac{\sqrt{2 g h}}{\sqrt{4 g h}}=\frac{1}{\sqrt{2}}$
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