Question

a body freely falling from a certain height h after striking a smooth floor rebounds to a height h/2 what is the coefficient of restitution between the floor and the body​

Answer 1

Answer:

• Coefficient of Restitution (e) is 1/√2.

Given:

1. Initial Height (H₁) = H m.
2. Final Height (H₂) = H/2 m.

Explanation:

$\rule{300}{1.5}$

Coefficient of Restitution (e) :-

• It is the Ratio of Relative velocities of separation after collision to Relative velocities of approach Before collision.
• It Depends on Nature of Colliding bodies.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

We know,

$\large\star \; {\boxed{\bold{e = \sqrt{\dfrac{H_2}{H_1}}}}}$

$\bold{Here}\begin{cases}\sf{H_1} = \text{Dropped Height} \\ \sf{H_2} = \text{Rebounded Height}\end{cases}$

$\large{\boxed{\tt e = \sqrt{\dfrac{H_2}{H_1}}}}$

Substituting The Values,

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{H/2}{H}}}$

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{H}{2 \times H}}}$

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{\cancel{H}}{2 \times \cancel{H}}}}$

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{1}{2}}}$

∵ [ √1 = 1]

This Becomes,

$\huge{\boxed{\boxed{\tt e = \dfrac{1}{\sqrt{2}}}}}$

∴ The Coefficient of Restitution (e) is 1/√2.

$\rule{300}{1.5}$

Answer 2

$\huge\star\mathfrak\orange{{HEY MATE...!!}}$

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