Question

A steel ball falls from a height of 1m on to a steel plate and jumps upto a height of 0.81m find the coefficient of restitution

Answer 1

Answer:

• The Coefficient of Restitution (e) is 0.9.

Given:

1. Initial Height (H₁) = 1 m.
2. Final Height (H₂) = 0.81 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.(I.e. Nature of collision).

We know,

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

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

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Formula we have,

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

Substituting the values,

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{0.81}{1}}}$

Multiplying 100 on Both sides.

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{0.81 \times 100}{1 \times 100}}}$

$\large{\tt \hookrightarrow e = \sqrt{\dfrac{81}{100}}}$

Now,

$\large{\tt \hookrightarrow e = \dfrac{9}{10}}$

$\huge{\boxed{\boxed{\tt e = 0.9}}}$

The Coefficient of restitution is 0.9 & The collision is Inelastic.

$\rule{300}{1.5}$