The first ball of mass m moving with the velocity v collides head… - Vidyadip

MCQ

The first ball of mass $m$ moving with the velocity $v$ collides head on with the second ball of mass $m$ at rest. If the coefficient of restitution is $e$, then the ratio of the velocities of the first and the second ball after the collision is:

✓
$\frac{1-\text{ e}}{1+\text{ e}}$
B
$\frac{1+\text{e}}{1-\text{e}}$
C
$\frac{1+ \text{ e}}{2}$
D
$\frac{1-\text{ e}}{2}$

✓

Answer

Correct option: A.

$\frac{1-\text{ e}}{1+\text{ e}}$

Here, $\text{m}_1=\text{m}_2=\text{m},\text{u}_1=\text{u},\text{u}_2=0$.
Let $\text{v}_1,\text{v}_2$ be their velocities after collision.
According to principle of conservation of linear momentum.
$\text{mu}+0=\text{m}(\text{v}_1+\text{v}_2)$ or $\text{ v}_1+\text{v}_2=\text{u}\dots\text{(i})$
By definition,
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{u}-0}$ or $\text{ v}_2-\text{v}_1=\text{eu}\dots\text{(ii)}$
Add $(i)$ and $(ii),$
$\text{v}_2=\frac{\text{u(1+ e)}}{2}$
$\therefore\frac{\text{v}_1}{\text{v}_2}=\frac{1-\text{ e}}{1+\text{ e}}$

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