Question

Two identical balls $A$ and $B$ of equal masses are lying on a smooth horizontal surface as shown in the fig. Ball $A$ hits ball $B$ with a velocity $16{\text{ms}}^{-1}$. What should be the minimum value of coefficient of restitution ${}^{\prime }{e}^{\prime }$ between $A$ and $B$ so that $B$ just reaches the highest point of the inclined plane? Take $g=10{\text{ms}}^{-2}$.

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{4}$

After collision, velocity of ball $B$ (called $v_2$) will become,
$v_2=\left[\frac{m_2-e{m}_1}{m_1+m_2}\right]\left(0\right)+\left[\frac{m_2+e{m}_1}{m_1+m_2}\right]v=\left[\frac{1+e}{2}\right]v$
$\Rightarrow v_2=\left[\frac{1+e}{2}\right]\left(16\right)=(1+e)8$
From conservation of energy, if ball $B$ rises to a height $h$,
$v_2=\sqrt{2gh}\Rightarrow \sqrt{2gh}=(1+e)8$
Solving, we get, $e=\frac{1}{4}$