Question

Two identical balls $\text{A}$ and $\text{B}$ having mass $m$ are lying on a smooth surface as shown in figure. Ball $\text{A}$ hits the ball $\text{B}$ (which is at rest) with a velocity $16{\text{ms}}^{-1}.$ What should be the minimum value of coefficient of restitution between $\text{A}$ and $\text{B}$ so that $\text{B}$ just reaches the highest point of inclined plane? [Take $g=10{\text{ms}}^{-2}$]

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

Answer 1

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




Given, $u_A=16{\text{ms}}^{-1};u_B=0{\text{ms}}^{-1}$

Let, velocity of $\text{A}$ and $\text{B}$ after the collision be $v_A$ and $v_B$ respectively.

Minimum velocity of $\text{B}$, such that it reaches the highest point of inclined plane is,

$v_B^2=u_B^2+2a{S}_B[\because u_B=0{\text{ms}}^{-1}]$

$v_B=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10{\text{ms}}^{-1}$

Now, using conservation of linear momentum,

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $m{u}_A+m{u}_B=m{v}_A+m{v}_B$

$m\times 16=m{v}_A+m{v}_B$

$16=v_A+10$

$\Rightarrow v_A=6{\text{ms}}^{-1}$

Now, using Newton's law of impact,

$\text{e}=\frac{\text{velocity of separation}}{\text{velocity of approach}}=\frac{(v_2-v_1)}{(u_1-u_2)}$

$\Rightarrow e=\frac{(v_B-v_A)}{(u_A-u_B)}=\frac{10-6}{16}=\frac{1}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{B}\right)$ is the correct answer.

Alternate solution: $v_1=\frac{(m_1-e{m}_2)u_1}{m_1+m_2}+\frac{m_2{u}_2(1+e)}{m_1+m_2}$ $v_2=\frac{m_1{u}_1(1+e)}{m_1+m_2}+\frac{(m_2-e{m}_1)u_2}{m_1+m_2}$ Here, $v_2=v_B=10\text{m/s};u_1=16\text{m/s};u_2=0$ $m_1=m_2=m$ $v_2=10=\frac{m\times 16(1+e)}{2m}+\frac{(m-em)\times 0}{2m}$ $\therefore e=\frac{20}{16}-1=\frac{1}{4}$