Question

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

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

Answer 1

Answer: (B)

$\frac{1}{4}$

For one dimensional collision between two particles, the equations are as below:

${v_1}^{\prime }=\left(\frac{m_1-e{m}_2}{m_1+m_2}\right)v_1+\left(\frac{m_2+e{m}_2}{m_1+m_2}\right)v_2$;; .....(1)

${v_2}^{\prime }=\left(\frac{m_2-e{m}_1}{m_1+m_2}\right)v_1+\left(\frac{m_1+e{m}_1}{m_1+m_2}\right)v_2$; .....(2)

Given:
$m_1=m_2=m$; (say)
$v_2=v=16$; : Ball A
$v_1=0$; : Ball B
Substituting the above values in (1) and (2)

${v_1}^{\prime }=\left(\frac{1+e}{2}\right)v$;

${v_2}^{\prime }=\left(\frac{1-e}{2}\right)v$;
${v_1}^{\prime }=\left(\frac{1+e}{2}\right)16=(1+e)8$;

From conservation of energy, if ball B rises to a height h,
$(1+e)8=\sqrt{(}2gh)$;

$\Rightarrow (1+e)8=\sqrt{(}2\times 10\times 5)=10))\Rightarrow e=\frac{10}{8}-1=\frac{2}{8}=0.25$;