Question

Two balls of masses $m_1$ and $m_2$ are moving towards each other with speeds $u_1$ and $u_2$ respectively. After they collide head–on, their speeds are $v_1$ and $v_2$ respectively. (Take $m_1=8\text{kg},m_2=2\text{kg},u_2=3\text{m/s}$) All quantities are in SI units and $e$ is the coefficient of restitution.
Match the statements in List $1$ with results in List $2$
$\begin{array}{cc}\text{List 1}& \text{List 2}\\ \text{A) The speed}{u}_1\text{so that both the balls are moving in the same direction after collision is (}e=0.5)& \text{p)}1/14\\ \text{B) The speed}{u}_1\text{so that maximum fraction of energy is transferred to mass}{m}_2\text{(Assume}e=1\text{) is}& \text{q)}1.5\\ \text{C) If}{u}_1=0.5m/s\text{and}{m}_2\text{stops, then the value of e is}& \text{r)}2\\ \text{D) If}{u}_1=3m/s\text{and e=0,}\text{then fractional loss of KE after collision is}& \text{s)}0.64\end{array}$

• A. $A–r,q,B–q,C–p,D–p$
• B. $A–r,B–q,C–p,D–s$
• C. $A–r,p,B–p,C–q,D–s$
• D. $A–q,r,B–r,C–p,D–s$

Answer 1

The correct option is D $A–q,r,B–r,C–p,D–s$


Velocities of particles after head on inelastic collision of coefficient of restitution $e$ is given by

$V_1=\frac{m_1-e{m}_2}{m_1+m_2}{u}_1+\frac{m_2(1+e)}{m_1+m_2}{u}_2{V}_2=\frac{m_1(1+e)}{m_1+m_2}{u}_1+\frac{m_2-e{m}_1}{m_1+m_2}{u}_2\text{a) Given e=0.5},V_1\text{and}{V}_2\text{are positive}{V}_1=\frac{8-\left(0.5\right)\left(2\right)}{10}{u}_1+\frac{2\left(1.5\right)}{10}(-3)V_1=\frac{7}{10}{u}_1-\frac{9}{10}{V}_2=\frac{8\left(1.5\right)}{10}{u}_1+\frac{2-\left(0.5\right)8}{10}(-3)V_2=1.2u_1+0.6$
$V_2$ is always positive. Hence $V_1$ should be positive.
$\Rightarrow V_1\ge 0\Rightarrow \frac{7}{10}{u}_1+\frac{2\left(1.5\right)}{10}(-3)\ge 0\Rightarrow u_1\ge \frac{9}{7}\text{m/s}\Rightarrow u_1\ge 1.285\text{m/s}$
Hence $q,r$

b) Given $e=1$
$V_1=\frac{m_1-m_2}{m_1+m_2}{u}_1+\frac{2m_2}{m_1+m_2}{u}_2$

For maximum energy to be transferred to $m_2$, final velocity of $m_1$ has to be zero.

$V_1=0\Rightarrow \frac{8-2}{10}{u}_1+\frac{2\left(2\right)}{10}(-3)=0\Rightarrow 0.6u_1=\frac{12}{10}\Rightarrow u_1=2\text{m/s}$
Hence $r$

$\text{c) Given}{u}_1=0.5m/s{V}_2=\frac{m_1(1+e)}{m_1+m_2}{u}_1+\frac{m_2-e{m}_1}{m_1+m_2}{u}_2\text{If}{m}_2\text{rests after collision}{V}_2=0\Rightarrow \frac{8(1+e)}{10}\left(0.5\right)+\left[\frac{2-8e}{10}\right](-3)=0\Rightarrow 4(1+e)=6-24e\Rightarrow 28e=2\Rightarrow e=1/14$
Hence $p$

d) If $e=0$; the two bodies will move together after collision.
Loss of Kinetic energy during inealstic collision
$\Delta KE=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(u_1-u_2{)}^2(1-e^2)=\frac{1}{2}\frac{8\times 2}{10}(3+3{)}^2=28.8JK{E}_i=\frac{1}{2}\times 8\times 3^2+\frac{1}{2}\times 2\times 3^2=\frac{1}{2}\times 10\times 9=45\text{J}\text{fractional loss}=\frac{\Delta KE}{K{E}_i}=0.64$
Hence $s$