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Solve the following :

Question:

A particle of mass $100 \mathrm{~g}$ moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes $0.2 \mathrm{~J}$ after the collision, what could be the minimum and the maximum value of $u$.

Solution:

Use C.O.L.M

$0.1(u)=0.1\left(v_{1}+v_{2}\right)$

$\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}=0.2$

$\left(v_{1}^{2}+v_{2}^{2}\right)=\frac{0.2 \times 2}{0.1}=4$

$u=v_{1}+v_{2}$

$=\sqrt{4-v_{2}^{2}}+v_{2}$

$\frac{d u}{d v_{2}}=\frac{1}{2} \frac{-2 v_{2}}{4-v_{2}^{2}}+1=0$

$\Rightarrow \sqrt{4-v_{2}^{2}}=v_{2}$

$\Rightarrow 4=2 v_{2}^{2} \Rightarrow v_{2}=\pm \sqrt{2}$

When $v_{2}=\sqrt{2}, u=\sqrt{4-2}+\sqrt{2}=2 \sqrt{2}$

$u_{\max }=2 \sqrt{2} \mathrm{~m} / \mathrm{s}$

$u=u_{\min }$ when $v_{1}=0$ or $v_{2}=0$

if $v_{1}=0, v_{2}=2$

if $v_{2}=0, v_{1}=2$

as $v_{1}^{2}+v_{2}^{2}=4$

$\therefore u_{\min }=0+2=2 \mathrm{~m} / \mathrm{s}$