A particle of mass m moving in the x direction with speed 2 v is hit by another particle of mass 2 m moving…

1. $62\%$
2. $44\%$
3. $50\%$
4. $56\%$

Solution

The initial momentum of system is $\overrightarrow{P_i}=m\left(2V\right)\widehat{i}+\left(2m\right)V \widehat{j}$

According to question as



On perfectly inelastic collision the particles stick to each other.

$\overrightarrow{P_f}=3m \overrightarrow{V_f}$

By conservation of linear momentum

$\overrightarrow{P_f}=\overrightarrow{P_i} \Rightarrow 3m \overrightarrow{V_f}=m2V \widehat{i}+2mV\widehat{j}$

$\Rightarrow \overrightarrow{V_f}=\frac{2V}{3}\left(\widehat{i}+\widehat{j}\right)\Rightarrow \left|V_f\right|=\frac{2\sqrt{2}}{3}V$

$\therefore$ loss in KE. of system $=K_{initial}-K_{final}$

$=\frac{1}{2}m{\left(2V\right)}^2+\frac{1}{2}\left(2m\right)V^2-\frac{1}{2}\left(3m\right){\left(\frac{2\sqrt{2}V}{3}\right)}^2$

$=2m{V}^2+m{V}^2-\frac{4}{3}m{V}^2=3m{V}^2-\frac{4m{V}^2}{3}=\frac{5}{3}m{V}^2$

% Loss in KE $=100\times \frac{∆K}{K_i}=\frac{\frac{5}{3}m{v}^2}{3m{V}^2}=\frac{5}{9}\times 100=56\%$

Asked in: JEE Main 2015 (04 Apr)

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