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  •  Two particles A and B of equal mass M are moving with the same speed v as shown in the figure.  They collide completely inelas

 Two particles A and B of equal mass M are moving with the same speed v as shown in the figure.  They collide completely inelastically and move as a single particle C.  The angle θ that the path of C makes with the X-axis is given by : 
Option: 1   \tan \Theta = \frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}

Option: 2  \tan \Theta= \frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}

Option: 3   \tan \Theta =\frac{1-\sqrt{2}}{\sqrt{2}\left ( 1+\sqrt{3} \right )}

Option: 4   \tan \Theta =\frac{1-\sqrt{3}}{1+\sqrt{2}}
 

Answers (1)

best_answer

As we learned

Inelastic Collision -

Law of conservation of momentum hold good but kinetic energy is not conserved

i.e 

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\neq \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}

but m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}

where 

m_{1},m_{2}: masses

u_{1},v_{1}: initional \: and \: final\: velocities\: of \: mass \: m_{1}

u_{2},v_{2}: initional \: and \: final\: velocities\: of \: mass \: m_{2}

 

So apply Law of conservation of momentum

(i) Along x-direction

mv sin 30^{o} - mvsin45^{o}=2mv^{'}cos\theta \Rightarrow v^{'}cos\theta=\frac{1}{2}(\frac{1}{2}-\frac{1}{\sqrt{2}})

(ii) Along y-direction

mv cos 30^{o} + mvcos45^{o}=2mv^{'}sin\theta \Rightarrow v^{'}sin\theta=\frac{1}{2}(\frac{\sqrt3}{2}+\frac{1}{\sqrt{2}})

Divide (i) with (ii):

tan\theta=\frac{\sqrt3+\sqrt2}{1-\sqrt2}

Posted by

vishal kumar

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