A particle of mass m moving in the x directio | vedclass

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Question

(a)

Initial momentum of the system

$p _{ i }=m 	imes 2 v \hat{i}+2 m 	imes v \hat{j}$

$=\sqrt{(m 	imes 2 v)^2+(2 m 	imes v)^2} \quad 	ex

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(a)

Initial momentum of the system

$p _{ i }=m 	imes 2 v \hat{i}+2 m 	imes v \hat{j}$

$=\sqrt{(m 	imes 2 v)^2+(2 m 	imes v)^2} \quad 	ext { (magnitude) }$

$=2 \sqrt{2}\,m v$

Final momentum of the system $=3\,mV$

By the law of conservation of momentum

$2 \sqrt{2} m v=3 mV$

$\Rightarrow  \frac{2 \sqrt{2} v}{3}= V _{	ext {combined }}$

Loss in energy

$\Delta E =\frac{1}{2} m _1 V _1^2+\frac{1}{2} m _2 V _2^2-\frac{1}{2}\left( m _1+ m _2ight) V _{	ext {combined }}^2$

$\Delta E =3 mv ^2-\frac{4}{3} mv ^2=\frac{5}{3} mv ^2=55.55 \%$

A particle of mass $m$ moving in the $x$ direction with speed $2 v$ is hit by another particle of mass $2 m$ moving in the $y$ direction with speed $v$. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to $..........\%$

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