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# A thing ''A" directly collided with an another staitonary thing "B" and for this Collison "A" stopped. If in this Collison the original Kinetic energy made half then what is the value of coefficient of restitution?

Eshan
3 years ago
Dear student,

From conservation of linear momentum,

$\dpi{80} m_Av_0=m_Bv\implies v=\dfrac{m_A}{m_B}v_0$

It is given that$\dpi{80} \dfrac{1}{2}m_Bv^2=\dfrac{1}{2}\implies \dfrac{1}{2}m_Bv^2=\dfrac{1}{2}\times \dfrac{1}{2}m_Av_0^2$
$\dpi{80} \implies m_Bv^2=\dfrac{1}{2}m_Av_0^2$

Solving the two relations gives,
$\dpi{80} \implies \dfrac{v}{v_0}=2\dfrac{v^2}{v_0^2} \implies v=\dfrac{v_0}{2}$
Hence ceffcient of restitution=$\dpi{80} \dfrac{v_{sep}}{v_{app}}=\dfrac{v_0/2}{v_0}=0.5$