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A body 'A' experiences perfectly elastic collision with a stationary body 'B'. If after collision the bodies fly apart in opposite directions with equal velocities , the mass ratio of 'A' and 'B' is

\[\begin {array} {1 1} (a)\;\frac{1}{2} & \quad (b)\;\frac{1}{3} \\ (c)\;\frac{1}{4} & \quad  (d)\;\frac{1}{5} \end {array}\]

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1 Answer

By momentum conservation
$m_1u_1= m_2v_2-m_1v_1$
$\large\frac{v_2+v_1}{0+u_1}$$=e=1$
$v_2+v_1=u_1$
$\therefore \; m_1(v_2+v_1)=m_2v_2-m_1v_1$
$m_1(v_2+2v_1)=m_2v_2$$\qquad v_1=v_2=v$
$\large\frac{m_1}{m_2}=\large\frac{v}{3v}$
$\qquad= \large\frac{1}{3}$
answered Nov 23, 2013 by meena.p
 

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