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In two separate collisions the coefficient the coefficient of restitution s $e_1$ and $e_2$ are in the ratio $3:1$ . In the first collision, the relative velocity of approach is twice the relative velocity of separation, then the ratio between the relative velocity of approach and relative velocity of separation in the second collision is

\[\begin {array} {1 1} (a)\;1:6 & \quad (b)\;2:3 \\ (c)\;3:2 & \quad  (d)\;6:1 \end {array}\]

1 Answer

$\large\frac{e_1}{e_2}=\frac{3}{1}$
$\large\frac{v_1-v_2}{u_1-u_2}=\frac{1}{2}$$=-e_1$
$\large\frac{v_1'-v_2'}{u_1'-u_2'}$$=-e_2$
$\large\frac{e_1}{e_2}=\frac{1}{2.e_2}$
$\qquad=\large\frac{3}{1}$
$e_2=\large\frac{1}{6}$
$\large\frac{u_1'-v_2'}{v_1'-v_2'}$$=6$
answered Nov 22, 2013 by meena.p
 

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