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A body 'X' with momentum 'p' collides with another identical stationary body ' y ' one dimensionally. During collision ' y' gives an impulse ' J ' to the body 'x'. Then the coefficient of restitution is

\[\begin {array} {1 1} (a)\;\frac{2J}{p}-1 & \quad (b)\;\frac{J}{p}+1 \\ (c)\;\frac{J}{p}-1 & \quad  (d)\;\frac{J}{4p}-1 \end {array}\]

Can you answer this question?
 
 

1 Answer

+1 vote
Impulse on A =-Impulse on B
Impulse = change in momentum
Let $u_1$ be the initial velocity of A
$p= mu_1$
After collision let $v_1$ and $v_2$ be the velocities of A and B in the same direction.
$J= mv_2$ ---- (1)
$-J=mv_1-mu_1$
$J=mu_1-mv_1$
$J= p-mv_1$ ----(2)
$2J= p+m(v_2-v_1)$
$2J-p= m(v_2-v_1)$
$\large\frac{v_2-v_1}{u_1}$$=e$
$\qquad= \large\frac{2J-p}{p}$
$\qquad= \large\frac{2J}{p}$$-1$
answered Nov 23, 2013 by meena.p
edited Jun 17, 2014 by lmohan717
 

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