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(B) For a head-on collision between two bodies of equal mass $m$,where the second body is initially at rest $(u_2 = 0)$ and the first body has initial

Vedclass · Accepted Answer

$v_1 = \frac{(1-e)v}{2}, v_2 = \frac{(1+e)v}{2}$

Vedclass · Answer

(B) For a head-on collision between two bodies of equal mass $m$,where the second body is initially at rest $(u_2 = 0)$ and the first body has initial velocity $u_1 = v$:$1$. Conservation of linear momentum: $mv + 0 = mv_1 + mv_2 \implies v = v_1 + v_2$.$2$. Coefficient of restitution definition: $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{v - 0} \implies ev = v_2 - v_1$.$3$. Adding the two equations: $v + ev = (v_1 + v_2) + (v_2 - v_1) = 2v_2 \implies v_2 = \frac{(1+e)v}{2}$.$4$. Subtracting the equations: $v - ev = (v_1 + v_2) - (v_2 - v_1) = 2v_1 \implies v_1 = \frac{(1-e)v}{2}$.Thus,the correct option is $B$.

$A$ ball of mass $m$ moving with velocity $v$ collides head-on with a second ball of mass $m$ at rest. If the coefficient of restitution is $e$,the velocity of the first ball after collision is $v_1$,and the velocity of the second ball after collision is $v_2$,then:

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