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(B) In an elastic collision,both linear momentum and kinetic energy are conserved.Since the collision is elastic,the total kinetic energy before the c

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$\sqrt{v^2-v_1^2}$

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(B) In an elastic collision,both linear momentum and kinetic energy are conserved.Since the collision is elastic,the total kinetic energy before the collision is equal to the total kinetic energy after the collision.Let $v_2$ be the speed of the second block after the collision.Initial kinetic energy $KE_i = \frac{1}{2}mv^2 + \frac{1}{2}m(0)^2 = \frac{1}{2}mv^2$.Final kinetic energy $KE_f = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$.Equating $KE_i = KE_f$:$\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$Dividing by $\frac{1}{2}m$:$v^2 = v_1^2 + v_2^2$$v_2^2 = v^2 - v_1^2$$v_2 = \sqrt{v^2 - v_1^2}$.

$A$ block of mass $m$ moving on a frictionless surface at speed $v$ collides elastically with a block of the same mass,initially at rest. Now,the first block moves at an angle $\theta$ with its initial direction and has speed $v_1$. The speed of the second block after the collision is

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