A particle of mass m moves with a velocity v_ | vedclass

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(B) Since the tension force acts along the radius,the torque about the center of the circle is zero. Therefore,the angular momentum of the particle re

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$4$

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(B) Since the tension force acts along the radius,the torque about the center of the circle is zero. Therefore,the angular momentum of the particle remains conserved.Initial angular momentum $L_i = m v_0 R_0$.Final angular momentum $L_f = m v_f (\frac{R_0}{2})$.By conservation of angular momentum,$L_i = L_f$,so $m v_0 R_0 = m v_f (\frac{R_0}{2})$.This gives $v_f = 2 v_0$.The initial kinetic energy is $K_i = \frac{1}{2} m v_0^2$.The final kinetic energy is $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 4 K_i$.Thus,the final kinetic energy is $4$ times the initial kinetic energy.

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