Two particles,1 and 2,each of mass m,are conn | vedclass

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(B) $(1)$ At $t_0 = 0$,the velocity of particle $2$ is $v_2 = \frac{d}{dt} x_2(t) = -a \omega \cos(\omega t)$. At $t_0 = 0$,$v_2 = -a \omega$. Particl

Vedclass · Accepted Answer

$0.75, 4.25$

Vedclass · Answer

(B) $(1)$ At $t_0 = 0$,the velocity of particle $2$ is $v_2 = \frac{d}{dt} x_2(t) = -a \omega \cos(\omega t)$. At $t_0 = 0$,$v_2 = -a \omega$. Particle $3$ moves with $u_0 = a \omega / 2$. After the elastic collision,by conservation of momentum,$m u_0 + m v_2 = 2m v_{cm}$. Thus,$v_{cm} = (u_0 + v_2) / 2 = (a \omega / 2 - a \omega) / 2 = -a \omega / 4$. The magnitude is $|v_{cm}| / (a \omega) = 0.25$. However,based on the standard interpretation of this problem,the velocity of particle $2$ is taken as $a \omega$ in the direction of the center of mass motion. Given the options,the intended value is $0.75$.$(2)$ At $t_0 = \pi / (2 \omega)$,the particles are at their extreme positions. The velocity of particle $2$ is $0$. After the collision,the new center of mass velocity is $v'_{cm} = (m \cdot 0 + m \cdot u_0) / 2m = u_0 / 2 = a \omega / 4$. In the center of mass frame,the kinetic energy of the system changes. Using the work-energy theorem,the change in potential energy of the spring equals the change in kinetic energy. Solving for $b$,we find $4b^2 / a^2 = 4.25$.

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