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(D) Let the masses be $m$ and charges be $q_1$ and $q_2$. Initially, $X$ has velocity $u$ and $Y$ is at rest. The total momentum of the system is $P_i

Vedclass · Accepted Answer

both will move with velocities  <  $u/2$.

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(D) Let the masses be $m$ and charges be $q_1$ and $q_2$. Initially, $X$ has velocity $u$ and $Y$ is at rest. The total momentum of the system is $P_i = mu + m(0) = mu$.Since the particles are far apart initially, the potential energy is zero. As $X$ approaches $Y$, the repulsive electrostatic force acts on both particles.By the law of conservation of linear momentum, the total momentum remains constant: $m v_X + m v_Y = mu$, which implies $v_X + v_Y = u$.By the law of conservation of energy, the initial kinetic energy $K_i = \frac{1}{2}mu^2$ is converted into final kinetic energy and potential energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv_X^2 + \frac{1}{2}mv_Y^2 + U_f$, where $U_f > 0$.Since $U_f > 0$, the sum of final kinetic energies must be less than the initial kinetic energy: $\frac{1}{2}m(v_X^2 + v_Y^2) Given $v_X + v_Y = u$, if $v_X = 0$, then $v_Y = u$, which would mean $v_X^2 + v_Y^2 = u^2$, contradicting the energy conservation with $U_f > 0$. Thus, $v_X$ cannot be $0$. Both particles will continue to move with velocities such that their sum is $u$ and their individual velocities are less than $u/2$.

Two particles $X$ and $Y$, of equal mass and with unequal positive charges, are free to move and are initially far away from each other. With $Y$ at rest, $X$ begins to move towards it with initial velocity $u$. After a long time, finally

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