Two balls of charge q_1 and q_2 initially hav | vedclass

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(D) Let $v$ be the initial velocity. The change in momentum is $\Delta \vec{p} = q \vec{E} \Delta t$. Since $\vec{E}$ is uniform,$\Delta \vec{p}$ is t

Vedclass · Accepted Answer

$\frac{4}{3} k_1$

Vedclass · Answer

(D) Let $v$ be the initial velocity. The change in momentum is $\Delta \vec{p} = q \vec{E} \Delta t$. Since $\vec{E}$ is uniform,$\Delta \vec{p}$ is the same for both balls if we assume the same $qE\Delta t$ or relate them via ratios. For the first ball,the final velocity $\vec{v}_1$ has magnitude $v/2$ and is at $60^{\circ}$ to $\vec{v}$. The change in momentum $\Delta \vec{p}_1 = m_1(\vec{v}_1 - \vec{v})$ must be perpendicular to $\vec{v}_1$ because the magnitude decreased. Using the law of cosines: $(v/2)^2 = v^2 + (\Delta p_1/m_1)^2 - 2v(\Delta p_1/m_1)\cos(120^{\circ})$. Solving this,$\Delta p_1 = m_1 v \sin(60^{\circ}) = m_1 v \frac{\sqrt{3}}{2}$. For the second ball,the velocity changes by $90^{\circ}$,so $\vec{v}_2 \perp \vec{v}$. Thus,$v_2 = v / \cos(90^{\circ})$ is not applicable; rather,$\Delta p_2 = m_2 v 	an(90^{\circ})$ implies the force is perpendicular to the initial velocity. The magnitude of the second ball's velocity is $v_2 = v / \cos(30^{\circ}) = 2v/\sqrt{3}$ is incorrect; the correct relation is $v_2 = v 	an(30^{\circ}) = v/\sqrt{3}$. The ratio $\frac{q_2/m_2}{q_1/m_1} = \frac{4}{3}$,so $k_2 = \frac{4}{3} k_1$.

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