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Question

(B) Let the velocity of the first particle be $\vec{v}_1 = v \hat{i}$ and the velocity of the second particle be $\vec{v}_2 = (at) \cos \alpha \hat{i}

Vedclass · Accepted Answer

$\frac{v}{a} \cos \alpha$

Vedclass · Answer

(B) Let the velocity of the first particle be $\vec{v}_1 = v \hat{i}$ and the velocity of the second particle be $\vec{v}_2 = (at) \cos \alpha \hat{i} + (at) \sin \alpha \hat{j}$.The relative velocity vector is $\vec{v}_r = \vec{v}_1 - \vec{v}_2 = (v - at \cos \alpha) \hat{i} - (at \sin \alpha) \hat{j}$.The square of the magnitude of relative velocity is $v_r^2 = (v - at \cos \alpha)^2 + (at \sin \alpha)^2$.Expanding this,we get $v_r^2 = v^2 + a^2 t^2 \cos^2 \alpha - 2vat \cos \alpha + a^2 t^2 \sin^2 \alpha$.Since $\sin^2 \alpha + \cos^2 \alpha = 1$,this simplifies to $v_r^2 = v^2 + a^2 t^2 - 2vat \cos \alpha$.To find the minimum relative velocity,we differentiate $v_r^2$ with respect to $t$ and set it to zero:$\frac{d}{dt}(v_r^2) = 2a^2 t - 2va \cos \alpha = 0$.Solving for $t$,we get $2a^2 t = 2va \cos \alpha$,which gives $t = \frac{v \cos \alpha}{a}$.

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