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Question: Three particles start moving simultaneously from a point on a horizontal smooth plane. First particl...

Three particles start moving simultaneously from a point on a horizontal smooth plane. First particle moves with speed v1 towards east, second particle moves towards north with speed v2 and third one moves towards north east. The velocity of the third particle, so that the three always lie on a straight line, is

A

v1+v22\frac{v_{1} + v_{2}}{2}

B

v1v2\sqrt{v_{1}v_{2}}s

C

v1v2v1+v2\frac{v_{1}v_{2}}{v_{1} + v_{2}}

D

2v1v2v1+v2\sqrt{2}\frac{v_{1}v_{2}}{v_{1} + v_{2}}

Answer

2v1v2v1+v2\sqrt{2}\frac{v_{1}v_{2}}{v_{1} + v_{2}}

Explanation

Solution

using y2y1x2x1=y1y3x1x3,\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{y_{1} - y_{3}}{x_{1} - x_{3}}, we have

v2t00v1t=0(v3/2)tv1t(v3/2)t\frac{v_{2}t - 0}{0 - v_{1}t} = \frac{0 - \left( v_{3}/\sqrt{2} \right)t}{v_{1}t - \left( v_{3}/\sqrt{2} \right)t}

⇒ v3 = 2v1v2v1+v2\frac{\sqrt{2}v_{1}v_{2}}{v_{1} + v_{2}},

Hence (4) is correct.