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Standard XII

Physics

Relative Velocity

Three particl...

Question

Three particles start from the origin at the same time, one with a velocity $a$ , alone $X - a x i s$ , the second along the $Y - a x i s$ with a velocity $b$ and the third along $x = y$ line. The velocity of third particle so that the three may always lie on the same plane is

\frac{a + b}{2}

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a + b

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\frac{\sqrt{2} a b}{(a + b)}

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\frac{a b}{(a + b)}

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Solution

The correct option is C $\frac{\sqrt{2} a b}{(a + b)}$
Let $v_{1} = a$ and $v_{1} = b, v_{1} = c$
Three points to be in a straight line
$\frac{y_{3} - y_{1}}{y_{2} - y_{1}} = \frac{x_{3} - x_{1}}{x_{2} - x_{1}}$ (slopes are equal)
$(a, 0) (0, b)$ and ${(c / \sqrt{2}), (c / \sqrt{2})}$ are the co ordinates of the three points after time $t = 1$
$\frac{\frac{c}{\sqrt{2}}}{b} = \frac{[(\frac{c}{\sqrt{2}}) - a]}{- a}$
$(\frac{c}{\sqrt{2}}) {(\frac{1}{b}) + (\frac{1}{a})} = 1$ sec
$∴ c = \frac{\sqrt{2} a b}{(a + b)}$

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Three particles start from the origin at the same time, one with a velocity a, alone X−axis, the second along the Y−axis with a velocity b and the third along x=y line. The velocity of third particle so that the three may always lie on the same plane is

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