\[(a)\;\frac{1}{3} \quad (b)\;\frac{2}{3} \quad (c)\;\frac{1}{4}\quad (d)\;\frac{2}{5}\]

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Let h be total height let x be desired fraction.

For Ball A

$(1-x)h=\large\frac{1}{2} $$gt^2$

$=>t=\sqrt {\large\frac{2(1-x)h}{g}}$

For Ball B

$xh=ut-\large\frac{1}{2}$$ gt^2$

$=>xh=u\bigg(\sqrt {\large\frac{2(1-x)h}{g}}\bigg)$$-(1-x)h$

Solving

$u=\sqrt {\large\frac{gh}{2(1-x)}}$

$v_A=2v_B$ at the time of collision

${v_A}^2=4 {v_B}^2$

$2g(1-x)h=4 \{u^2-2gxh\}$

$(1-x)=\large\frac{1}{1-x}$$-4x$

$x=\large\frac{2}{3}$

Hence b is the correct answer.

Nice and easy solution.

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