A) \[1/3\]
B) \[2/3\]
C) \[1/4\]
D) \[2/5\]
Correct Answer: B
Solution :
[b] Let h be the total height and x the desired fraction. Initial velocity of ball B is u, at time of collision it is \[{{v}_{B}}\]. Then \[(1-x)h=\frac{1}{2}g{{t}^{2}}\] ??(1) or \[t=\sqrt{\frac{2(1-x)h}{g}}\] .....(2) \[xh=ut-\frac{1}{2}g{{t}^{2}}\] or \[xh=u\sqrt{\frac{2(1-x)h}{g}}-(1-x)h\] or \[u=\sqrt{\frac{gh}{2(1-x)}}\] ??(3) Now \[{{v}_{A}}=2{{v}_{B}}\] (at the time of collision) Or \[v_{A}^{2}=4v_{B}^{2}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2g(1-x)h=4\left\{ {{u}^{2}}-2gxh \right\}\] or \[2g(1-x)=4\left\{ \frac{gh}{2(1-x)}-2gxh \right\}\] or \[(1-x)\,=\frac{1}{1-x}-4x\] or \[1+{{x}^{2}}-2x=1-4x+4{{x}^{2}}\] or \[x=\frac{2}{3}\]You need to login to perform this action.
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