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5.Work, Energy, Power and Collision
normal
A disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v$, the height to which the disc will rise will be
A
$\frac {3v^2}{2g}$
B
$\frac {3v^2}{4g}$
C
$\frac {v^2}{4g}$
D
$\frac {v^2}{2g}$
Solution
$\frac{1}{2}M{v^2} + \frac{1}{2}\frac{{M{R^2}}}{2} \times \frac{{{v^2}}}{{{R^2}}} = mgh \Rightarrow h = \frac{{3{v^2}}}{{4g}}$
Standard 11
Physics
