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5.Work, Energy, Power and Collision
normal
A mass $m$ moving horizontally with velocity $v_0$ strikes a pendulum of mass $m$. If the two masses stick together after the collision, then the maximum height reached by the pendulum is
A
$v_0^2/8g$
B
$v_0^2/2g$
C
$\sqrt {2{v_0}g} $
D
$\sqrt {{v_0}g} $
Solution
$\mathrm{mv}_{0}=2 \mathrm{mv}$
$\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{g}}=\frac{\left(\frac{\mathrm{v}_{0}}{2}\right)^{2}}{2 \mathrm{g}}=\frac{\mathrm{v}_{0}^{2}}{8 \mathrm{g}}$
Standard 11
Physics
