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Two identical bodies are allowed to fall from different heights h1 and h2. What is the ratio of their momentum when they reach the ground?

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h1 : h2

h12 : h22

\sqrt{h_{1}} : \sqrt{h_{2}}

answer is C.

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Detailed Solution

The ratio of their momentums will be

\sqrt{h_{1}} : \sqrt{h_{2}}

.
The displacement of an object with a velocity v is given by

s = \frac{v^{2}}{2 a}

When the object is allowed to fall from a height the above expression becomes

h = \frac{v^{2}}{2 g} or v = \sqrt{2 gh}

When two identical bodies are allowed to fall from heights h1 and h2, the ratio of their velocities will be:

\frac{v_{1}}{v_{2}} = \frac{\sqrt{h_{1}}}{\sqrt{h_{2}}}

The momentum of the given bodies will be in the ratio of:

\frac{P_{1}}{P_{2}} = \frac{m v_{2}}{m v_{2}} = \frac{v_{1}}{v_{2}} = \frac{\sqrt{h_{1}}}{\sqrt{h_{2}}}

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