(A) When a body of mass $m$ is dropped from a height $h$,its velocity $v$ just before striking the ground is given by the equation of motion $v^2 = u^

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(A) When a body of mass $m$ is dropped from a height $h$,its velocity $v$ just before striking the ground is given by the equation of motion $v^2 = u^2 + 2gh$. Since it is dropped,initial velocity $u = 0$,so $v = \sqrt{2gh}$.The momentum $p$ of the body is given by $p = mv$.Substituting the value of $v$,we get $p = m\sqrt{2gh}$.Rearranging the equation to solve for mass $m$,we get $m = \frac{p}{\sqrt{2gh}}$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

Medium

$A$ body is dropped from a height $h$. If it acquires a momentum $p$ just before striking the ground,then the mass of the body is

A
$\frac{p}{\sqrt{2gh}}$
B
$\frac{p^2}{2gh}$
C
$\frac{2gh}{p}$
D
$\sqrt{\frac{2gh}{p}}$

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Motion in Straight Line

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Motion Under Gravity

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$A$ body is dropped from a height $h$. If it acquires a momentum $p$ just before striking the ground,then the mass of the body is

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