Question

Two stones having masses in the ratio of 3 : 2 are dropped from the heights in the ratio of 4:9. The ratio of the magnitudes of their linear momentum just before reaching the ground is (neglect air resistance)
(A) 4 : 9
(B) 2 : 3
(C) 3 : 2
(D) 1 : 1

Answer 1

The velocity is directly proportional to the square root of the height. For the ratio of linear momentum, multiply the ratio of the velocity to the ratio of the velocity.

Formula used: In this solution we will be using the following formulae;
$p = mv$ where $p$ is the linear momentum of a body, $m$ is the mass of a body, and $v$ is the linear velocity of the body.
${v^2} = {u^2} + 2gh$ where $v$ is the final velocity of a body falling from a height $h$ , $u$ is initial velocity and $g$ is acceleration due to gravity.

Complete step by step answer:
Generally, when air resistance is neglected, the velocity of a body being dropped from a height is independent of mass and, if the initial velocity is zero, is directly proportional to the square root of the height at which the ball was dropped. Hence, we can write that the ratio of the velocities is
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {{s_1}} }}{{\sqrt {{s_2}} }} = \sqrt {\dfrac{{{s_1}}}{{{s_2}}}}$
The ratio of the height of the balls is said to be 4 : 9 which can be written as
$\dfrac{{{s_1}}}{{{s_2}}} = \dfrac{4}{9}$
Then from above, the ratio of their velocities would be
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{4}{9}} = \dfrac{2}{3}$
Also, the ratio of their masses is said to be 3 : 2, which can be written as
$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{3}{2}$
Hence, the ratio of their linear momentum can be gotten from
$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{{{m_1}{v_1}}}{{{m_2}{v_2}}} = \dfrac{{{m_1}}}{{{m_2}}} \times \dfrac{{{v_1}}}{{{v_2}}}$
Since $p = mv$ where $p$ is the linear momentum of a body, $m$ is the mass of a body, and $v$ is the linear velocity of the body.
Hence, by inserting know values, we get
$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{3}{2} \times \dfrac{2}{3} = \dfrac{1}{1}$
Hence, the ratio is 1 : 1
The correct option is D.

Note:
For clarity, the fact that velocity can be given as the square root of height can be gotten from ${v^2} = {u^2} + 2gh$ where $v$ is the final velocity of a body falling from a height $h$ , $u$ is initial velocity and $g$ is acceleration due to gravity.
Hence, if initial velocity is zero, we have
${v^2} = 2gh$ or $v = \sqrt {2gh}$
Hence, we can say $v \propto \sqrt h$ .