Question

A mass m = 0.7 kg is released from rest at the origin 0. The mass falls under the influence of gravity. When the mass reaches point A, it is a distance x below the origin 0; when the mass reaches point B it is a distance of 3 x below the origin 0. What is vB/vA?

Answer 1

Answer:

$v_B/v_A=\sqrt{3}$

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

$v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2$

Therefore, if we consider $t_A$ the time for the object to reach point A, and $t_B$ the time for it to reach point B, then:

$v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}$

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

$x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3$

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

$(\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}$