Question

When a body is projected vertically up from the ground with certain velocity, its potential energy and kinetic energy at a point $A$ are in the ratio $2$ : $3$. If the same body is projected with double the previous velocity, then at the same point $A$ the ratio of its potential energy to kinetic energy is
A. $9$ : $1$
B. $2$ : $9$
C. $1$ : $9$
D. $9$ : $2$
E. $3$ : $2$

Answer 1

Hint: The total mechanical energy of a system remains conserved throughout the motion. When a body is projected with a velocity from the ground it acquires a kinetic energy. This kinetic energy is the total mechanical energy of the system. You need to get this total mechanical energy conserved.

Formulae Used:
If a body of mass $m$ moves with a velocity $v$ then the kinetic energy $K$ of the system can be defined as
$K = \dfrac{1}{2}m{v^2}$
If a body of mass $m$ reaches a height of $h$ then the total potential energy $P$ of the system is
$P = mgh$

Complete step by step solution:
The potential energy $P$ and kinetic energy $K$ at a point $A$ are in the ratio $2$ : $3$.
The velocity with which the body is projected is doubled in the scenario.
To get: The changed ratio of the potential energy and kinetic energy at the same point $A$.
Step 1:
Let the body has a mass $m$ and was projected vertically upwards with a velocity $v$ from the ground.
Calculate the kinetic energy acquired by the body from eq (1)
${K_T} = \dfrac{1}{2}m{v^2}$
This kinetic energy ${K_T}$ is the total mechanical energy of the system.
Step 2:
Let the point $A$ is at a height $h$ from the ground.
Hence, calculate the potential energy of the body at the point $A$ from eq (2).
${P_A} = mgh$
Let the body acquires a velocity ${u_A}$ at the point $A$.
Hence calculate the kinetic energy from eq (1).
${K_A} = \dfrac{1}{2}m{u_A}^2$
Step 2:
 Now by the problem, you have ${P_A}$ : ${K_A} = 2$ : $3$ .
$
  \therefore \dfrac{{{P_A}}}{{{K_A}}} = \dfrac{2}{3} \\
   \Rightarrow \dfrac{{mgh}}{{\dfrac{1}{2}m{u_A}^2}} = \dfrac{2}{3} \\
   \Rightarrow \dfrac{{2gh}}{{{u_A}^2}} = \dfrac{2}{3} \\
   \Rightarrow 3gh = {u_A}^2 \\
 $
$\therefore {u_A}^2 = 3gh$
Step 3:
By the conservation of the mechanical energy you have
$
  {P_A} + {K_A} = {K_T} \\
   \Rightarrow mgh + \dfrac{1}{2}m{u_A}^2 = \dfrac{1}{2}m{v^2} \\
   \Rightarrow {v^2} = {u_A}^2 + 2gh \\
 $
Now use the relation in eq (3).
$ \Rightarrow {v^2} = 3gh + 2gh = 5gh$
$\therefore {v^2} = 5gh$
Step 4:
Now for the scenario when the body is initially projected with a velocity $2v$ you have the total kinetic energy aquired that is the total mechanical energy of the system as
${K_T}' = \dfrac{1}{2}m{\left( {2v} \right)^2} = 2m{v^2}$
Step 5:
Now the point $A$ is at the same position, hence the potential energy will remain same.
Let the body acquires a velocity \[{v_A}\] at the point $A$.
Hence calculate the kinetic energy from eq (1).
${K_A}' = \dfrac{1}{2}m{v_a}^2$
Step 6:
By the conservation of the mechanical energy you have
$
  {P_A} + {K_A}' = {K_T}' \\
   \Rightarrow mgh + \dfrac{1}{2}m{v_A}^2 = 2m{v^2} \\
   \Rightarrow 4{v^2} = {v_A}^2 + 2gh \\
 $
Now from eq (4) you have ${v^2} = 5gh$ hence,
$
   \Rightarrow 4{v^2} = {v_A}^2 + 2gh \\
   \Rightarrow 4 \times \left( {5gh} \right) = {v_A}^2 + 2gh \\
   \Rightarrow 20gh - 2gh = {v_A}^2 \\
   \Rightarrow {v_A}^2 = 18gh \\
 $
Step 7:
 Now calculate the ratio ${P_A}$ : ${K_A}'$.
$
  \therefore \dfrac{{{P_A}}}{{{K_A}'}} = \dfrac{{mgh}}{{\dfrac{1}{2}m{v_A}^2}} \\
   \Rightarrow \dfrac{{{P_A}}}{{{K_A}'}} = \dfrac{{mgh}}{{\dfrac{1}{2}m\left( {18gh} \right)}} \\
   \Rightarrow \dfrac{{{P_A}}}{{{K_A}'}} = \dfrac{{gh}}{{9gh}} \\
   \Rightarrow \dfrac{{{P_A}}}{{{K_A}'}} = \dfrac{1}{9} \\
 $
So in the changed case the potential energy and kinetic energy at the point $A$ are in the ratio $1$ : $9$ .

If the body is projected with double the previous velocity, then at the same point $A$ the ratio of its potential energy to kinetic energy is (C) Increased by $1$ : $9$.

Note:
The height of the point $A$ will be unchanged. So, you will have the potential energy unchanged. So, you must need to get the relation with the initial velocity in terms of the height of the point. The initial projection gives the body a kinetic energy. This kinetic energy is the total mechanical energy. This is to be conserved even at the point $A$.