Question

A toy gun uses a spring of force constant K. Before being triggered in the upward direction, the spring is compressed by a distance x. If the mass of the shot is m, on being triggered, it will go up to a height of:
A. $\dfrac{{K{x^2}}}{{2mg}}$
B. $\dfrac{{{x^2}}}{{Kmg}}$
C. $\dfrac{{K{x^2}}}{{mg}}$
D. $\dfrac{{{{\left( {Kx} \right)}^2}}}{{mg}}$

Answer 1

Hint: In the question, they’ve given us a spring of force constant K which is compressed giving rise to elastic potential energy. This results in the shot of mass m moving upwards. The problem is simply based on the law of conservation of energy. That is the potential energy in spring is converted into potential energy of the shot.
Formula used:
$\eqalign{
  & P{E_{spring}} = \dfrac{1}{2}K{x^2} \cr
  & P{E_{shot}} = mgh \cr} $

Complete step-by-step solution:
If we draw a simple figure, from the given situation, it would look something like this.

Initially, when the spring is compressed by x, it will store this as potential energy in the spring. By the law of conservation of energy, this elastic potential energy will impart energy to the shot of mass m allowing it to move upwards.
Mathematically this can be written as,
$P{E_{spring}} = P{E_{shot}}$
The elastic potential energy of the spring is given by
$P{E_{spring}} = \dfrac{1}{2}K{x^2}$
Where,
K is the spring constant of the spring
x is the compression
The potential energy of the shot is given by
$P{E_{shot}} = mgh$
Where,
m is the mass of the shot
g is the acceleration due to gravity
h is the height to which the mass rises
On equating these equations we’ll have,
$\eqalign{
  & P{E_{spring}} = P{E_{shot}} \cr
  & \Rightarrow \dfrac{1}{2}K{x^2} = mgh \cr
  & \Rightarrow h = \dfrac{{K{x^2}}}{{2mg}} \cr} $
Therefore, the correct option is A.

Note: Keep in mind that the sign of the potential energy is taken to be positive at the extreme positions of the string and zero at the mean position. The law of conservation of energy states that energy can neither be created nor destroyed. It only changes forms. As of our problem, it changes from elastic potential energy to potential energy.