Question

A toy gun consists of a spring and rubber dart of mass $25\,g$. When the spring is compressed by $4\,cm$, and the dart is fired vertically, it projects the dart to a height of $2\,m$. If the spring is compressed by $8\,cm$, and the same dart is projected vertically, the dart will rise to a height of
(A) $16\,m$
(B) $8\,m$
(C) $4\,m$
(D) $2\,m$

Answer 1

Hint
The height of the dart can be determined by equating the elastic potential energy of the spring with the potential energy of the dart, from that equation, the height of the ball and the spring compression is equated, then the height is determined.
The potential energy of the dart is given by,
$\Rightarrow PE = mgh$
Where, $PE$ is the potential energy of the ball, $m$ is the mass of the dart, $g$ acceleration due to gravity on the dart and $h$ is the height travelled by the dart.
The elastic potential energy of the dart is given by,
$\Rightarrow PE = \dfrac{1}{2}k{x^2}$
Where, $PE$ is the elastic potential energy of the spring, $k$ is the spring constant and $x$ is the extension or compression of the spring.

Complete step by step answer
Given that, The mass of the dart is, $m = 25\,g$
The compression of the spring for first time, ${x_1} = 4\,cm = 0.04\,m$
The height of the dart is, ${h_1} = 2\,m$
The compression of the spring for second time, ${x_2} = 8\,cm = 0.08\,m$
Now, By equating the both the potential energy, then
$\Rightarrow \dfrac{1}{2}k{x^2} = mgh\,..................\left( 1 \right)$
The process is done two times, then from the above equation
$\Rightarrow \dfrac{1}{2}k{x_1}^2 = mg{h_1}\,..................\left( 2 \right)$
Then,
$\Rightarrow \dfrac{1}{2}k{x_2}^2 = mg{h_2}\,..................\left( 3 \right)$
On dividing the equation (2) and equation (3), then
$\Rightarrow \left( {\dfrac{{\dfrac{1}{2}k{x_1}^2}}{{\dfrac{1}{2}k{x_2}^2}}} \right) = \dfrac{{mg{h_1}}}{{mg{h_2}}}$
By cancelling the same terms, then the above equation is written as,
$\Rightarrow \dfrac{{{x_1}^2}}{{{x_2}^2}} = \dfrac{{{h_1}}}{{{h_2}}}$
By keeping the term ${h_2}$ in one side and the other terms in other side, then
$\Rightarrow {h_2} = {h_1} \times \dfrac{{{x_2}^2}}{{{x_1}^2}}$
By substituting the height of the dart, compression of the spring for the first and second time in the above equation, then,
$\Rightarrow {h_2} = 2 \times \dfrac{{{{\left( {0.08} \right)}^2}}}{{{{\left( {0.04} \right)}^2}}}$
By using the square on both numerator and denominator, then
$\Rightarrow {h_2} = 2 \times \dfrac{{6.4 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 3}}}}$
By cancelling the same terms, then
$\Rightarrow {h_2} = 2 \times \dfrac{{6.4}}{{1.6}}$
On dividing the terms, then
$\Rightarrow {h_2} = 2 \times 4$
On multiplying the terms, then
$\Rightarrow {h_2} = 8\,m$
The dart will rise to a height of $8\,m$
Hence, the option (B) is the correct answer.

Note
The equation (2) and the equation (3), both are divided because the spring constant value is not given in the question, to cancel the spring constant, so the equation (2) and the equation (3) are divided. The spring compression values are given in the centimetre, it is converted to meters because the height is given in meters.