Question

A 15 g ball is shot from a spring gun whose spring has a force constant of 600 N/m. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression is (\[g = 10\,{\text{m/}}{{\text{s}}^2}\])
A. 6.0 m
B. 10.0 m
C. 12.0 m
D. 8.0 m

Answer 1

Hint: Using the law of conservation of energy, the elastic potential energy of the spring is converted into the kinetic energy. Therefore, calculate the velocity of the ball. Recall the expression for the horizontal range. The projectile attains the maximum range when the angle of projection is equal to \[45^\circ \].

Formula used:
Potential energy stored in the spring, \[U = \dfrac{1}{2}k{x^2}\]
Here, k is the force constant and x is the displacement of the spring.
Range, \[R = \dfrac{{{v^2}\sin 2\theta }}{g}\]
Here, v is the initial velocity of the ball, \[\theta \] is the angle of projection and g is the acceleration due to gravity.

Complete step by step answer:
As the spring is compressed by a distance 5cm, the energy is stored in the spring in the form of potential energy. We have the expression for the elastic potential energy of the spring,
\[U = \dfrac{1}{2}k{x^2}\]
Here, k is the force constant and x is the displacement of the spring.
Substituting \[k = 600\,{\text{N/m}}\] and \[x = 0.05\,{\text{m}}\] in the above equation, we get,
\[U = \dfrac{1}{2}\left( {600} \right){\left( {0.05} \right)^2}\]
\[ \Rightarrow U = \dfrac{1}{2}\left( {600} \right){\left( {0.05} \right)^2}\]
\[ \Rightarrow U = 0.75\,{\text{J}}\]

Using the law of conservation of energy, the elastic potential energy of the spring is converted into the kinetic energy. Therefore, we can write,
\[\dfrac{1}{2}m{v^2} = \dfrac{1}{2}k{x^2}\]
\[ \Rightarrow \dfrac{1}{2}m{v^2} = 0.75\,{\text{J}}\]
Substituting \[m = 0.015\,{\text{kg}}\] in the above equation, we get,
\[\dfrac{1}{2}\left( {0.015} \right){v^2} = 0.75\]
\[ \Rightarrow {v^2} = 100\]
\[ \Rightarrow v = 10\,{\text{m/s}}\]

We have the expression for the horizontal range of the projectile,
\[R = \dfrac{{{v^2}\sin 2\theta }}{g}\]
Here, v is the initial velocity of the ball, \[\theta \] is the angle of projection and g is the acceleration due to gravity.
We know that the projectile attains the maximum range when the angle of projection is equal to \[45^\circ \].
Substituting \[v = 10\,{\text{m/s}}\], \[\theta = 45^\circ \] and \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[R = \dfrac{{{{\left( {10} \right)}^2}\sin 2\left( {45} \right)}}{{10}}\]
\[ \Rightarrow R = \dfrac{{{{\left( {10} \right)}^2}\sin 90}}{{10}}\]
\[ \therefore R = 10\,{\text{m}}\]

So, the correct answer is option B.

Note: For the greatest possible horizontal range, the term \[\sin 2\theta \] should be equal to 1. For this, the angle \[\theta \] should be \[45^\circ \]. If you don’t remember the formula for the horizontal range, use the kinematic equation in the horizontal direction.