A toy gun uses a spring with a force constant of \(300{\rm{ N}}/{\rm{m}}\) to propel a \(10.0 - {\rm{g}}\) steel ball. If the spring is compressed \(7.00{\rm{ cm}}\) and friction is negligible:

(a) How much force is needed to compress the spring?

(b) To what maximum height can the ball be shot?

(c) At what angles above the horizontal may a child aim to hit a target \(3.00{\rm{ m}}\) away at the same height as the gun?

(d) What is the gun’s maximum range on level ground?

The range of the projectile is,

\(R = \frac{{{v^2}\sin \left( {2\theta } \right)}}{g}\)

Here, \(R\) is the range, v is the initial velocity, \(\theta \) is the angle of projection, and \(g\) is the acceleration due to gravity.

(a)

The force needed to compress the spring is,

\(F = kx\)

Here, \(k\) is the spring constant \(\left( {k = 300{\rm{ N}}/{\rm{m}}} \right)\), \(x\) is the compression in the spring \(\left( {x = 7.00{\rm{ cm}}} \right)\).

Putting all known values,

\(\begin{array}{c}F = \left( {300{\rm{ N}}} \right) \times \left( {7.00{\rm{ cm}}} \right)\\ = \left( {300{\rm{ N}}} \right) \times \left( {7.00{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ m}}}}} \right)\\ = 21{\rm{ N}}\end{array}\)

Therefore, the required force needed to compress the spring is \(21{\rm{ N}}\).

(b)

When the ball is shot, the spring potential energy stored in the spring gets converted into potential energy of the ball.

\(\frac{1}{2}k{x^2} = mgh\)

Here, \(k\) is the spring constant \(\left( {k = 300{\rm{ N}}/{\rm{m}}} \right)\), \(x\) is the compression in the spring \(\left( {x = 7.00{\rm{ cm}}} \right)\), \(m\) is the mass of the ball \(\left( {m = 10{\rm{ g}}} \right)\), \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and \(h\) is the maximum height attained.

The expression for the maximum height attained is,

\(h = \frac{{k{x^2}}}{{2mg}}\)

Putting all known values,

\(\begin{array}{c}h = \frac{{\left( {300{\rm{ N}}/{\rm{m}}} \right) \times {{\left( {7.00{\rm{ cm}}} \right)}^2}}}{{2 \times \left( {10{\rm{ g}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = \frac{{\left( {300{\rm{ N}}/{\rm{m}}} \right) \times {{\left[ {\left( {7.00{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right]}^2}}}{{2 \times \left( {10{\rm{ g}}} \right) \times \left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 7.5{\rm{ m}}\end{array}\)

Therefore, the required maximum height the ball will reach is \(7.5{\rm{ m}}\).

(c)

When the ball is shot, the potential energy stored in the spring gets converted into kinetic energy of the ball.

\(\frac{1}{2}m{v^2} = \frac{1}{2}k{x^2}\)

Here, \(k\) is the spring constant \(\left( {k = 300{\rm{ N}}/{\rm{m}}} \right)\), \(x\) is the compression in the spring \(\left( {x = 7.00{\rm{ cm}}} \right)\), \(m\) is the mass of the ball \(\left( {m = 10{\rm{ g}}} \right)\), and \(v\) is the velocity of the ball.

The expression for the velocity is,

\(v = \sqrt {\frac{{k{x^2}}}{m}} \)

Putting all known values,

\(\begin{array}{c}v = \sqrt {\frac{{\left( {300{\rm{ N}}/{\rm{m}}} \right) \times {{\left( {7.00{\rm{ cm}}} \right)}^2}}}{{\left( {10{\rm{ g}}} \right)}}} \\ = \sqrt {\frac{{\left( {300{\rm{ N}}/{\rm{m}}} \right) \times {{\left[ {\left( {7.00{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right]}^2}}}{{\left( {10{\rm{ g}}} \right) \times \left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right)}}} \\ = \sqrt {147} {\rm{ m}}/{\rm{s}}\end{array}\)

The range of the projectile is,

\(R = \frac{{{v^2}\sin \left( {2\theta } \right)}}{g}\)

Here, \(R\) is the range \(\left( {R = 3.00{\rm{ m}}} \right)\), \({v^2}\) is the square of the velocity \(\left( {v = \sqrt {147} {\rm{ m}}/{\rm{s}}} \right)\), \(\theta \) is the angle of the projection, and \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\).

The expression for the angle of projection is,

\(\theta = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{Rg}}{{{v^2}}}} \right)\)

Putting all known values,

\(\begin{array}{c}\theta = \frac{1}{2}{\sin ^{ - 1}}\left[ {\frac{{\left( {3.00{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}{{{{\left( {\sqrt {147} {\rm{ m}}/{\rm{s}}} \right)}^2}}}} \right]\\ = 5.77^\circ \end{array}\)

Therefore, the angle above the horizontal is \(5.77^\circ \).

(d)

The range of the projectile is,

\(R = \frac{{{v^2}\sin \left( {2\theta } \right)}}{g}\)

Here, \(R\) is the range, \({v^2}\) is the square of the velocity \(\left( {v = \sqrt {147} {\rm{ m}}/{\rm{s}}} \right)\), \(\theta \) is the angle of the projection (for maximum range \(\theta = 45^\circ \)), and g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\).

Putting all known values,

\(\begin{array}{c}R = \frac{{{{\left( {\sqrt {147} {\rm{ m}}/{\rm{s}}} \right)}^2} \times \sin \left( {2 \times 45^\circ } \right)}}{{\left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 15{\rm{ m}}\end{array}\)

Therefore, the required maximum range is \(15{\rm{ m}}\).