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College Physics (Urone)
Found in: Page 264
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

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?

(a) The force needed to compress the spring is \(21{\rm{ N}}\).

(b) The maximum height the ball will reach is \(7.5{\rm{ m}}\).

(c) The angle above the horizontal is \(5.77^\circ \).

(d) The maximum range is \(15{\rm{ m}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of Concepts 

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.

Step 2: Calculate the force needed to compress the spring

(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 \(21{\rm{ N}}\).

Step 3: Find the maximum height the ball will reach

(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 \(7.5{\rm{ m}}\).

Step 4: Find the angle above the horizontal

(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 \(5.77^\circ \).

Step 5: Find the maximum range

(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 \(15{\rm{ m}}\).

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