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Expert-verified Found in: Page 264 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 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 $$15{\rm{ m}}$$. ### Want to see more solutions like these? 