91Ó°ÊÓ

A spring-loaded toy gun exerts a variable force on a plastic ball as the spring expands. Consider a horizontal spring and a ball of mass m whose position when barely touching a fully expanded spring is $x=0$. The ball is pushed to the left, compressing the spring. You’ll learn in Chapter $9$that the spring force on the ball, when the ball is at position $x$(which is negative), can be written as ${\left(F_{{\mathrm{Sp}}_{\mathrm{p}}}\right)}_x=-kx$, where k is called the spring constant. The minus sign is needed to make the $x$-component of the force positive. Suppose the ball is initially pushed to $x_0=-L$, then released and shot to the right.

a. Use what you’ve learned in calculus to prove that

$a_x=v_x\frac{d{v}_x}{dx}$

b. Find an expression, in terms of $m$, $k$, and $L$, for the speed of the ball as it comes off the spring at$x=0$.

Step by step solution

01

Prove the equation (part a)

Calculus is that the study of change in mathematics, the same as how geometry is that the analysis of shape while algebra is that the research of operations and the way they're applied to unravel problems.

Using calculus,

$a_x=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

Note,

$a_x=\frac{\mathrm{d}{V}_x}{\mathrm{d}t}$

$V_x=\frac{\mathrm{d}x}{\mathrm{d}t}$

Derive,

$a_x=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

$\frac{\mathrm{d}{V}_x}{\mathrm{d}t}=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

$\frac{V_x}{V_x}×\frac{\mathrm{d}{V}_x}{\mathrm{d}t}=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

$V_x×\frac{1}{V_x}×\frac{\mathrm{d}{V}_x}{\mathrm{d}t}=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

$V_x×\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}×\frac{\mathrm{d}{V}_x}{\mathrm{d}t}=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

$V_x×\frac{\mathrm{d}t}{\mathrm{d}x}×\frac{\mathrm{d}{V}_x}{\mathrm{d}t}=V_x\frac{\mathrm{d}{V}_x}{\mathrm{d}x}$

${\mathit{V}}_{\mathit{x}}\mathbf{×}\frac{\mathrm{d}{\mathit{V}}_{\mathit{x}}}{\mathrm{d}\mathit{x}}=V_{\mathit{x}}\frac{\mathrm{d}{\mathit{V}}_{\mathit{x}}}{\mathrm{d}\mathit{x}}$

Hence proved

02

Find the expression (part b)

The expression of ball velocity when spring leaves,

${\left(F_{sp}\right)}_x=-kx$

$m{a}_x=-kx$

$a_x=\frac{-kx}{m}$

Expand equations using values,

$V_x×\frac{\mathrm{d}{V}_x}{\mathrm{d}x}=\frac{-kx}{m}$

$V_x×\mathrm{d}{V}_x=\frac{-kx}{m}\mathrm{d}x$

Integrate the equations,

$V_x{∫}_0^V\mathrm{d}{V}_x=\frac{-k}{m}{∫}_{-L}^{0}x\mathrm{d}x$

${\left[\frac{V_x^2}{2}\right]}_0^V={\left[\frac{-k}{m}×\frac{x^2}{2}\right]}_{-L}^0$

$V^2=\left[\frac{-k(0{)}^2}{m}-\frac{-k(-L{)}^2}{m}\right]$

$V^2=\frac{k{L}^2}{m}$

$V=\sqrt{\frac{k{L}^2}{m}}$

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