Chapter 10, Problem 112P

(a)

To determine

The velocity of pan just after the landing of oranges.

Answer to Problem 112P

The velocity is $2.18\text{m}/\text{s}$.

Explanation of Solution

The spring constant is $450\text{N}/\text{m}$, mass of pan is $0.250\text{kg}$, mass of oranges is $2.20\text{kg}$, and height of drop is $30.0\text{cm}$.

Write the equation for law of conservation of energy for the fall of oranges.

$\frac{1}{2}m{v}_{oi}^2=mgh$

Here, $m$ is the mass of oranges, $v_{oi}$ is the speed of oranges at a moment just before it lands on the pan, $g$ is the gravitational acceleration, and $h$ is the height of fall.

Rewrite the above equation in terms of $v$.

$v_{oi}=\sqrt{2gh}$ (I)

Write the equation for law of conservation of momentum for before and after the landing of oranges on pan.

$m{v}_{oi}=\left(m+m^{\prime }\right)v$ (II)

Here, $v_{oi}$ is the velocity of oranges before landing, $m^{\prime }$ is the mass of pan, and $v$ is the speed of pan with oranges.

Conclusion:

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30.0\text{cm}$ for $h$ in equation (I) to find $v_{oi}$.

$\begin{array}{c}{v}_{oi}=\sqrt{2\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(30.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}\\ =2.4249\text{m}/\text{s}\end{array}$

Substitute $2.4249\text{m}/\text{s}$ for $v_{oi}$, $2.20\text{kg}$ for $m$, and $0.250\text{kg}$ for $m^{\prime }$ in equation (II) to find $v$.

$\begin{array}{c}\left(2.20\text{kg}\right)\left(2.4249\text{m}/\text{s}\right)=\left(2.20\text{kg}+0.250\text{kg}\right)v\\ v=\frac{5.33478\text{kg}\cdot \text{m}/\text{s}}{2.45\text{kg}}\\ =2.177\text{m}/\text{s}\\ \approx 2.18\text{m}/\text{s}\end{array}$

Therefore, the velocity is $2.18\text{m}/\text{s}$.

(b)

To determine

The new equilibrium point of pan with oranges from the position before the oranges was dropped.

Answer to Problem 112P

The distance to new equilibrium point is $4.8\text{cm}$.

Explanation of Solution

The spring constant is $450\text{N}/\text{m}$, mass of pan is $0.250\text{kg}$, mass of oranges is $2.20\text{kg}$, and height of drop is $30.0\text{cm}$.

Write the equilibrium condition for the old equilibrium point.

$mg=kd$

Here, $k$ is the spring constant and $d$ is the stretched distance.

Write the equation for new equilibrium on adding certain amount of mass.

$\left(\Delta m\right)g=k\left(\Delta d\right)$

Here, $\Delta m$ is the additional mass and the new equilibrium distance.

Conclusion:

Substitute $2.20\text{kg}$ for $\Delta m$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $450\text{N}/\text{m}$ for $k$ in the above equation to find $\Delta d$.

$\begin{array}{c}\left(2.20\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)=\left(450\text{N}/\text{m}\right)\left(\Delta d\right)\\ \Delta d=\frac{21.56\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{450\text{N}/\text{m}}\\ =0.0479\text{m}\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)\\ \approx 4.8\text{cm}\end{array}$

Therefore, the distance to new equilibrium point is $4.8\text{cm}$.

(c)

To determine

The amplitude of the oscillations.

Answer to Problem 112P

The amplitude is $17\text{cm}$.

Explanation of Solution

The spring constant is $450\text{N}/\text{m}$, mass of pan is $0.250\text{kg}$, mass of oranges is $2.20\text{kg}$, and height of drop is $30.0\text{cm}$.

Write the equation for law of conservation of energy.

$K_i+U_i=K_f+U_f$ (I)

Here, $K_i$ is the kinetic energy at the mean position of oscillation, $U_i$ is the potential energy at the mean position.$K_f$ is the kinetic energy at the extreme position, and $U_f$ is the potential energy at the extreme position.

Write the equation for $K_i$.

$K_i=\frac{1}{2}\left(m+m^{\prime }\right)v^2$ (II)

Write the equation for $U_i$.

$U_i=\frac{1}{2}k{\left(\Delta d\right)}^2$ (III)

Write the equation for $U_f$.

$U_f=\frac{1}{2}k{A}^2$ (IV)

Rewrite equation (I) by substituting equations (II), (III), (IV) and $0$ for $K_f$.

$\frac{1}{2}\left(m+m^{\prime }\right)v^2+\frac{1}{2}k{\left(\Delta d\right)}^2=0+\frac{1}{2}k{A}^2$

Eliminate the common terms and rewrite the above expression in terms of $A$.

$A=\sqrt{\frac{\left(m+m^{\prime }\right)v^2}{k}+{\left(\Delta d\right)}^2}$

Conclusion:

Substitute $2.20\text{kg}$ for $m$, $0.250\text{kg}$ for $m^{\prime }$, $2.17\text{m}/\text{s}$ for $v$, $450\text{N}/\text{m}$ for $k$, and $4.8\text{cm}$ for $\Delta d$ in the above equation to find $A$.

$\begin{array}{c}A=\sqrt{\frac{\left(2.20\text{kg}+0.250\text{kg}\right){\left(2.17\text{m}/\text{s}\right)}^2}{450\text{N}/\text{m}}+{\left(4.8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\\ =\sqrt{0.0289}\text{m}\\ =0.17\text{m}\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)17\text{cm}\end{array}$

Therefore, the amplitude is $17\text{cm}$.

(d)

To determine

The frequency of oscillations of the system.

Answer to Problem 112P

The frequency is $2.2\text{Hz}$.

Explanation of Solution

The spring constant is $450\text{N}/\text{m}$, mass of pan is $0.250\text{kg}$, mass of oranges is $2.20\text{kg}$, and height of drop is $30.0\text{cm}$.

Write the equation for the oscillations of a spring.

$f=\frac{1}{2\pi }\sqrt{\frac{k}{\left(m+m^{\prime }\right)}}$

Here, $f$ is the frequency of oscillations of spring.

Conclusion:

Substitute $3.14$ for $\pi$, $2.20\text{kg}$ for $m$, and $0.250\text{kg}$ for $m^{\prime }$, and $450\text{N}/\text{m}$ for $k$ in the above equation to find $\Delta d$.

$\begin{array}{c}f=\frac{1}{2\left(3.14\right)}\sqrt{\frac{450\text{N}/\text{m}}{\left(2.20\text{kg}+0.250\text{kg}\right)}}\\ =\left(0.159\right)\left(13.55\text{N}/\text{m}\cdot \text{kg}\right)\\ =2.15\text{Hz}\\ \approx 2.2\text{Hz}\end{array}$

Therefore, the frequency is $2.2\text{Hz}$.