Chapter 12, Problem 63P

(a)

To determine

The new amplitude of the vibration system after collision.

Answer to Problem 63P

The new amplitude of the vibration system after collision is $1.26\text{m}$.

Explanation of Solution

Section 1:

To determine: The angular frequency of the system.

Answer: The angular frequency of the system is $5\text{rad}/\text{s}$.

Given information: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the displacement amplitude is $2.00\text{m}$ and the mass of another object is $6.00\text{kg}$.

The formula for the angular frequency is,

$\omega =\sqrt{\frac{k}{m}}$

$k$ is the force constant of the spring.

$m$ is the mass to be hanged.

Substitute $100\text{N}/\text{m}$ for $k$ and $4.00\text{kg}$ for $m$ in above equation to find $\omega$.

$\begin{array}{c}\omega =\sqrt{\frac{100\text{N}/\text{m}}{4.00\text{kg}}}\\ =5\text{rad}/\text{s}\end{array}$

Section 2:

To determine: The maximum speed of the system.

Answer: The maximum speed of the system is $10\text{m}/\text{s}$.

Given information: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The formula to calculate maximum speed is,

$v_{\text{max}}=A\omega$

$A$ is the amplitude.

Substitute $2.00\text{m}$ for $A$ and $5\text{rad}/\text{s}$ for $\omega$ in above equation to find $v_{\text{max}}$.

$\begin{array}{c}{v}_{\text{max}}=\left(2.00\text{m}\right)\left(5\text{rad}/\text{s}\right)\\ =10\text{m}/\text{s}\end{array}$

Section 3:

To determine: The speed of the system when the objects stick together after the collision.

Answer: The speed of the system when the objects stick together after the collision is $4\text{m}/\text{s}$.

Given information: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The formula to calculate speed after the collision is,

$v=\frac{m}{m+M}{v}_{\text{max}}$

$M$ is the another mass to be placed on the hanged object.

Substitute $4.00\text{kg}$ for $m$, $6.00\text{kg}$ for $M$ and $10\text{m}/\text{s}$ for $v_{\text{max}}$ in above equation to find $v$.

$\begin{array}{c}v=\left(\frac{4.00\text{kg}}{4.00\text{kg}+6.00\text{kg}}\right)\left(10\text{m}/\text{s}\right)\\ =4\text{m}/\text{s}\end{array}$

Section 4:

To determine: The new amplitude of the vibration system after collision.

Answer: The new amplitude of the vibration system after collision is $1.26\text{m}$.

Given info: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The law of conservation of energy is,

$\frac{1}{2}\left(m+M\right)v^2=\frac{1}{2}k{A}_{\text{new}}^2$

$A_{\text{new}}$ is the new amplitude after collision.

Rearrange the above equation for $A_{\text{new}}$.

$\begin{array}{c}\frac{1}{2}\left(m+M\right)v^2=\frac{1}{2}k{A}_{\text{new}}^2\\ A_{\text{new}}^2=\frac{\left(m+M\right)v^2}{k}\\ A_{\text{new}}=v\sqrt{\frac{\left(m+M\right)}{k}}\end{array}$ (I)

Substitute $4.00\text{kg}$ for $m$, $6.00\text{kg}$ for $M$, $100\text{N}/\text{m}$ for $k$ and $4\text{m}/\text{s}$ for $v$ in equation (I) to find $A_{\text{new}}$.

$\begin{array}{c}{A}_{\text{new}}=\left(4\text{m}/\text{s}\right)\sqrt{\frac{\left(4.00\text{kg}+6.00\text{kg}\right)}{100\text{N}/\text{m}}}\\ =1.26\text{m}\end{array}$

Conclusion:

Therefore, the new amplitude of the vibration system after collision is $1.26\text{m}$.

(b)

To determine

The factor by which the period of system changed.

Answer to Problem 63P

The factor by which the period of system changed is $1.58$.

Explanation of Solution

Section 1:

To determine: The initial period of system.

Answer: The initial period of system is $1.25\text{s}$.

Given info: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The formula for the period of the system before collision is,

$T_{\text{initial}}=\text{2π}\sqrt{\frac{m}{k}}$

Substitute $4.00\text{kg}$ for $m$ and $100\text{N}/\text{m}$ for $k$ in above equation to find $T_{\text{initial}}$.

$\begin{array}{c}{T}_{\text{initial}}=\text{2π}\sqrt{\frac{\left(4.00\text{kg}\right)}{100\text{N}/\text{m}}}\\ =1.25\text{s}\end{array}$

Section 2:

To determine: The final period of system.

Answer: The final period of system is $1.98\text{s}$.

Given info: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The formula for the period of the system after collision is,

$T_{\text{final}}=\text{2π}\sqrt{\frac{m+M}{k}}$

Substitute $4.00\text{kg}$ for $m$, $6.00\text{kg}$ for $M$ and $100\text{N}/\text{m}$ for $k$ in above equation to find $T_{\text{final}}$.

$\begin{array}{c}{T}_{\text{final}}=\text{2π}\sqrt{\frac{\left(4.00\text{kg}+6.00\text{kg}\right)}{100\text{N}/\text{m}}}\\ =1.98\text{s}\end{array}$

Section 3:

To determine: The factor by which the period of system changed.

Answer: The factor by which the period of system changed is $1.58$.

Given info: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The factor by which period is changed calculated as,

$\text{Factor}=\frac{T_{\text{final}}}{T_{\text{initial}}}$

Substitute $1.98\text{s}$ for $T_{\text{final}}$ and $1.25\text{s}$ for $T_{\text{initial}}$ in above equation.

$\begin{array}{c}\text{Factor}=\frac{1.98\text{s}}{1.25\text{s}}\\ =1.58\end{array}$

Conclusion:

Therefore, the factor by which the period of system changed is $1.58$.

(c)

To determine

The energy changed of the system after the collision.

Answer to Problem 63P

The energy of the system after the collision is decreased by factor $120\text{J}$.

Explanation of Solution

Given info: The mass of the particle is $4.00\text{kg}$, the force constant of spring is $100\text{N}/\text{m}$, the amplitude is $2.00\text{m}$ and the another mass is $6.00\text{kg}$.

The formula for the energy of the system before collision is,

$K{E}_{\text{initial}}=\frac{1}{2}m{v}_{\text{max}}^2$

The formula for the energy of the system after collision is,

$K{E}_{\text{final}}=\frac{1}{2}\left(m+M\right)v^2$

The chance in the energy is calculated as,

$KE=K{E}_{\text{final}}-K{E}_{\text{initial}}$

Substitute $\frac{1}{2}\left(m+M\right)v^2$ for $K{E}_{\text{final}}$ and $\frac{1}{2}m{v}_{\text{max}}^2$ for $K{E}_{\text{initial}}$ in above expression.

$KE=\frac{1}{2}\left(m+M\right)v^2-\frac{1}{2}m{v}_{\text{max}}^2$

Substitute $4.00\text{kg}$ for $m$, $6.00\text{kg}$ for $M$, $10\text{m}/\text{s}$ for $v_{\text{max}}$ and $4\text{m}/\text{s}$ for $v$ in above equation to find $KE$.

$\begin{array}{c}KE=\frac{1}{2}\left(4.00\text{kg}+6.00\text{kg}\right){\left(4\text{m}/\text{s}\right)}^2-\frac{1}{2}\left(4.00\text{kg}\right){\left(10\text{m}/\text{s}\right)}^2\\ =-120\text{J}\end{array}$

Conclusion:

Therefore, the energy of the system after the collision is decreased by factor $120\text{J}$.

(d)

To determine

To explain: The change in the energy.

Explanation of Solution

The energy of the system is defined as the capacity to do any work. The energy is the sum of potential and the kinetic energy of the system.

The type of the collision of the system is inelastic due to this the kinetic energy does not remains conserved. The mechanical energy of the system is transformed into the internal energy. So there are energy losses due to conversion of energy.

Conclusion:

Therefore, the mechanical energy of the system is transformed into the internal energy in the perfectly inelastic collision.