Chapter 6, Problem 1TC

The mass and speed of three vehicles is shown below.

(a) Compare the momentum, from most to least.

(b) Compare the kinetic energy, from most to least.

To determine

The ranking form most to least in terms of momentum.

Answer to Problem 1TC

The ranking form most to least in terms of momentum is $A>C>B$.

Explanation of Solution

Given Info: Mass of the vehicle in the case (A) is $90\text{kg}$, speed of the vehicle in the case (A) is $10\text{m}/\text{s}$, mass of the vehicle in the case (B) is $110\text{kg}$, speed of the vehicle in case (B) is $5\text{m}/\text{s}$, mass of the vehicle in the case (C) is $700\text{kg}$, and speed of the vehicle in case (C) is $2\text{m}/\text{s}$.

Write the expression for the momentum.

$p=mv$......(I)

Here,

$m$ is the mass

$v$ is the speed

Substitute $90\text{kg}$ for $m$ and $10\text{m}/\text{s}$ for $v$ in the expression (I) to get $p$.

$\begin{array}{c}{p}_A=90\text{kg}\times 10\text{m}/\text{s}\\ p_A=900\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum of the vehicle in case (A) is $900\text{kg}\cdot \text{m}/\text{s}$.

Substitute $110\text{kg}$ for $m$ and $5\text{m}/\text{s}$ for $v$ in the expression (I) to get $p$.

$\begin{array}{c}{p}_B=110\text{kg}\times 5\text{m}/\text{s}\\ p_B=550\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum of the vehicle in case (B) is $550\text{kg}\cdot \text{m}/\text{s}$.

Substitute $700\text{kg}$ for $m$ and $2\text{m}/\text{s}$ for $v$ in the expression (I) to get $p$.

$\begin{array}{c}{p}_C=700\text{kg}\times 2\text{m}/\text{s}\\ p_C=700\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum of the vehicle in case (C) is $700\text{kg}\cdot \text{m}/\text{s}$.

Conclusion:

Therefore, the ranking form most to least in terms of momentum is $A>C>B$.

To determine

The ranking from most to least in terms of kinetic energy.

Answer to Problem 1TC

The ranking from most to least in terms of kinetic energy is $A>C>B$.

Explanation of Solution

Write the expression for the kinetic energy.

$KE=\frac{1}{2}m{v}^2$ ...... (II)

Here,

$m$ is the mass

$v$ is the velocity

Substitute $90\text{kg}$ for $m$ and $10\text{m}/\text{s}$ for $v$ in the expression (II) to get $KE$.

$\begin{array}{c}K{E}_A=\frac{1}{2}\times \left(90\text{kg}\right)\times {\left(10\text{m}/\text{s}\right)}^2\\ K{E}_A=4500\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ K{E}_A=4500\text{J}\end{array}$

Thus, the kinetic energy of the vehicle in the case (A) is $4500\text{J}$.

Substitute $110\text{kg}$ for $m$ and $5\text{m}/\text{s}$ for $v$ in the expression (II) to get $KE$.

$\begin{array}{c}K{E}_B=\frac{1}{2}\times \left(110\text{kg}\right)\times {\left(5\text{m}/\text{s}\right)}^2\\ K{E}_B=1375\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ K{E}_B=1375\text{J}\end{array}$

Thus, the kinetic energy of the vehicle in the case (B) is $1375\text{J}$.

Substitute $110\text{kg}$ for $m$ and $5\text{m}/\text{s}$ for $v$ in the expression (II) to get $KE$.

$\begin{array}{c}K{E}_C=\frac{1}{2}\times \left(\text{700}\text{kg}\right)\times {\left(2\text{m}/\text{s}\right)}^2\\ K{E}_C=1400\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ K{E}_C=1400\text{J}\end{array}$

Thus, the kinetic energy of the vehicle in the case (C) is $1400\text{J}$.

Conclusion:

Therefore, the ranking from most to least in terms of kinetic energy is $A>C>B$.