Chapter 5, Problem 1TC

The balls have different masses and speeds.

(a) Rank them in terms of momentum, from greatest to least.

(b) Rank them in terms of the impulse needed to stop them, from greatest to least.

To determine

The ranking of balls in term of momentum from greatest to least.

Answer to Problem 1TC

The ranking of balls in term of momentum from greatest to least is $p_b>p_d>p_c>p_a$.

Explanation of Solution

Given Info: Mass of the ball in the case (a) is $1.0\text{kg}$, speed of the ball in the case (a) is $9.0\text{m}/\text{s}$, mass of the ball in the case (b) is $1.2\text{kg}$, speed of the ball in the case (b) is $8.5\text{m}/\text{s}$, mass of the ball in the case (c) is $0.8\text{kg}$, speed of the ball in the case (c) is $12.0\text{m}/\text{s}$, mass of the ball in the case (d) is $5.0\text{kg}$, and speed of the ball in the case (d) is $2.0\text{m}/\text{s}$.

Write the expression for the momentum.

$p=mv\mathrm{......}(\text{I})$

Here,

$m$ is the mass

$v$ is the speed

Substitute $1.0\text{kg}$ for $m$ and $9.0\text{m}/\text{s}$ for $v$ in equation (I) to get $p$.

$\begin{array}{l}{p}_a=1.0\text{kg}\times 9.0\text{m}/\text{s}\\ p_a=9.0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum of ball in case (a) is $9.0\text{kg}\cdot \text{m}/\text{s}$.

Substitute $1.2\text{kg}$ for $m$ and $8.5\text{m}/\text{s}$ for $v$ in equation (I) to get $p$.

$$\begin{array}{l}{p}_b=1.2\text{kg}\times 8.5\text{m}/\text{s}\\ p_b=10.2\text{kg}\cdot \text{m}/\text{s}\end{array}$$

Thus, the momentum of ball in case (b) is $10.2\text{kg}\cdot \text{m}/\text{s}$.

Substitute $0.8\text{kg}$ for $m$ and $12.0\text{m}/\text{s}$ for $v$ in equation (I) to get $p$.

$$\begin{array}{l}{p}_c=0.8\text{kg}\times 12.0\text{m}/\text{s}\\ p_c=9.6\text{kg}\cdot \text{m}/\text{s}\end{array}$$

Thus, the momentum of ball in case (c) is $9.6\text{kg}\cdot \text{m}/\text{s}$.

Substitute $5.0\text{kg}$ for $m$ and $2.0\text{m}/\text{s}$ for $v$ in equation (I) to get $p$.

$$\begin{array}{l}{p}_d=5.0\text{kg}\times 2.0\text{m}/\text{s}\\ p_d=10.0\text{kg}\cdot \text{m}/\text{s}\end{array}$$

Thus, the momentum of ball in case (d) is $10.0\text{kg}\cdot \text{m}/\text{s}$.

Conclusion:

Therefore, the ranking of balls in term of momentum from greatest to least is $p_b>p_d>p_c>p_a$.

To determine

The ranking of the balls in term of impulse needed to stop them, from greatest to least.

Answer to Problem 1TC

The ranking of the balls in term of impulse needed to stop them, from greatest to least is $p_b>p_d>p_c>p_a$.

Explanation of Solution

Given Info: Momentum of the ball in the case (a) is $9.0\text{kg}\cdot \text{m}/\text{s}$, momentum of the ball in the case (b) is $10.2\text{kg}\cdot \text{m}/\text{s}$, momentum of the ball in the case (c) is $9.6\text{kg}\cdot \text{m}/\text{s}$, momentum of the ball in the case (d) is $10.0\text{kg}\cdot \text{m}/\text{s}$.

Write the expression for the impulse.

$\begin{array}{l}I=\Delta p\\ I=p_f-p_i\mathrm{......}(\text{II})\end{array}$

Here,

$p_f$ is the final momentum

$p_i$ is the initial momentum

The final momentum of a body will be zero if it is bought to rest.

Substitute $9.0\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $0$ for $p_i$ in equation (II) to get $I$.

$\begin{array}{l}{I}_a=p_a-0\\ I_a=9.0\text{kg}\cdot \text{m}/\text{s}-0\\ I_a=9.0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the impulse required to stop the ball in case (a) is $9.0\text{kg}\cdot \text{m}/\text{s}$.

Substitute $10.2\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $0$ for $p_i$ in equation (II) to get $I$.

$$\begin{array}{l}{I}_b=p_b-0\\ I_b=10.2\text{kg}\cdot \text{m}/\text{s}-0\\ I_b=10.2\text{kg}\cdot \text{m}/\text{s}\end{array}$$

Thus, the impulse required to stop the ball in case (b) is $10.2\text{kg}\cdot \text{m}/\text{s}$.

Substitute $9.6\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $0$ for $p_i$ in equation (II) to get $I$.

$\begin{array}{l}{I}_c=p_b-0\\ I_c=9.6\text{kg}\cdot \text{m}/\text{s}-0\\ I_c=9.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the impulse required to stop the ball in case (c) is $9.6\text{kg}\cdot \text{m}/\text{s}$.

Substitute $10.0\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $0$ for $p_i$ in equation (II) to get $I$.

$\begin{array}{l}{I}_d=p_b-0\\ I_d=10.0\text{kg}\cdot \text{m}/\text{s}-0\\ I_d=10.0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the impulse required to stop the ball in case (b) is $10.0\text{kg}\cdot \text{m}/\text{s}$

Conclusion:

Therefore, the ranking of the balls in term of impulse needed to stop them, from greatest to least is $p_b>p_d>p_c>p_a$.