Chapter 5, Problem 2TC

Evan pushes crate starting at rest across a floor for $4\text{seconds}$ with a net force as shown (including friction). Rank the three situations from biggest to smallest.

(a) Change in momentum

(b) Final speed

(c) Momentum in $4\text{seconds}$

To determine

The ranking from biggest to smallest in terms of change in momentum.

Answer to Problem 2TC

The ranking from biggest to smallest in terms of change in momentum is A > B > C.

Explanation of Solution

Given Info: Net force in situation (a) is $100\text{N}$, mass of the crate in situation (a) is $30\text{kg}$ net force in situation (b) is $75\text{N}$, mass of the crate in situation (b) is $20\text{kg}$, net force in situation (c) is $50\text{N}$, mass of the crate in situation (c) is $10\text{kg}$, and time interval is $4\text{s}$.

Write the expression for the change in momentum.

$\Delta p=F\times t\mathrm{......}(I)$

Here,

$F$ is the force

$t$ is the time

Substitute $100\text{N}$ for $F$ and $4\text{s}$ for $t$ in the above expression to get $\Delta p$.

$\begin{array}{l}\Delta p_a=100\text{N}\times 4\text{s}\\ \Delta p_a=400\text{N}\cdot \text{s}\end{array}$

Thus, the change in momentum in situation (a) is $400\text{N}\cdot \text{s}$.

Substitute $75\text{N}$ for $F$ and $4\text{s}$ for $t$ in the above expression to get $\Delta p$.

$\begin{array}{l}\Delta p_b=75\text{N}\times 4\text{s}\\ \Delta p_b=300\text{N}\cdot \text{s}\end{array}$

Thus, the change in momentum in situation (b) is $300\text{N}\cdot \text{s}$.

Substitute $50\text{N}$ for $F$ and $4\text{s}$ for $t$ in the equation (I) to get $\Delta p$.

$\begin{array}{l}\Delta p_c=50\text{N}\times 4\text{s}\\ \Delta p_c=200\text{N}\cdot \text{s}\end{array}$

Thus, the change in momentum in situation (c) is $200\text{N}\cdot \text{s}$.

Conclusion:

Therefore, the ranking from biggest to smallest in terms of change in momentum is A > B > C.

To determine

The ranking from biggest to smallest in terms final speed.

Answer to Problem 2TC

The ranking from biggest to smallest in terms final speed is C > B > A.

Explanation of Solution

Write the expression for change in momentum.

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

Here,

$m$ is the mass

$v_i$ is the initial velocity

$v_f$ is the final velocity

As the crate was at rest initially, so its initial velocity will be $0$.

Substitute $400\text{N}\cdot \text{s}$ for $\Delta p$ and $30\text{kg}$ for $m$ in the above expression to get $v_f$.

$\begin{array}{c}400\text{N}\cdot \text{s}=\left(30\text{kg}\times v_f\right)-\left(30\text{kg}\times 0\right)\\ 400\text{N}\cdot \text{s}=30\text{kg}\times v_f\\ v_f=\frac{400\text{N}\cdot \text{s}}{30\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2\cdot \text{s}}{1\text{N}\cdot \text{s}}\\ v_f=13.4\text{m}/\text{s}\end{array}$

Thus, the final speed of crate in situation (a) is $13.4\text{m}/\text{s}$.

Substitute $300\text{N}\cdot \text{s}$ for $\Delta p$ and $20\text{kg}$ for $m$ in the equation (II) to get $v_f$.

$\begin{array}{c}300\text{N}\cdot \text{s}=\left(20\text{kg}\times v_f\right)-\left(20\text{kg}\times 0\right)\\ 300\text{N}\cdot \text{s}=20\text{kg}\times v_f\\ v_f=\frac{300\text{N}\cdot \text{s}}{20\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2\cdot \text{s}}{1\text{N}\cdot \text{s}}\\ v_f=15\text{m}/\text{s}\end{array}$

Thus, the final speed of crate in situation (b) is $15\text{m}/\text{s}$.

Substitute $200\text{N}\cdot \text{s}$ for $\Delta p$ and $10\text{kg}$ for $m$ in the above expression to get $v_f$.

$\begin{array}{c}200\text{N}\cdot \text{s}=\left(10\text{kg}\times v_f\right)-\left(30\text{kg}\times 0\right)\\ 200\text{N}\cdot \text{s}=10\text{kg}\times v_f\\ v_f=\frac{200\text{N}\cdot \text{s}}{10\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2\cdot \text{s}}{1\text{N}\cdot \text{s}}\\ v_f=20\text{m}/\text{s}\end{array}$

Thus, the final speed of crate in situation (a) is $20\text{m}/\text{s}$.

Conclusion:

Therefore, the ranking from biggest to smallest in terms final speed is C > B > A.

To determine

The ranking from biggest to smallest in terms of momentum in $4\text{seconds}$.

Answer to Problem 2TC

The ranking from biggest to smallest in terms of momentum in $4\text{seconds}$ is A > B > C.

Explanation of Solution

Given Info: Final velocity in situation (a) is $13.4\text{m}/\text{s}$, final velocity in situation (b) is $15\text{m}/\text{s}$, and final velocity in situation (c) is $20\text{m}/\text{s}$.

Write the expression for the momentum.

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

Here,

$m$ is the mass

$v$ is the velocity

Substitute $30\text{kg}$ for $m$ and $13.4\text{m}/\text{s}$ for $v$ in the above expression to get $p$.

$\begin{array}{l}{p}_a=30\text{kg}\times 13.4\text{m}/\text{s}\\ p_a=402\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum in 4 seconds in case (a) is $402\text{kg}\cdot \text{m}/\text{s}$.

Substitute $20\text{kg}$ for $m$ and $15\text{m}/\text{s}$ for $v$ in the equation (III) to get $p$.

$\begin{array}{l}{p}_b=20\text{kg}\times 15\text{m}/\text{s}\\ p_b=300\text{kg}\cdot \text{m}/\text{s}\end{array}$

Thus, the momentum in 4 seconds in case (b) is $300\text{kg}\cdot \text{m}/\text{s}$.

Substitute $10\text{kg}$ for $m$ and $20\text{m}/\text{s}$ for $v$ in the equation (III) to get $p$.

$$\begin{array}{l}{p}_c=10\text{kg}\times 200\text{m}/\text{s}\\ p_c=200\text{kg}\cdot \text{m}/\text{s}\end{array}$$

Thus, the momentum in 4 seconds in case (c) is $200\text{kg}\cdot \text{m}/\text{s}$.

Conclusion:

Therefore, the ranking from biggest to smallest in terms of momentum in $4\text{seconds}$ is A > B > C.