Chapter 10, Problem 75PQ

(a)

To determine

Final speed of the sled and driver relative to the ground.

Answer to Problem 75PQ

The final speed of the sled and driver relative to the ground is $\underset{\_}{3.36\text{m}/\text{s}}$ in the positive x direction.

Explanation of Solution

Apply law of conservation of momentum to the system. The momentum of combination of sled, backpack and driver is equal to the momentum of backpack plus the momentum of sled and driver.

Write the equation to find the combined momentum of backpack, driver and sled.

  ${\overrightarrow{P}}_s=M{v}_{si}\widehat{i}$                                                                                                                (I)

Here, ${\overrightarrow{P}}_s$ is the momentum of sled initially, $M$ is the total mass of sled, and $v_{si}$ is the initial velocity of sled.

After some time harness of the sled breaks. In order to stop the harness the driver throws away his back pack in the opposite direction of motion of sled. Thus the final momentum of system is equal to the sum of momentum of back pack plus the momentum of driver and sled.

Write the equation to find the momentum after the breakage of harness.

  ${\overrightarrow{P}}_f=\left(M-m\right)v_{sf}\widehat{i}+m\left(-v_{bf}\right)\widehat{i}$                                                                                 (II)

Here, ${\overrightarrow{P}}_f$ is the momentum after the breakage of harness, $M-m$ is the mass of sled without back pack, $v_{sf}$ is the final velocity of sled, and $v_{bf}$ is the final velocity of back pack.

Apply conservation of momentum.

  ${\overrightarrow{P}}_s={\overrightarrow{P}}_f$                                                                                                                   (III)

Substitute (II) and (I) in (III).

  $M{v}_{si}\widehat{i}=\left(M-m\right)v_{sf}\widehat{i}+m\left(-v_{bf}\right)\widehat{i}$                                                                        (IV)

The relative velocity of back pack with respect to the sled is equal to the sum of velocity of sled and velocity of back pack.

Write the equation to find the relative speed of back pack with respect to the sled.

  $v_{bs}=v_{sf}+v_{bf}$                                                                                                         (V)

Here, $v_{bs}$ is the relative speed of back pack with respect to sled.

Rearrange equation (V) to find $v_{sf}$.

  $v_{sf}=v_{sb}-v_{bf}$                                                                                                          (VI)

Substitute $v_{sb}-v_{bf}$ for $v_{sf}$ in equation (IV).

  $M{v}_{si}\widehat{i}=\left(M-m\right)\left(v_{sb}-v_{bf}\right)\widehat{i}+m\left(-v_{bf}\right)\widehat{i}$                                                              (VII)

Simplify equation (VII) to get $v_{bf}$.

  $v_{bf}=\frac{v_{sb}\left(M-m\right)-M{v}_{si}}{M}$                                                                                    (VIII)

Conclusion:

Substitute $5.00\text{m}/\text{s}$ for $v_{sb}$, $275\text{kg}$ for $M$, $3.00\text{m}/\text{s}$ for $v_{si}$, and $20\text{kg}$ for $m$, in equation (VIII) and solve for $v_{bf}$.

  $\begin{array}{c}{v}_{bf}=\frac{\left(5.00\text{m}/\text{s}\right)\left(275\text{kg}-\text{20}\text{kg}\right)-275\text{kg}\left(3.00\text{m}/\text{s}\right)}{275\text{kg}}\\ =1.64\text{m}/\text{s}\end{array}$

Therefore substitute $1.64\text{m}/\text{s}$ for $v_{bf}$ and $5.00\text{m}/\text{s}$ in equation (VI) to get $v_{sf}$.

  $\begin{array}{c}{v}_{sf}=5.00\text{m}/\text{s}\widehat{i}-1.64\text{m}/\text{s}\widehat{i}\\ =3.36\text{m}/\text{s}\widehat{i}\end{array}$

Therefore, the final speed of the sled and driver relative to the ground is $\underset{\_}{3.36\text{m}/\text{s}}$ in the positive x direction.

(b)

To determine

Final speed of the back pack with respect to the ground.

Answer to Problem 75PQ

The final speed of back pack with respect to the ground is $\underset{\_}{1.64\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the $v_{bf}$.

  $v_{bf}=\frac{v_{sb}\left(M-m\right)-M{v}_{si}}{M}$

Conclusion:

Substitute $5.00\text{m}/\text{s}$ for $v_{sb}$, $275\text{kg}$ for $M$, $3.00\text{m}/\text{s}$ for $v_{si}$, and $20\text{kg}$ for $m$, in above equation and solve for $v_{bf}$.

  $\begin{array}{c}{v}_{bf}=\frac{\left(5.00\text{m}/\text{s}\right)\left(275\text{kg}-\text{20}\text{kg}\right)-275\text{kg}\left(3.00\text{m}/\text{s}\right)}{275\text{kg}}\\ =1.64\text{m}/\text{s}\end{array}$

Therefore, final speed of the back pack with respect to the ground is $\underset{\_}{1.64\text{m}/\text{s}}$ in the opposite direction of velocity of sled.