Q29E - University Physics with Modern Physics

Question

You (mass $\mathbf{55}\mathbf{kg}$) are riding a frictionless skateboard (mass $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{kg}$) in a straight line at a speed of $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{ }\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.$. A friend standing on a balcony above you drops a $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\text{ kg}}$ sack of flour straight down into your arms. (a) What is your new speed while you hold the sack? (b) Since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Now you try to rid yourself of the extra weight by throwing the sack straight up. What will be your speed while the sack is in the air? Explain.

Step-by-Step Solution

Answer

(a) The person’s speed after holding the sack is $4.32 \raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.$. 

(b) The sack does not have a horizontal motion.

(c) The your speed (person’s speed) while the sack is in the air is $4.32 \raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.$.

Step 1: Conservation of momentum:

The law of conservation of momentum states that in an isolated system, the total momentum of two or more bodies acting on each other remains constant unless an external force acts. Therefore, momentum can neither be created nor destroyed.

Step 2: A concept:

Assume that the person is moving in the positive direction of the x-axis.

Before dropping the bag:

Use the equation for the relationship between the momentum, the mass of the person and the skateboard, and the speed of the person on the skateboard to calculate the momentum of the person on the skateboard before throwing the sake.

 

After dropping the bag:

Use the equation for the relationship between the momentum, and the mass of the person on the skateboard holding the sake, and obtain an expression for the velocity of the person on the skateboard holding the sake after the sake is thrown.

 

Apply the law of conservation of momentum in the absence of an external horizontal force; calculate the speed of the person on the skateboard holding the bag after the bag is thrown.

Step 3: (a) Person’s new speed while you hold the sack:

Before the sack is dropped:

Determine the momentum of the person on the skateboard before the sack is dropped by using the following relation,

$p_{i,x}=\left(m_p+m_{sb}\right)v_{i,x}$                                                                                      ..... (1)

Here, 

The mass of the person, $m_p=55\text{ kg}$

The mass of the skateboard, $m_{sb}=5\text{ kg}$

The speed of the person on the skateboard before the sack is dropped is $v_{i,x}=4.5 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Substitute these above values into equation (1), and you get

$\begin{array}{rcl}{p}_{i,x}& =& \left(55\text{ kg}+5.0\text{ kg}\right)4.5 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\\ & =& \left(\text{60 kg}\right)4.5 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\\ & =& 270 \raisebox{1ex}{$\text{kg·m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

After the sack is dropped:

Define the momentum of the person on the skateboard after the sack is dropped by the relation,

$p_{f,x}=\left(m_p+m_{sb}+m_s\right)v_{f,x}$                                                                            ..... (2)

Here, 

Mass of the sack,  $m_s=2.5\text{ kg}$

The speed of the person (you) on the skateboard after the sack is dropped is $v_{f,x}$.  

The momentum of the person on the skateboard after the sack is dropped is $p_{f,x}$.

 

Putting known values into equation (2), you obtain

$\begin{array}{rcl}{p}_{f,x}& =& \left(55\text{ kg}+5.0\text{ kg}+2.5\text{ kg}\right)v_{f,x}\\ & =& 62.5\text{ kg}\times v_{f,x}\end{array}$

Step 4: Law of Conservation of Momentum:

Apply the law of conservation for equations (1) and (2), you have

$p_{f,x}=p_{i,x}$

Substitute known values in the above equation.

$$\begin{gathered} 62.5\text{ kg}\times v_{f,x}=270\raisebox{1ex}{${\displaystyle \text{kg·m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right. \\ {v}_{f,x}=\frac{270{\displaystyle \raisebox{1ex}{$\text{kg·m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}{62.5\text{ kg}} \\ {v}_{f,x}=4.32 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right. \end{gathered}$$

Hence, the person’s speed after holding the sack is $4.32 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Step 5: (b) Since the sack was dropped vertically, how can it affect your horizontal motion:

The sack has no horizontal movement. To take the ack at your speed, (the person) must exert a horizontal force on the bag and slow the person down.

Step 6: (c) Your speed while the sack is in the air:

Throwing the sack in the air requires vertical force; its horizontal component is zero. Hence, the person after throwing the sack in the air remains the same, that is $4.32 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

 

Hence, your speed (person’s speed) while the sack is in the air is $4.32 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.