Chapter 9, Problem 70A

(a)

To determine

Tosketch:The given situation.

Explanation of Solution

Given:

Mass of the marble $C$ is $m_C=5.0\text{ g = 5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $C$ is $v_{Ci}=20.0\text{ cm/s = 20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Mass of the marble $D$ is $m_D=10.0\text{ g = 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $D$ is $v_{Di}=10.0\text{ cm/s = 10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Both the marbles moving in the same direction before collision.

After collision marble $C$ continues in the same direction.

Final velocity of the marble $C$ is $v_{Cf}=8.0\text{ cm/s = 8}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Before collision:

  $m_C=\text{5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

  $v_{Ci}=\text{ 20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

  $m_D=\text{ 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

  $v_{Di}=\text{ 10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

After collision:

  $m_C=\text{5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

  $v_{Cf}=\text{ 8}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

  $m_D=\text{ 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

The situation before and after the collision between the two marbles are as shown in figure $\text{1}$ and figure $\text{2}$ .

Figure 1

Figure 2

(b)

To determine

The marbles momentum before collision.

Answer to Problem 70A

  $p_{Ci}=p_{Di}=1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Explanation of Solution

Given:

Mass of the marble $C$ is $m_C=5.0\text{ g = 5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $C$ is $v_{Ci}=20.0\text{ cm/s = 20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Mass of the marble $D$ is $m_D=10.0\text{ g = 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $D$ is $v_{Di}=10.0\text{ cm/s = 10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Formula used:

Momentum $\left(p\right)$ is the product of mass $\left(m\right)$ and velocity $\left(v\right)$ of a moving body. That is,

  $p=mv$

Initial momentum of the marble $C$ can be calculated using the equation,

  $p_{Ci}=m_C{v}_{Ci}$  $\left(1\right)$

Where,

$m_C$ is the mass of the marble $C$

$v_{Ci}$ is initial velocity of the marble $C$

Initial momentum of the marble $D$ can be calculated using the equation,

  $p_{Di}=m_D{v}_{Di}$  $\left(2\right)$

Where,

$m_D$ is the mass of the marble $D$

$v_{Di}$ is initial velocity of the marble $D$

Calculation:

Initial momentum of the marble $C$ can be calculated by substituting the numerical values in equation $\left(1\right)$ ,

  $p_{Ci}=\left(\text{5}\text{.0}\times {\text{10}}^{-3}\text{ kg}\right)\left(\text{20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}\right)$

  $=1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Initial momentum of the marble $D$ can be calculated by substituting the numerical values in equation $\left(2\right)$ ,

  $p_{Di}=\left(\text{10}\text{.0}\times {\text{10}}^{-3}\text{ kg}\right)\left(\text{10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}\right)$

  $=1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Conclusion:

Initial momentum of both the marbles is same which is equal to $1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$ .

(c)

To determine

The momentum of marble $C$ after collision.

Answer to Problem 70A

  $p_{Cf}=4\times {10}^{-4}\text{ kg}.\text{m/s}$

Explanation of Solution

Given:

Mass of the marble $C$ is $m_C=5.0\text{ g = 5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $C$ is $v_{Ci}=20.0\text{ cm/s = 20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Mass of the marble $D$ is $m_D=10.0\text{ g = 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $D$ is $v_{Di}=10.0\text{ cm/s = 10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Formula used:

Final momentum of the marble $C$ can be calculated using the equation,

  $p_{Cf}=m_C{v}_{Cf}$  $\left(3\right)$

Where,

$m_C$ is the mass of the marble $C$

$v_{Cf}$ is final velocity of the marble $C$

Calculation:

Final momentum of the marble $C$ can be calculated by substituting the numerical values in equation $\left(3\right)$ ,

  $p_{Cf}=\left(\text{5}\text{.0}\times {\text{10}}^{-3}\text{ kg}\right)\left(\text{8}\text{.0}\times {\text{10}}^{-2}\text{ m/s}\right)$

  $=4\times {10}^{-4}\text{ kg}.\text{m/s}$

Conclusion:

The momentum of marble $C$ after collision is $4\times {10}^{-4}\text{ kg}.\text{m/s}$ .

(d)

To determine

The momentum of marble $D$ after collision.

Answer to Problem 70A

  $p_{Df}=1.6\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Explanation of Solution

Given:

Mass of the marble $C$ is $m_C=5.0\text{ g = 5}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $C$ is $v_{Ci}=20.0\text{ cm/s = 20}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Mass of the marble $D$ is $m_D=10.0\text{ g = 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

Initial velocity of the marble $D$ is $v_{Di}=10.0\text{ cm/s = 10}\text{.0}\times {\text{10}}^{-2}\text{ m/s}$

Formula used:

Since the momentum is conserved, initial momentum of the system is equal to the final momentum of that system. In this problem, total momentum of the marbles $C$ and $D$ before collision is equal to the total momentum the same marbles after collision. That is,

  $p_{Ci}+p_{Di}=p_{Cf}+p_{Df}$

  $\Rightarrow p_{Df}=p_{Ci}+p_{Di}-p_{Cf}$  $\left(4\right)$

Where,

$p_{Ci}$ and $p_{Di}$ are the initial momentum of the marbles $C$ and $D$ respectively,

$p_{Cf}$ and $p_{Df}$ are the final momentum of the marbles $C$ and $D$ respectively,

Calculation:

Final momentum of the marble $D$ can be calculated by substituting the numerical values of $p_{Ci}$ , $p_{Di}$ , and $p_{Cf}$ obtained in the part (a), (b) and (c) respectively, in equation $\left(4\right)$ ,

  $p_{Df}=1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}+1\times {\text{10}}^{-3}\text{ kg}.\text{m/s}-4\times {10}^{-4}\text{ kg}.\text{m/s}$

  $\text{= 2}\times {\text{10}}^{-3}\text{ kg}.\text{m/s - 0}\text{.4}\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

  $=1.6\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Conclusion:

The momentum of marble $D$ after collision is $1.6\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$ .

(e)

To determine

The speed of marble $D$ after collision.

Answer to Problem 70A

  $v_{Df}=\text{ 0}\text{.16 m/s}$

Explanation of Solution

Given:

Mass of the marble $D$ is $m_D=10.0\text{ g = 10}\text{.0}\times {\text{10}}^{-3}\text{ kg}$

From part (d), final momentum of the marble $D$ is $p_{Df}=1.6\times {\text{10}}^{-3}\text{ kg}.\text{m/s}$

Formula used:

The speed of marble $D$ after collision can be calculated using the equation,

  $v_{Df}=\frac{p_{Df}}{m_D}$  $\left(5\right)$

Calculation:

The speed of marble $D$ after collision can be calculated by substituting the numerical values in equation $\left(5\right)$ ,

  $v_{Df}=\frac{1.6\times {\text{10}}^{-3}\text{ kg}.\text{m/s}}{\text{10}\text{.0}\times {\text{10}}^{-3}\text{ kg}}=\text{ 0}\text{.16 m/s}$

Conclusion:

The speed of marble $D$ after collision is $\text{0}\text{.16 m/s}$ .