Chapter 8, Problem 92P

(a)

To determine

The velocity of the center of mass.

Answer to Problem 92P

The speed of center of mass is $-3\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of first block is $m_3=3.0\text{-kg}$ .

The mass of second block is $m_1=1.0\text{-kg}$ .

The speed of first block is $v_3=5.0\text{m}/\text{s}$ .

The speed of second block is $v_1=3.0\text{m}/\text{s}$ .

Formula used:

The expression for velocity of center of mass is given by,

  $V_{\text{cm}}=\frac{m_3{v}_3+m_1{v}_1}{m_3+m_1}$

Calculation:

The velocity of center of mass is calculated as,

  $\begin{array}{c}{V}_{\text{cm}}=\frac{m_3{v}_3+m_1{v}_1}{m_3+m_1}\\ =\frac{\left(3.0\text{-kg}\right)\left(-5.0\text{m}/\text{s}\right)+\left(1.0\text{-kg}\right)\left(3.0\text{m}/\text{s}\right)}{\left(3.0\text{-kg}\right)+\left(1.0\text{-kg}\right)}\\ =-3\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of center of mass is $-3\text{m}/\text{s}$ .

(b)

To determine

The velocity of each block in center of mass reference frame.

Answer to Problem 92P

The velocity of first block is $-2\text{m}/\text{s}$ and second block is $6\text{m}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for velocity of first block is given by,

  $u_3=v_3-v_{\text{cm}}$

The expression for velocity of second block is given by,

  $u_1=v_1-v_{\text{cm}}$

Calculation:

The expression for velocity of first block is calculated as,

  $\begin{array}{c}{u}_3=v_3-v_{\text{cm}}\\ =\left[-5.0\text{m}/\text{s}-\left(-3\text{m}/\text{s}\right)\right]\\ =-2\text{m}/\text{s}\end{array}$

The expression for velocity of second block is calculated as,

  $\begin{array}{c}{u}_1=v_1-v_{\text{cm}}\\ =\left[3\text{m}/\text{s}-\left(-3\text{m}/\text{s}\right)\right]\\ =6\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of first block is $-2\text{m}/\text{s}$ and second block is $6\text{m}/\text{s}$ .

(c)

To determine

The velocity of each block in center of mass reference frame after collision.

Answer to Problem 92P

The velocity of first block is $2\text{m}/\text{s}$ and second block is $-6\text{m}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for velocity of first block is given by,

  $u_3{}^{\prime }=-u_3$

The expression for velocity of second block is given by,

  $u_1{}^{\prime }=-u_1$

Calculation:

The expression for velocity of first block is calculated as,

  $\begin{array}{c}{u}_3{}^{\prime }=-u_3\\ =-\left(-2\text{m}/\text{s}\right)\\ =2\text{m}/\text{s}\end{array}$

The expression for velocity of second block is calculated as,

  $\begin{array}{c}{u}_1{}^{\prime }=-u_1\\ =-6\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of first block is $2\text{m}/\text{s}$ and second block is $-6\text{m}/\text{s}$ .

(d)

To determine

The velocity in original frame.

Answer to Problem 92P

The velocity of first block is $-9\text{m}/\text{s}$ and second block is $-1\text{m}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for velocity of first block is given by,

  $v_3{}^{\prime }=u_3{}^{\prime }+v_{\text{cm}}$

The expression for velocity of second block is given by,

  $v_1{}^{\prime }=u_1{}^{\prime }+v_{\text{cm}}$

Calculation:

The expression for velocity of first block is calculated as,

  $\begin{array}{c}{v}_3{}^{\prime }=u_3{}^{\prime }+v_{\text{cm}}\\ =\left(-6\text{m}/\text{s}\right)+\left(-3\text{m}/\text{s}\right)\\ =-9\text{m}/\text{s}\end{array}$

The expression for velocity of second block is calculated as,

  $\begin{array}{c}{v}_1{}^{\prime }=u_1{}^{\prime }+v_{\text{cm}}\\ =\left(2\text{m}/\text{s}\right)+\left(-3\text{m}/\text{s}\right)\\ =-1\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of first block is $-9\text{m}/\text{s}$ and second block is $-1\text{m}/\text{s}$ .

(e)

To determine

The initial and final kinetic energies.

Answer to Problem 92P

The initial kinetic energy is $42\text{J}$ and final kinetic energy is $42\text{J}$ . When the energies are compared it is found that both the initial and final energy in original frame of reference are same.

Explanation of Solution

Formula used:

The expression for initial kinetic energy is given by,

  $K_{\text{i}}=\frac{1}{2}{m}_3{v}_3^2+\frac{1}{2}{m}_1{v}_1^2$

The expression for final kinetic energy is given by,

  $K_{\text{f}}=\frac{1}{2}{m}_3{v}_3^2{}^{\prime }+\frac{1}{2}{m}_1{v}_1^2{}^{\prime }$

Calculation:

The initial kinetic energy is calculated as,

  $\begin{array}{c}{K}_{\text{i}}=\frac{1}{2}{m}_3{v}_3^2+\frac{1}{2}{m}_1{v}_1^2\\ =\frac{1}{2}\left(3.0\text{-kg}\right){\left(-5.0\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(1.0\text{-kg}\right){\left(3.0\text{m}/\text{s}\right)}^2\\ =42\text{J}\end{array}$

The final kinetic energy is calculated as,

  $\begin{array}{c}{K}_{\text{f}}=\frac{1}{2}{m}_3{v}_3^2{}^{\prime }+\frac{1}{2}{m}_1{v}_1^2{}^{\prime }\\ =\frac{1}{2}\left(3.0\text{-kg}\right){\left(-1.0\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(1.0\text{-kg}\right){\left(-9.0\text{m}/\text{s}\right)}^2\\ =42\text{J}\end{array}$

Conclusion:

Therefore, the initial kinetic energy is $42\text{J}$ and final kinetic energy is $42\text{J}$ . When the energies are compared it is found that both the initial and final energy in original frame of reference are same.