Chapter 7, Problem 13P

(a)

To determine

The speed and direction of each section of rocket.

Answer to Problem 13P

  $v_1=6900\text{ m/s}$ away from the earth.

  $v_2=4700\text{ m/s}$ away from the earth.

Explanation of Solution

Given:

Mass of rocket, $M_R=975\text{ kg}$

Speed of rocket with respect to earth, $v_R=5.8\times {10}^3\text{m/s}$

Mass of each section, $M_e=\frac{M_R}{2}$

Velocity after explosion is $2.2\times {10}^3\text{m}/\text{s}$

Formula used:

  $P_i=P_f$

Where, $P_i$ = initial momentum

  $P_f$ = final momentum

Calculation:

The relative velocity is:

  $v_1-v_2=2.2\times {10}^3\text{m/s (1)}$

Applying principle of conservation,

  $\begin{array}{l}{M}_R=975\text{ kg}\\ v_R=5.8\times {10}^3\text{m/s}\\ P_i=P_f\\ M_{R}v=M_e{v}_1+M_e{v}_2\text{ (2)}\\ M_{R}v=\frac{M_R}{2}\left(v_1+v_2\right)\\ 2v=v_1+v_2\\ v_1+v_2=2\left(5.8\times {10}^3\right)\\ v_1+v_2=11600\text{ (3)}\end{array}$

From equation (1) and (3),

  $v_1=6900\text{ m/s}$ , away from the earth

  $v_2=4700\text{ m/s}$ , away from the earth

(b)

To determine

The energy supplied by explosion.

Answer to Problem 13P

The energy supplied by the explosion is $5.89\times {10}^8\text{ J}$ .

Explanation of Solution

Given:

  $v_1=6900\text{ m/s}$

  $v_2=4700\text{ m/s}$

  $v=5.8\times {10}^3\text{ m/s}$

Formula used:

  $\Delta E=E_f-E_i$

Where, $E_i$ = initial energy

  $E_f$ = final energy

Calculation:

  $\begin{array}{l}\Delta E=E_f-E_i\\ \Delta E=\left(\frac{M_e{v}_1^2}{2}+\frac{M_e{v}_2^2}{2}\right)-\left(\frac{M_R{v}^2}{2}\right)\\ \Delta E=\frac{M}{2}\left[\left(\frac{v_1^2}{2}+\frac{v_2^2}{2}\right)-v^2\right]\\ \Delta E=\frac{975}{2}\left[\left(\frac{{\left(6900\right)}^2}{2}+\frac{{\left(4700\right)}^2}{2}\right)-{\left(5.8\times {10}^3\right)}^2\right]\\ \Delta E=5.89\times {10}^8\text{J}\end{array}$