Chapter 39, Problem 70AP

(a)

To determine

The speed of the composite object.

Answer to Problem 70AP

The speed of the composite object after collision is $\underset{\_}{0.467c}$.

Explanation of Solution

Apply law of conservation of energy. Energy before collision is equal to the energy after collision.

Write the expression for the energy of colliding object.

  $E_c=\frac{m_c{c}^2}{\sqrt{1-\left(\frac{v_c{}^2}{c^2}\right)}}$                                                                                                         (I)

Here, $E_c$ is the energy of the colliding object before collision, $m_c$ is the mass of the colliding object, $c$ is the speed of light, and $v_c$ is the speed of colliding object.

Write the expression for the energy of the stationary object.

  $E_s=\frac{m_s{c}^2}{\sqrt{1-\left(\frac{v_s{}^2}{c^2}\right)}}$                                                                                                        (II)

Here, $E_s$ is the energy of the stationary object, $m_s$ is the mass of stationary object, and $v_s$ is the speed of stationary object.

Write the expression for the energy of the composite object.

  $E=\frac{M{c}^2}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$                                                                                                        (III)

Here, $E$ is the energy of the composite object, $M$ is the mass of the composite object, and $v$ is the speed of the composite object.

The sum of energy of colliding object and the stationary object is equal to the energy of the composite object.

  $E_s+E_c=E$                                                                                                             (IV)

Use expressions (I), (II), and (III) in (IV).

  $\frac{m_c{c}^2}{\sqrt{1-\left(\frac{v_c{}^2}{c^2}\right)}}+\frac{m_s{c}^2}{\sqrt{1-\left(\frac{v_s{}^2}{c^2}\right)}}=\frac{M{c}^2}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$                                                                        (V)

Write the expression for the momentum of the colliding particle.

  $P_c=\frac{m_c{v}_c}{\sqrt{1-\left(\frac{v_c^2}{c^2}\right)}}$                                                                                                      (VI)

Here, $P_c$ is the momentum of colliding particle.

Write the expression for the momentum of the stationary object.

  $P_s=\frac{m_s{v}_s}{\sqrt{1-\left(\frac{v_s^2}{c^2}\right)}}$                                                                                                    (VII)

Here, $P_s$ is the momentum of stationary object.

Write the expression for the momentum of the composite object.

  $P=\frac{Mv}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$                                                                                                     (VIII)

Here, $P$ is the momentum of the composite object.

Apply law of conservation of momentum. The sum of momentum of the stationary object and the colliding particle is equal to the momentum of composite object.

  $P_s+P_c=P$                                                                                                           (IX)

Use expressions (VI), (VII), and (VIII) in (IX).

  $\frac{m_c{v}_c}{\sqrt{1-\left(\frac{v_c^2}{c^2}\right)}}+\frac{m_s{v}_s}{\sqrt{1-\left(\frac{v_s^2}{c^2}\right)}}=\frac{Mv}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$                                                                  (X)

Conclusion:

Substitute $1400\text{kg}$ for $m_s$, $900\text{kg}$ for $m_c$, $0.850c$ for $v_c$, and $0$ for $v_s$ in equation (V) to find $M$.

  $\begin{array}{c}\frac{\left(1400\text{kg}\right)c^2}{\sqrt{1-\left(0\right)}}+\frac{\left(900\text{kg}\right)c^2}{\sqrt{1-{\left(\frac{0.850c}{c}\right)}^2}}=\frac{M{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}\\ M=\left(3108\text{kg}\right)\sqrt{1-\frac{v^2}{c^2}}\end{array}$                                               (XI)

Substitute $1400\text{kg}$ for $m_s$, $900\text{kg}$ for $m_c$, $0.850c$ for $v_c$, and $0$ for $v_s$ in equation (X).

  $\begin{array}{c}\frac{\left(900\text{kg}\right)\left(0.850c\right)}{\sqrt{1-\left(\frac{{\left(0.850c\right)}^2}{c^2}\right)}}+0=\frac{Mv}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}\\ \left(1452\text{kg}\right)\sqrt{1-\frac{v^2}{c^2}}=\frac{Mv}{c}\end{array}$                                                                  (XII)

Divide expression (XII) by (XI).

  $\begin{array}{c}\frac{\left(1452\text{kg}\right)\sqrt{1-\frac{v^2}{c^2}}}{\left(3108\text{kg}\right)\sqrt{1-\frac{v^2}{c^2}}}=\frac{\frac{Mv}{c}}{M}\\ \frac{v}{c}=\frac{1452\text{kg}}{3108\text{kg}}\\ v=0.467c\end{array}$

Therefore, the speed of the composite object after collision is $\underset{\_}{0.467c}$.

(b)

To determine

The mass of the composite object.

Answer to Problem 70AP

The mass of the composite object is $\underset{\_}{2.75\times {10}^3\text{kg}}$.

Explanation of Solution

In part (a), equation for the mass of the composite object is found.

  $M=\left(3108\text{kg}\right)\sqrt{1-\frac{v^2}{c^2}}$

Conclusion:

Substitute $0.467c$ for $v$ in the above equation to find $M$.

  $\begin{array}{c}M=\left(3108\text{kg}\right)\sqrt{1-\frac{{\left(0.467c\right)}^2}{c^2}}\\ =2.75\times {10}^3\text{kg}\end{array}$

Therefore, the mass of the composite object is $\underset{\_}{2.75\times {10}^3\text{kg}}$.