Question

The last stage of a rocket, which is traveling at a speed of 7600 m/s , consists of two parts that are clamped together: a rocket case with a mass of 290.0 kgand a payload capsule with a mass of 150.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0 m/s. What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.

Short answer

a) The speed of the rocket case after separation is 7290 m/s .

b) Speed of the payload after separation is 8200 m/s.

c) The total kinetic energy of the two parts before separation is $1.271\times {10}^{10}\mathrm{J}$.

d) The total kinetic energy of the two parts after separation is $1.275\times {10}^{10}\mathrm{J}$.

d) The difference between total kinetic energy before and after the collision is due to the initially stored energy in the spring.

Step-by-step solution

Listing the given quantities

Mass of the rocket case is,$m_1=290.0\mathrm{kg}$.

Mass of the payload capsule is,$m_2=150\mathrm{kg}$.

The speed of the last stage of the rocket is,$v=910.0\frac{\mathrm{m}}{s}$.

The relative speed of the two parts of the rocket is, $v_r=910.0\frac{\mathrm{m}}{s}$.

Understanding the concept of the law of conservation energy

We can find the Speed of the rocket case after separation and the speed of the payload after separation using the law of conservation of energy. Substituting these values in the formula for K.E, we can find the total K.E of the system before and after separation.

Formula:

The momentum of the system before separation = Momentum of the system after separation.

(a) Calculation of speed of rocket case after separation

The total momentum of the system is conserved if no external force is acting on it.

According to the law of conservation of momentum,

The momentum of the system before separation = Momentum of the system after separation

$\left(m_1+m_2\right)v=m_1{v}_1+m_2{v}_2$ (1)

Since the mass of the payload is lesser than that of the case, it has a greater speed than that of the case. Hence,

$v_2=v_1+v_r$ (2)

Inserting it into the momentum equation (1), we get,

$$\begin{gathered} \left(m_1+m_2\right)v=m_1{v}_1+m_2\left(v_1+v_r\right) \\ \left(m_1+m_2\right)v=m_1{v}_1+m_2{v}_1+m_2{v}_r \\ {m}_1{v}_1+m_2{v}_1=\left(m_1+m_2\right)v=m_2{v}_r \end{gathered}$$

Solving further, and we get,

$$\begin{gathered} \left(m_1+m_2\right)v_1=(m_1+m_2)v-m_2{v}_r \\ {v}_1=\frac{(m_1+m_2)v-m_2{v}_r}{(m_1+m_2)} \\ {v}_1=\frac{m_2{v}_r}{(m_1+m_2)} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v_1=7600\mathrm{m}/\mathrm{s}-\frac{\left(150\mathrm{kg}\right)\left(910\mathrm{m}/\mathrm{s}\right)}{290\mathrm{kg}+150\mathrm{kg}} \\ =7290\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the rocket case after separation is 7290 m/s .

(b) Calculation of speed of payload after separation

Substitute the values in equation 2, and we get,

$$\begin{gathered} v_2=7290\mathrm{m}/\mathrm{s}+910\mathrm{m}/\mathrm{s} \\ =8200\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the payload after separation is 8200 m/s.

(c) Calculation of total kinetic energy of the two parts before separation

The total kinetic energy of the two parts before separation can be calculated as,

$K_i=\frac{1}{2}\left(m_1+m_2\right)v^2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} K_j=\frac{1}{2}\left(290\mathrm{kg}+150\mathrm{kg}\right){\left(7600\mathrm{m}/\mathrm{s}\right)}^2 \\ =1.271\times {10}^{10}.\left(1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.271\times {10}^{10}\mathrm{J} \end{gathered}$$

Therefore, the total kinetic energy of the two parts before separation is $1.271\times {10}^{10}\mathrm{J}$.

(d) Calculation of total kinetic energy of the two parts after separation

The total kinetic energy of the two parts after separation can be calculated as,

$K_f=\frac{1}{2}\left(m_1{v}_1^2+m_2{v}_2^2\right)$

Substitute the values in the above expression, and we get,

$$\begin{gathered} K_f=\frac{1}{2}\left(290\mathrm{kg}\right){\left(7290\mathrm{m}/\mathrm{s}\right)}^2+\left(150\mathrm{kg}\right){\left(8200\mathrm{m}/\mathrm{s}\right)}^2 \\ =1.275\times {10}^{10}.\left(1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.275\mathrm{x}{10}^{10}\mathrm{J} \end{gathered}$$

Therefore, the total kinetic energy of the two parts before separation is$1.275\times {10}^{10}\mathrm{J}$ .

(e) Calculations of difference between total kinetic energy before and after collision

The difference between the kinetic energy can be calculated as,

$$\begin{gathered} =1.275\times {10}^{10}\mathrm{J}-1.271\times {10}^{10}\mathrm{J} \\ =4\times {10}^7\mathrm{J} \end{gathered}$$

Thus, total kinetic energy after separation is greater than that of before separation. This comes from the initially stored energy in the spring.