Question

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are 1200 kgand 1800 kg; the magnitude of the impulse on each part from the bolts is 300 N. With what relative speed do the two parts separate because of the detonation?

Short answer

The relative speed with which two parts separate because of the detonation is 0417 m/s .

Step-by-step solution

Listing the given quantities

Masses of the parts are,$m_1=1200kg,m_2=1800kg$

The magnitude of the impulse on each part of from the bolts is, J = 300 N .s .

Understanding the terms impulse, force, kinetic energy

We can find the speed of each part after separation from the impulse on them from the bolt using the relation between impulse, speed, and mass. Then using the concept of relativity, we can find the relative speed with which two parts separate because of the detonation.

Calculations  

Impulse on the first part can be calculated as,

$$\begin{gathered} J=P \\ J=m_1\left(v_{1f}-v_{1i}\right) \end{gathered}$$

Here $v_{1i}$is the initial velocity of part 1.

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

$$\begin{gathered} J=m_1\left(v_{1f}-0\right) \\ {v}_{1f}=\frac{J}{m_1} \end{gathered}$$

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

$$\begin{gathered} v_{1f}=\frac{300N.s}{1200kg} \\ =0.25.\left(\frac{1N.s}{1kg}\times \frac{1kg.m/s^2}{1N}\right) \\ =0.25m/s \end{gathered}$$

Impulse on the second part will be,

$$\begin{gathered} J=P \\ J=m_2\left(v_{2f}-v_{2i}\right) \end{gathered}$$

Here $v_{2i}$is the initial velocity of part 2.

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

$$\begin{gathered} J=m_2\left(v_{2f}-0\right) \\ {v}_{2f}=\frac{J}{m_2} \end{gathered}$$

Since the impulse onthesecond part is intheopposite direction to that onthefirst part.

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

$$\begin{gathered} v_{2f}=\frac{-300N.s}{1800kg} \\ =-0.167.\left(\frac{1N.s}{1kg}\times \frac{1kg.m/s^2}{1N}\right) \\ =-0.167m/s \end{gathered}$$

Thus, the relative speed of the parts is,

$v=v_{1f}-v_{2f}$

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

$$\begin{gathered} v=0.25-\left(-0.167\right) \\ =0.417m/s \end{gathered}$$

Therefore, the relative speed with which two parts separate because of the detonation is equal to$0.417\frac{m}{s}$ .