91Ó°ÊÓ

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 .

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.

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,

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

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,

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

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$0.417\frac{m}{s}$ .