Question

Short answer

The velocity of the new asteroid is \(v' = 0.20\;{\rm{km/s}}\).

Step-by-step solution

Understanding the conservation of momentum

In this case, an inelastic collision takes place between two asteroids. Here, consider the direction of asteroid A as positive and the direction of asteroid B as negative.

Given data

Given data:

The mass of one of the asteroids is\({m_{\rm{A}}} = 7.5 \times {10^{12}}\;{\rm{kg}}\).

The mass of the other asteroid is\({m_{\rm{B}}} = 1.45 \times {10^{13}}\;{\rm{kg}}\).

The speed of asteroid A is\({v_{\rm{A}}} = 3.3\;{\rm{km/s}}\).

The speed of asteroid B is \({v_{\rm{B}}} = - 1.4\;{\rm{km/s}}\).

Calculate the speed of the new asteroids

The law of conservation of momentum is given by the following equation:

\(\begin{array}{c}P = P'\\{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)v'\\v' = \frac{{{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}}}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)}}\end{array}\)

Here,\(v'\)is the speed of the new asteroid.

On plugging the values in the above equation,

\(\begin{array}{l}v' = \left[ {\frac{{\left( {7.5 \times {{10}^{12}}\;{\rm{kg}}} \right)\left( {3.3\;{\rm{km/s}}} \right) + \left( {1.45 \times {{10}^{13}}\;{\rm{kg}}} \right)\left( { - 1.4\;{\rm{km/s}}} \right)}}{{\left( {7.5 \times {{10}^{12}}\;{\rm{kg}} + 1.45 \times {{10}^{13}}\;{\rm{kg}}} \right)}}} \right]\\v' = 0.20\;{\rm{km/s}}\end{array}\).

Thus, \(v' = 0.20\;{\rm{km/s}}\) is the required speed, and the direction is towards the initial direction of asteroid A.