Two asteroids strike head-on: before the collision, asteroid A\(\left( {{m_{\rm{A}}} = 7.5 \times {{10}^{12}}\;{\rm{kg}}} \right)\)has velocity 3.3 km/s and asteroid B\(\left( {{m_{\rm{B}}} = 1.45 \times {{10}^{13}}\;{\rm{kg}}} \right)\)has velocity 1.4 km/s in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?

Short Answer

Expert verified

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

Step by step solution

01

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.

02

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 \({v_{\rm{B}}} = - 1.4\;{\rm{km/s}}\).

03

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.

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