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The linear momentum \[P\] of an object is given by

\[P = mv\] …… (1)

Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.

We have given that the mass of the car A is \[2000\,{\text{kg}}\] and the velocity of the car A before collision is \[10\,{\text{m/s}}\].

\[{m_A} = 2000\,{\text{kg}}\]

\[\Rightarrow{v_A} = 10\,{\text{m/s}}\]

We have also given that the mass of the car B is \[500\,{\text{kg}}\].

\[{m_B} = 500\,{\text{kg}}\]

We are asked to calculate the velocity of car B before collision.The two cars after the head-on collision stops. Hence, the final velocities of both the cars A and B after collision are zero.Hence, the linear momentum of both the cars A and B after the collision is zero.

According to the law of conservation of linear momentum, the sum of the linear momenta of the car A and B before collision is equal to the linear momenta of the cars A and B after collision.

\[{m_A}{v_A} + {m_B}{v_B} = 0\]

Rearrange the above equation for velocity \[{v_B}\] of the car B before collision.

\[{v_B} = - \dfrac{{{m_A}{v_A}}}{{{m_B}}}\]

Substitute \[500\,{\text{kg}}\] for \[{m_A}\], \[500\,{\text{kg}}\] for \[{m_B}\] and \[10\,{\text{m/s}}\] for \[{v_A}\] in the above equation.

\[{v_B} = - \dfrac{{\left( {2000\,{\text{kg}}} \right)\left( {10\,{\text{m/s}}} \right)}}{{500\,{\text{kg}}}}\]

\[ \therefore {v_B} = - 40\,{\text{m/s}}\]

Hence, the velocity of car B before collision was \[40\,{\text{m/s}}\].

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