# A heavy car A of mass \[2000\,{\text{kg}}\] travelling at \[10\,{\text{m/s}}\] has a head-on collision with a sports car B of mass \[500\,{\text{kg}}\]. If both cars stop dead on colliding, what was the velocity of car B?

Answer

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**Hint:**Use the formula for linear momentum of an object. This formula gives the relation between the linear momentum of an object, mass of the object and velocity of an object. Use the law of conservation of linear momentum for the system of the cars A and B before and after collision. The momentum of the system of the two cars is zero after collision as they stop after collision.

**Formula used:**

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.

**Complete step by step answer:**

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 \[40\,{\text{m/s}}\].

**The negative sign indicates that the direction of motion of car B was opposite to that of the car A.**

**Note:**The students should keep in mind that the two cars dead stop after the head-on collision. Hence, their final velocities are zero after collision. If this concept is not used in the solution, then one will not be able to solve the question and calculate the velocity of the car B before collision with the car A. One can also use the negative value of the velocity of the car A before collision as both the cars are moving in the opposite direction and hence, the velocity of the car B will be positive.

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