Two cars are involved in a head-on collision and come to a complete stop. Car 1 has a mass of 750kg, car 2 has a mass of 500kg Car 2 was travelling at 15m/s what was the velocity of car 1 before the impact? square

To solve this problem, we can use the principle of conservation of momentum. The total momentum of the system before the collision must be equal to the total momentum of the system after the collision.<br /><br />Let's denote the velocity of car 1 before the impact as $v_1$.<br /><br />The momentum of car 1 before the collision is given by:<br />$P_1 = m_1 \cdot v_1$<br /><br />The momentum of car 2 before the collision is given by:<br />$P_2 = m_2 \cdot v_2$<br /><br />The total momentum of the system before the collision is:<br />$P_{total} = P_1 + P_2$<br /><br />After the collision, both cars come to a complete stop, so their velocities are 0. Therefore, the total momentum of the system after the collision is 0.<br /><br />Using the conservation of momentum, we can write:<br />$P_{total} = 0$<br /><br />Substituting the expressions for $P_1$ and $P_2$, we get:<br />$m_1 \cdot v_1 + m_2 \cdot v_2 = 0$<br /><br />Given that the mass of car 1 is 750 kg and the mass of car 2 is 500 kg, and the velocity of car 2 is 15 m/s, we can substitute these values into the equation:<br />$750 \cdot v_1 + 500 \cdot 15 = 0$<br /><br />Simplifying the equation, we get:<br />$750 \cdot v_1 + 7500 = 0$<br /><br />Solving for $v_1$, we get:<br />$750 \cdot v_1 = -7500$<br />$v_1 = -10 \, \text{m/s}$<br /><br />Therefore, the velocity of car 1 before the impact was $-10 \, \text{m/s}$.