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Question: An 1470 kg car stopped at a traffic light is struck from the rear by a 935 kg car and the two become entangled. If the smaller car was moving at 11.8 m/s before the collision, what is the speed of the entangled mass after the collision?

The situation described above can be solved using the law of conservation of momentum which states that in a closed system the total momentum before and after the event should remain constant if no external forces act on the system. The formula for momentum (p) is given as: p = m*v Where: 'm' is mass 'v' is velocity Before the collision, the total momentum (P1) would be the sum of the momentums of both cars. After the collision, the total momentum (P2) would be the combined mass of both cars times their new, shared velocity. Let's calculate: Before collision: - The 1470 kg car is stationary, thus its momentum is 0 kg*m/s. - The 935 kg car's momentum is its mass times velocity = 935 kg * 11.8 m/s = 11033 kg*m/s. So, P1 = 0 kg*m/s + 11033 kg*m/s = 11033 kg*m/s. After the collision, the two entangled cars move together with a velocity that we will denote as 'V'. So, their combined momentum will be: - (1470 kg + 935 kg) * V = 2405 kg * V. According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision. Therefore, - P1 = P2 - 11033 kg*m/s = 2405 kg * V. To find V, we divide 11033 kg*m/s by 2405 kg: - V = 11033 kg*m/s / 2405 kg = 4.59 m/s (rounded to two decimal places) Therefore, the speed of the entangled mass after the collision is approximately 4.59 m/s.

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