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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?

This is a case of a completely inelastic collision where the two cars stick together and move as one mass afterwards. According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision. The formula is: Initial Momentum = Final Momentum Which we can write as: m1*u1 + m2*u2 = (m1 + m2)*v Where: - m1 & m2 are the masses of the two cars, - u1 & u2 are the initial speeds of the cars, and - v is the final speed of the entangled mass. Given: - m1 (the mass of the large car which is stationary) = 1470 kg, - u1 (initial speed of the large car) = 0 m/s (as it's stationary), - m2 (the mass of the small car) = 935 kg, - u2 (initial speed of the small car) = 11.8 m/s. So, we are required to find v (final speed of the combined mass) Substitute these values into the equation, we get: 1470 kg * 0 m/s + 935 kg * 11.8 m/s = (1470 kg + 935 kg)*v 0 + 11033 kg m/s = 2405 kg * v From this, we can solve for v: v = 11033 kg m/s / 2405 kg = 4.59 m/s Therefore, the speed of the entangled mass after the collision is approximately 4.59 m/s.

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