Question

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.5 m/s and that of the trailing car is 5.9 m/s. Hint b. Assuming that the total mass of leading bumper car is 40% greater than that of the trailing bumper car, what are their final speeds? The speed of the leading bumper car (now 40% more massive) is m/s, and the speed of the trailing bumper car after collision is m/s. License Question 2. Points possible: 15 This is attempt 1 of 2.

Answer 1

Ug = 5.9m/s v z 5. 5m/s mi mia = 400kg A Mass of trailing care a me = 400 kg Mass of the leading car = m = (400 81.4) : = 560 kg. If relocity offer collision- taailing car z ut , leading car = a Then from momentum conservation. 1. maig, i mere = m ut + meri ir From kinetic energy conservation at ang mt ut + h mive = 1 / me of them, vt ③ Solving we get can get a ./-me - 12 NEE! 2 Aj 2 mi Vi T mttma - I met me | 22560 5.9 + x 5.5 1400-560 214 - 4.00+560) 7067 642 40o , . La + 6999 = 5.44 mls. -0.98 Ivelocity of trailing car car = 544 mis - ...at ther collision a l m - me ( 2 X 400 L 400-550 MB 1 400+560 1 \ 400 + 560) § 4.92+.92 -5.84 mis. velocity of the leading care . 584 mis ofter collision