Three cars having masses: m = 1000 kg, m = 2000 kg, and m = 3000 kg, are travell
ID: 1982611 • Letter: T
Question
Three cars having masses: m = 1000 kg, m = 2000 kg, and m = 3000 kg, are travelling due north, due east, and due south, respectively. The northbound car is travelling at 50 m/s, the eastbound car is travelling at 40 m/s and the southbound car is traveling at 30 m/s. All three cars then collide at an intersection. After the collision, the initially northbound and southbound cars stick together and moves in a direction of 30 degree south-of-east while the initially eastbound car moves off in a direction 30 degree north-of-east. Determine the speed of each car after the collision. Determine the fraction of kinetic energy lost due to the collision.Explanation / Answer
a) using momentum conservation ,
1000 x 50j + 2000 X 40i + 3000 x -30j = (1000 + 3000) x (v1cos30i - v1sin30j) + 2000 x (v2cos30i + v2sin30j)
80i - 40j = (4v1 + 2v2)cos30 i + (2v2 - 4v1)sin30 j
4v1 + 2v2 = 80/cos30 = 92.38
-4v1 + 2v2 = -40/sin30 = - 80
ading these two ,
v2 = 3.095 m/s
v1 = 21.55 m/s
speed of northbound and southbound car's is 3.095 m/s
speed of eastbound car is 21.55 m/s
b) fraction that lost = 1000 x 502 /2 + 2000 x 402 /2 + 3000 x302 /2 - ( (1000 + 3000) x 3.0952 /2 +2000 x 21.552/2) / 1000 x 502 /2 + 200 x 402 /2 + 3000 x302 /2
= 4200000 - 483560.55 / 4200000 = 0.8849 fraction of total K.E. or 88.49 % part
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