Two approaching cars, coasting frictionless on a horizontal track, collide and c
ID: 1455772 • Letter: T
Question
Two approaching cars, coasting frictionless on a horizontal track, collide and couple. Take the + direction to the right. Find the velocity of the coupled cars by taking momenta relative to the earth. Find the velocity of the coupled cars by taking momenta relative to B (initially). Show that the velocity of the CM of the system initially, relative to earth, is +4 cm/s. then find the velocity of the coupled cars by taking the momenta relative to the CM of the system. Why does this result hold for any isolated system, regardless of v's and m's? What is the total K.E. of cars A and B before and after the collision, and the heat lost in the collision, calculated relative to the center of the earth? What is the total K.E. of cars A and B before and after the collision, and the heat lost, calculated relative to the CM of the system?Explanation / Answer
(a) we use the conservation of momentum
ma Va + mb Vb = (ma + mb) Vab
100(10) + 50 (-8) =(100+50) Vab
Vab= 4 cm/s
(b) Vab relative to b is given by Vab/b = Vab/e - Vb/e = 4 - (-8) = 12 cm/s
(c) Vm= (ma Va + mb Vb) / (ma + mb) = [100(10) + 50 (-8)] =(100+50) = 4 cm/s
Va/cm =10-4 = 6 cm/s
Vb/cm=8-(-4)= 12 cm/s
we use the conservation of momentum with respect to centre of mass
ma Va/cm+ mb Vb/cm = (ma + mb) Vab/cm
100(6) + 50 (-12) = (100 + 50) Vab/cm
Vab/cm = 0 . the results remains same regardless beacsue the syetem is isolated and the net force =0
(d) KE before = 1/2 (100)(10)^2 + 1/2 (50) (8)^ = 6600 ergs
KE after = 1/2 (100+ 50)(4)^2= 1200 ergs
Energy lost as heat= 6600- 1200= 5400 ergs
(e)
KE before = 1/2 (100)(6)^2 + 1/2 (50) (12)^ = 5400 ergs
KE after = 1/2 (100+ 50)(0)^2= 0 ergs
Energy lost as heat= 5400-0 = 5400 ergs