No, not hard, conservation of momentum
Initial north momentum =
(14000 N/9.81 m/s^2) (40000m/3600 s)
Initial east momentum =
(7000/9.81)(62000/3600)
Final north and east momentums are the same
total mass = m = 21000/9.81
m v cos T = initial east momentum
m v sin T = initial north momentum
solve for T, the angle up from east axis
then get v, the speed
Vnorth = v sin T
Veast = v cos T
Final ke = (1/2) m v^2
initial ke = (1/2)m1 initial speed1 ^2 + (1/2) m2 initial speed2^2
A 14000 N automobile travels at a speed of 40 km/h northward along a street, and a 7000 N sports car travels at a speed of 62 km/h eastward along an intersecting street.(a) If neither driver brakes and the cars collide at the intersection and lock bumpers, what will the velocity of the cars be immediately after the collision?
(b) What percentage of the initial kinetic energy will be lost in the collision?
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