A 3000 kg airplane (plane A) flying north at an altitude of 1600m over the jungl

Question

A 3000 kg airplane (plane A) flying north at an altitude of 1600m over the jungles of the Congo at 75 m/s collided with a 7000 kg cargo plane (plane B) flying 35 degrees north of west with a speed of 100 m/s. As measured by a point on the ground directly below the collision, plane B wreckage was found 1000 m away at an angle of 25 degrees south of west. Plane B broke into two pieces. Rescuers located a 4000 kg piece 1800 m away from the same point at an angle of 22 degrees east of north.
Where should they look for the other piece? Give a distance and direction from the point directly below the collision.

Explanation / Answer

total momentum (let north by j, and east by i). 3000*75*j+7000*100*sin35*j-7000*100*cos35*i =0.63e6*j-0.57e6*i. -------------------- after collision. plane A have velocity of v=1000/t where t=sqrt(2*1600/9.8). so vA=55.3(m/s). vB1=100(m/s). so that we have total momentum. -55.3*3000*cos25*i-55.3*3000*sin25*j+4000*100*cos22*j+4000*100*sin22*i+v*3000 -513*i+0.3e6*j+v*3000=0.63e6*j-0.57e6*i. so v=-190i+110j so magnitude of v=220(m/s). so distance=220*sqrt(2*1600/9.8)=3970(m). angle 30 degrees south to west.

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