A shell is shot with an initial velocity Vq of 13 m/s, at an angle of = 60degree

Question

A shell is shot with an initial velocity Vq of 13 m/s, at an angle of = 60degree with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible? Where would the shell (or the shell's com) have landed had the shell not exploded? The flight of the com of an object cannot be affected by internal forces within the object, such as due to an explosion. Draw an x axis. You know where on the axis the com "lands" and where one piece lands, and you want where the second piece lands.

Explanation / Answer

1. lets find the point of explosion x: x=vcos60 *t y: y=vsin60t-gt^2/2 vy=vsin60-gt --> at the maximum point vy=0 --> vsin60=gt --> t=1.15 sec --> y=6.46m --> x=7.475 m 2. now after the exlosion, use the conservation of momentum in x direction x: mv=m1u1+m2u2 v=vcos60=6.5 u1 is zero m2=0.5*m1 --> u2=6.5/0.5=13 m/sec 3. now lets calculate how much time will it take to fall to the ground y: y=6.46 - gt^2/2 =0 --> t=1.15 sec by this time the fragment will travel 1.15*13=14.93 meters ---> total distance is: 7.475+14.93=22.4 meters

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