A 40 kg artillery shell was fired from a cannon with an initial speed of 84 m/s

Question

A 40 kg artillery shell was fired from a cannon with an initial speed of 84 m/s at a 45° angle above the ground. Unfortunately, the shell was poorly made and exploded into two pieces 4 s after it was launched. One piece was five times heavier than the other. The heavier piece landed 600 m from the cannon. How far from the cannon (in m) did the lighter piece land? HINT #1: the range of an artillery shell is given by the equation, where v0 is the initial speed of the shell and q is the launch angle. HINT #2: treat this problem in terms of what happens to the center of mass of the shell.

Explanation / Answer

intial velocity u = 84 m/s angle ? = 45 degrees Range of shell ( without considering explosion) = u^2 sin(2?) / g = 84^2 sin(2*45) / 9.81 = 719.2661 m even after explosion the centre of mass of the two pieces follows the same trajectory as the unexploded mass would have therefore postion of centre of mass after landing will be 719.2661 m let heavier mass is 5m kg at x1 metres an lighter mass is m kg at x2 metres then centre of mass will be at (5m*x1 + m*x2)/ (5m + m ) metres that is (5/6)x1 + (1/6)x2 meters = 719.2661 metres (as stated above) now x1 = 600 m given (5/6)*600 + (1/6)*x2 = 719.2661 500 + (x2 /6 ) = 719.2661 x2 = 219.2661 * 6 position of lighter piece = 1315.597 metres

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