You have constructed a simple catapult which basically consists of a 4 meter lon

Question

You have constructed a simple catapult which basically consists of a 4 meter long rod of negligible mass attached to a fulcrum point offset from the center of mass by 1.5 meters. So, in reference to the figure below, the labeled distances are a = 3.5 m and b = 0.5 m. You decide to launch a m2= 0.100 kg mass from the long end of the rod by dropping a m1= 20.0 kg ball of putty from a height of h = 5 m onto the short side of the rod. The putty sticks to the rod when it hits it, causing the rod to rotate about the fulcrum until the short end of the rod hits the ground. At this point the small mass is launched into the air at an angle of 45 degrees. You may model the mass of putty and the small block as point particles. How far away from the launch point does the mass land when it hits the ground? Give your answer in meters.

Explanation / Answer

let velocity of m2 = u.......so vertical velocity = usin45...........doing angular momentum conservation: V(of m1 at the bottom) = (2*9.8*5)^0.5 = 9.89 m/s..........20*9.89*0.5 = 0.1*usin45 *3.5.........u = 40 m/s............R = u^2sin2t/2g = 1600*1 / 19.6 = 81.63 m............answer...........

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