Two 2.3 kg balls are attached to the ends of a thin rod of negligible mass, 65 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 44 gm wad of wet putty drops onto one of the balls with a speed of 3.7 m/sec and sticks to it.

Question

Two 2.3 kg balls are attached to the ends of a thin rod of negligible mass, 65 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 44 gm wad of wet putty drops onto one of the balls with a speed of 3.7 m/sec and sticks to it. 
A) What is the ratio of the kinetic energy of the entire system just after the collision to just before the collision? 
B)Through what angle will the system rotate until it momentarily stops?

Anonymous · Accepted Answer

a) m=putty M=balls Li=Lf mvr=Iw w=mvr/I Ki=1/2*m*v^2 Kf=Iw^2=1/2*m^2*v^2*r^2/I Things that cancel: 1/2 m v^2 Left over: m*r^2/I << THIS is the answer. Plug in I I=M*(L/2)+(M+m)*(L/2)=L/2*(2M+m)