A bullet of mass 3.6 g strikes a ballistic pendulum of mass 3.0 kg. The center o

Question

A bullet of mass 3.6 g strikes a ballistic pendulum of mass 3.0 kg. The center of mass of the pendulum rises a vertical distance of 16 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed. ?

Explanation / Answer

Initial speed of pendulum = 0; mass of pendulum = 2 kg Hence, initial momentum of bullet-pendulum system p(i) = 0.006*u + 2*0 = (0.006u) kgm/s Just after collision Final speed of bullet = final speed of pendulum = v m/s Total mass of bullet and pendulum = 2 + 0.006 = 2.006 kg Hence, final momentum of bullet-pendulum system p(f) = 2.006*v = (2.006v) kgm/s The string attached to the pendulum will not have the time to react and hence there is no external force on the bullet-pendulum system. Thus, apply conservation of linear momentum. p(f) = p(i) => 2.006v = 0.006u => u = 334.33 v ------------- (1) Now apply principle of conservation of mechanical energy. KEi + PEi = KEf + PEf => (0.5)(2.006)(v^2) + 0 = 0 + (2.006)(9.81)(0.1) => v = sqrt(1.962) = 1.40 m/s Substitute this value into equation (1), u = 334.33*1.40 = 468.3 m/s Hence, the bullet's initial speed is 468.3 m/s

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