Question

A bullet of mass 3.1 g strikes a ballistic pendulum of mass 2.7 kg. The center of mass of the pendulum rises a vertical distance of 6.1 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

intial speed of bullet - umis mass of bullet = 0.0031 +2.7. 2.70031 final momentum of bullet-pendulum System & PLEJ = 207003 x V = 2.7003 V v PCF) = PO 2-700 3V = 0.0031 u -Yu 381106 V - ential speed of pendulum ao mass = 207 kg ential momentum of bullet-pendulen PO 0.0031 Ut 9,7x0 = 0000311 Final speed of bullet & final speed total mass of bullet and pendulum = 27+0003 =9070031 final momentum of bullet -pendulum System PLA - 90 7003 Xu. 39.900 3V PCE) = Pen 9.7003V = 0.003/4 u : 871.06U - 0 by energy conservation KEit RE; - KEGA PER loos) ( .7003) V2 H = 0+ (2.7003)(99B1(0261) V = 3.46 mis U = 301 3.44 mis