Question

A firearm with a 200 mm-long barrel shoots a 20 g bullet with a muzzle velocity of 1067 m/s into a 1 kg block of a ballistic pendulum. How high in m will the block swing? Assume light wires.

Answer 1

Step-by-step explanation:

The KE of the bullet is converted to PE of the block and bullet combo

Initial KE = 1/2 * .020 * 1067^2 = 11 384.89 J

this is converted to PE of the block and bullet

PE = mgh

11 384.89 J = ( 1 + .020 kg ) (9.81 m/s^2) * h

h =~ 1138 m ( that seems wrong ....but I re-checked the computations)

Someone tell me if I am incorrect ..........