A bullet of mass 10 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 12 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
p = mv, pi = pf, KEi + PEi = KEf + PEf
The Attempt at a Solution
m1 = 10 g = 0.01 kg, m2 = 2 kg, total mass M = 2.01 kg, h = 12 cm = 0.12 m
p1 = m1v1 = 0.01v1
p2 = Mv2 = 2.01v2
conservation of linear momentum pi = pf: 0.01v1 = 2.01v2
i would never have figured out how to use this otherwise -- i just don't know how people figure out to use equations from previous chapters (if someone has advice, please tell me), but apparently i apply conservation of mechanical energy:
KEi + PEi = KEf + PEf
0.5(2.01)vf^2 + 0 = 0 = (2.01)(9.8)(0.12)
1.005vf^2 = 2.36
vf = 1.53 m/s
pi = pf = 0.01v1 = 2.01v2
v1i = 201vf
v1i = 201(1.53) = 307.5 m/s
this is the correct answer, but i don't understand how to arrive at it. the part written in red is what confuses me. i would think initial KE means to use only the mass of the bullet, not the total, and also that when we figured out the velocity there, that would be the INITIAL velocity of the bullet, not the final velocity of bullet and pendulum.