Question

A 5 gram bullet is shot through a 1 kg wood block suspended on a 2m long massless rope. The center of mass rises a distance of .38 cm. What's the final speed of the bullet as it emerges from the block if its initial speed is 450 m/s?

Answer 1

Initial kinetic energy= (1/2) m v^2) = final kinetic energy + change in block potential energy, Mgh

Answer 2

This is a conservation of energy problem:\[E _{total}=E _{initial}=E _{final}\]The initial energy is the kinetic energy of the bullet; and the final energy is the kinetic energy of the bullet after it passes through the wood plus the potential energy the wood gains:\[\frac{ 1 }{ 2 }m _{b}v _{bi}^{2}=m _{w}gh+\frac{ 1 }{ 2 } m _{b}v _{b f}^{2}\]where mb is the mass of the bullet; vbi is the initial velocity of the bullet; mw is the mass of the wood; g is the acceleration of gravity; h is the height gained by the wood; and vbf is the final velocity of the bullet.

Answer 3

I got 449.98, but the answer is 395.4.