Question

1 bullet of mass 10.5 grams is fired into a block of wood with a mass of 6.5 kg. The length of the pendulum is 6 meters. The block swings back 8.8000cm. How fast was the bullet traveling before striking the block? How much thermal energy was produced when the bullet struck the block? What % of the bullet's kinetic energy was transformed into thermal energy?

Answer 1

after the bullet is struck

total energy of the bullet + block Ei = (1/2)*(M+m)*v^2

the bullet + block system reaches a height h = 8.8 cm = 0.088 m

at the height h energy

Ef = (M + m)*g*h

total energy is conserved

Ef = Ei

(M+m)*g*h = (1/2)*(M+m)*v^2

v = sqrt(2*g*h)

momentum before bullet struck the block Pi = m*vo

momentum after bullet struck the block Pf = ( M +m)v

totla moemtum is conserved

momentum before collision = moemntum after collision

Pi = Pf

m*vo = (M+m)*v

10.5*10^-3*vo = ( 6.5 + (10.5*10^-3) ) *sqrt(2*9.8*0.088)

speed of bullet vo = 814.32 m/s <<<---answer

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thermal energy produced = Ki - Kf

Ki = (1/2)*m*vo^2 = (1/2)*10.5*10^-3*814.32^2 = 3481.36 J

Kf = (1/2)*(m+M)*v^2

Kf = (1/2)*( (10.5*10^-3) + 6.5)*(2*9.8*0.088) = 5.614 J

thermal enrgy = 3475.746 J

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percentage = (Ki - Kf)/Ki = 99.84 %