The ballistic pendulum of mass m p = 0.125kg catches a horizontally fired bullet

Question

The ballistic pendulum of mass mp= 0.125kg catches a horizontally fired bullet of mass mb= 0.00805kg in a perfectly inelastic collision, where the bullet and pendulum arc upward together to a maximum angle of ? = 32.8o. Knowing that the arm length to center of mass of the pendulum is L = 0.236m, (a) then what was kinetic energy of the bullet/pendulum system immediately after impact? (b) What was the velocity of the bullet/pendulum system immediately after impact? (c) Knowing that momentum is conserved, what was the initial velocity of the bullet? (d) Find the value for the kinetic energy of the bullet/pendulum system just after impact divided by the kinetic energy of the bullet just before impact, i.e., find K(b+p)/Kb.

Answers for (a) J (b) m/s (c) m/s

Thanks.

Explanation / Answer

a)

K.E = ((0.125+0.00805)*9.8*0.236*(1-cos(32.8 degrees))) = 0.049 J

b)

velocity = sqrt(2*9.8*0.236*(1-cos(32.8 degrees))) = 0.859 m/s

c)

initial velocity = (0.125+0.00805)*0.859/0.00805 = 14.2 m/s

d)

K.E after/K.E before = (0.125+0.00805)*0.859^2/(0.00805*14.2^2) = 0.06

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