A bullet of mass .05 kg, enters and passes through a grapefruit of mass 0.25 kg

Question

A bullet of mass .05 kg, enters and passes through a grapefruit of mass 0.25 kg hanging from a length of string forming a pendulum with a length of 0.5 meters. The bullet emerges from the grapefruit at a velocity equal to 1/3 of it's initial velocity before entering the delicious piece of fruit. At what minimum initial velocity for the bullet does the grapefruit make it to the highest point in the pendulum's possible rotation 1 meter above the starting point? Problems 4-6: Chapter 9 - Problems 4, 54, 68

Explanation / Answer

Let the initial velocity of bullet = Vi
Final velocity of bullet = Vi/3
Mass of bullet mb = .05 Kg
Mass of Grapefruit mg= .25 Kg

Using Momentum Conservation,

mb * Vi + Mg * 0 = mb * vi/3 + mg*vf
.05*Vi = .05*Vi/3 + .25*Vf
Vf = 0.133Vi

Kinetic Emergy of Grapefruit = 0.5*mg*Vf^2
Kinetic Emergy of Grapefruit = 0.5*0.25*(0.133Vi)^2

At Highest Point, Kinetic Emergy will be converted to Potential Energy

Potential Energy = m*g*h
Potential Energy = .25*9.8*1 J
Potential Energy = 2.45 J

K.E = P.E
0.5*0.25*(0.133Vi)^2 = 2.45
Solving for Vi
Vi = 33.3 m/s

Minimum Initial Velocity of Bullet, Vi = 33.3 m/s

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