As shown in the figure below, a bullet of mass m = 0.1kg and speed v passes comp

Question

As shown in the figure below, a bullet of mass m = 0.1kg and speed v passes completely through a pendulum bob of mass M = 2kg. The bullet emerges with a speed of 100m/s. The pendulum bob is suspended by a stiff rod of length l = 50 cm, and negligible mass. Assuming that the velocity of the bullet was 200m/s, what is the velocity of the bob just after the bullet passes through it. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Now suppose the original speed of the bullet was 250m/s, what would be the tension the rod, just as it reaches the highest point?

Explanation / Answer

Conservation of energy for the pendulum bob gives:

Ki + Ui = Kf + Uf
(1/2)Mv^2 + 0 = Mg(2l) + 0 <- Pendulum reaches the height of twice its radius at the top

v^2 = 4gl
v = Sqrt(4gl)



mv1 + Mv2 = mv1 +Mv2
mv1 + 0 = m(v/2) + M(Sqrt(4gl))
v = v/2 + M*Sqrt(4gl)/m
v/2 = M*Sqrt(4gl)/m
v = 2*MSqrt(4gl)/m = 4Sqrt(gl)*M/m

so v= 2*2sqrt(4*10*0.5)/0.1=113.13 m/s

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